NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

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As CBSE board exam paper is designed entirely based on NCERT textbooks and most of the questions in CBSE board exam are directly asked from NCERT textbook, students must know the NCERT very well to perform well in the exam. Only knowing the answer it not enough to perform well in the exam. In the NCERT solutions students will get know how best to write answer in the board exam in order to get good marks.

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Q: What are the important topics in chapter Differential Equations ?

Basic concepts of differential equation, o rder and degree of the differential equation, gen eral and particular solutions of a differential equation, f ormation of a differential equation, m ethods of solving first order,first degree differential equations, ho mogeneous differential equations and l inear differential equations are the important topics of this chapter.

Q: Are the answers to all the textbook questions available in the differential equations ncert solutions maths class 12 chapter 9?

The NCERT class 12 maths differential equations are available in PDF format and have been created by subject experts in line with the textbook questions. These solutions for ch 9 maths class 12 adhere to the latest CBSE Syllabus for 2023 and encompass all the significant concepts for the board exam. The problems in the textbook are solved step by step in accordance with the marks weightage in the CBSE Board exams. Careers360 website offers both chapter-wise and exercise-wise PDF links that can be used by students to instantly clarify their doubts.

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

Edited By Ramraj Saini | Updated on Sep 14, 2023 10:22 PM IST | #CBSE Class 12th

NCERT Differential Equations Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations are provided here. In class 11th, you have already learned how to differentiate a given function (f) with respect to an independent variable. In this article, you will get NCERT solutions for class 12 maths chapter 9 for all major topics of NCERT Class 12 maths syllabus. The equation of function and its one or more derivatives is called a differential equation.

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NCERT Differential Equations Class 12 Questions And Answers
NCERT Differential Equations Class 12 Questions And Answers PDF Free Download
NCERT Differential Equations Class 12 Questions And Answers - Important Formulae
NCERT Differential Equations Class 12 Questions And Answers (Intext Questions and Exercise)
More About NCERT Solutions for Class 12 Maths Chapter 9
Differential Equations Class 12 - Topics
NCERT Exemplar Class 12 Solutions - Subject Wise
NCERT solutions for class 12 maths - Chapter Wise
NCERT solutions for class 12 subject wise
NCERT Solutions class wise
NCERT Books and NCERT Syllabus

In this differential equations class 12 questions and answers, some basic concepts related to the differential equations solutions, particular solutions, and general solutions of differential equations class 12 will be comprehensively discussed. In NCERT solutions for chapter 9 class 12 maths, questions from all these topics are covered in this article. If you are interested in other subjects then you can refer to NCERT solutions for class 12

You will also learn some methods to find the differential equations solutions, the formation of differential equations class 12, and applications of differential equations in different areas in this NCERT class 12 maths chapter 9 question answer are also explained in details. Questions related to these topics are also covered in the NCERT solutions for class 12 maths ch 9 differential equations article. You can refer to NCERT solutions from classes 6 to 12 to learn CBSE maths and science.

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NCERT Differential Equations Class 12 Questions And Answers PDF Free Download

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NCERT Differential Equations Class 12 Questions And Answers - Important Formulae

>> Ordinary Differential Equations (ODEs): Ordinary Differential Equations involve derivatives of a function concerning a single independent variable. They are commonly used to model dynamic systems and phenomena.

>> Partial Differential Equations (PDEs): Partial Differential Equations involve derivatives of a function concerning multiple independent variables. They are frequently used in physics to describe phenomena like heat diffusion, wave propagation, and fluid dynamics.

>> Types of Differential Equations: Differential equations can be categorised based on their order, linearity, and specific properties. Common types include:

First-Order Differential Equations
Second-Order Differential Equations
Linear Differential Equations
Nonlinear Differential Equations
Homogeneous Differential Equations
Non-Homogeneous Differential Equations

>> Methods for Solving Differential Equations: Various techniques can be employed to solve differential equations, including:

Separation of Variables
Integrating Factors
Exact Differential Equations
Linear Differential Equations with Constant Coefficients
Method of Undetermined Coefficients
Variation of Parameters
Laplace Transforms

>> Applications of Differential Equations: Differential equations have widespread applications in science and engineering. Some examples include modelling population growth, describing electrical circuits, predicting radioactive decay, and simulating fluid flow.

Free download NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations for CBSE Exam.

NCERT Differential Equations Class 12 Questions And Answers (Intext Questions and Exercise)

NCERT differential equations class 12 solutions - Exercise: 9.1

Question:1 Determine order and degree (if defined) of differential equation $\frac{\mathrm{d} ^4y}{\mathrm{d} x^4} +\sin(y''')=0$

Answer:

Given function is
$\frac{\mathrm{d} ^4y}{\mathrm{d} x^4} +\sin(y''')=0$
We can rewrite it as
$y^{''''}+\sin(y''') =0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''''}$

Therefore, the order of the given differential equation $\frac{\mathrm{d} ^4y}{\mathrm{d} x^4} +\sin(y''')=0$ is 4
Now, the given differential equation is not a polynomial equation in its derivatives
Therefore, it's a degree is not defined

Question:2 Determine order and degree (if defined) of differential equation $y' + 5y = 0$

Answer:

Given function is
$y' + 5y = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'}$
Therefore, the order of the given differential equation $y' + 5y = 0$ is 1
Now, the given differential equation is a polynomial equation in its derivatives and its highest power raised to y ' is 1
Therefore, it's a degree is 1.

Question:3 Determine order and degree (if defined) of differential equation $\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0$

Answer:

Given function is
$\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0$
We can rewrite it as
$(s^{'})^4+3s.s^{''} =0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $s^{''}$

Therefore, the order of the given differential equation $\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0$ is 2
Now, the given differential equation is a polynomial equation in its derivatives and power raised to s '' is 1
Therefore, it's a degree is 1

Question:4 Determine order and degree (if defined) of differential equation.

$\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0$

Answer:

Given function is
$\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0$
We can rewrite it as
$(y^{''})^2+\cos y^{''} =0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, the order of the given differential equation $\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0$ is 2
Now, the given differential equation is not a polynomial equation in its derivatives
Therefore, it's a degree is not defined

Question:5 Determine order and degree (if defined) of differential equation.

$\frac{d^2y}{dx^2} = \cos 3x + \sin 3x$

Answer:

Given function is
$\frac{d^2y}{dx^2} = \cos 3x + \sin 3x$
$\Rightarrow \frac{d^2y}{dx^2}- \cos 3x - \sin 3x = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}\left ( \frac{d^2y}{dx^2} \right )$

Therefore, order of given differential equation $\frac{d^2y}{dx^2}- \cos 3x - \sin 3x = 0$ is 2
Now, the given differential equation is a polynomial equation in it's dervatives $\frac{d^2y}{dx^2}$ and power raised to $\frac{d^2y}{dx^2}$ is 1
Therefore, it's degree is 1

Question:6 Determine order and degree (if defined) of differential equation $(y''')^2 + (y'')^3 + (y')^4 + y^5= 0$

Answer:

Given function is
(y''')^2 + (y'')^3 + (y')^4 + y^5= 0 Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'''}$

Therefore, order of given differential equation is 3 Now, the given differential equation is a polynomial equation in it's dervatives $y^{'''} , y^{''} \ and \ y^{'}$ and power raised to $y^{'''}$ is 2
Therefore, it's degree is 2

Question:7 Determine order and degree (if defined) of differential equation

$y''' + 2y'' + y' =0$

Answer:

Given function is
$y''' + 2y'' + y' =0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'''}$

Therefore, order of given differential equation $y''' + 2y'' + y' =0$ is 3
Now, the given differential equation is a polynomial equation in it's dervatives $y^{'''} , y^{''} \ and \ y^{'}$ and power raised to $y^{'''}$ is 1
Therefore, it's degree is 1

Question:8 Determine order and degree (if defined) of differential equation

$y' + y = e^x$

Answer:

Given function is
$y' + y = e^x$
$\Rightarrow$ $y^{'}+y-e^x=0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'}$

Therefore, order of given differential equation $y^{'}+y-e^x=0$ is 1
Now, the given differential equation is a polynomial equation in it's dervatives $y^{'}$ and power raised to $y^{'}$ is 1
Therefore, it's degree is 1

Question:9 Determine order and degree (if defined) of differential equation

$y'' + (y')^2 + 2y = 0$

Answer:

Given function is
$y'' + (y')^2 + 2y = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, order of given differential equation $y'' + (y')^2 + 2y = 0$ is 2
Now, the given differential equation is a polynomial equation in it's dervatives $y^{''} \ and \ y^{'}$ and power raised to $y^{''}$ is 1
Therefore, it's degree is 1

Question:10 Determine order and degree (if defined) of differential equation

$y'' + 2y' + \sin y = 0$

Answer:

Given function is
$y'' + 2y' + \sin y = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, order of given differential equation $y'' + 2y' + \sin y = 0$ is 2
Now, the given differential equation is a polynomial equation in it's dervatives $y^{''} \ and \ y^{'}$ and power raised to $y^{''}$ is 1
Therefore, it's degree is 1

Question:11 The degree of the differential equation $\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0$ is

(A) 3

(B) 2

(D) not defined

Answer:

Given function is
$\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0$
We can rewrite it as
$(y^{''})^3+(y^{'})^2+\sin y^{'}+1=0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, order of given differential equation $\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0$ is 2
Now, the given differential equation is a not polynomial equation in it's dervatives
Therefore, it's degree is not defined

Therefore, answer is (D)

Question:12 The order of the differential equation $2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$ is

(A) 2

(B) 1

(D) Not Defined

Answer:

Given function is
$2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$
We can rewrite it as
$2x.y^{''}-3y^{'}+y=0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, order of given differential equation $2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$ is 2

Therefore, answer is (A)

NCERT differential equations class 12 solutions - Exercise: 9.2

Question:1 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = e^x + 1 \qquad :\ y'' -y'=0$

Answer:

Given,

$y = e^x + 1$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d}e^x }{\mathrm{d} x} = e^x$

Again, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y' }{\mathrm{d} x} = \frac{\mathrm{d}e^x }{\mathrm{d} x} = e^x$

$\implies y'' = e^x$

Substituting the values of y’ and y'' in the given differential equations,

y'' - y' = e ^x - e ^x = 0 = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

Question:2 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = x^2 + 2x + C\qquad:\ y' -2x - 2 =0$

Answer:

Given,

$y = x^2 + 2x + C$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(x^2 + 2x + C) = 2x + 2$

Substituting the values of y’ in the given differential equations,

$y' -2x - 2 =2x + 2 - 2x - 2 = 0= RHS$ .

Therefore, the given function is the solution of the corresponding differential equation.

Question:3. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = \cos x + C\qquad :\ y' + \sin x = 0$

Answer:

Given,

$y = \cos x + C$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(cosx + C) = -sinx$

Substituting the values of y’ in the given differential equations,

$y' - \sin x = -sinx -sinx = -2sinx \neq RHS$ .

Therefore, the given function is not the solution of the corresponding differential equation.

Question:4. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = \sqrt{1 + x^2}\qquad :\ y' = \frac{xy}{1 + x^2}$

Answer:

Given,

$y = \sqrt{1 + x^2}$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(\sqrt{1 + x^2}) = \frac{2x}{2\sqrt{1 + x^2}} = \frac{x}{\sqrt{1 + x^2}}$

Substituting the values of y in RHS,

$\frac{x\sqrt{1+x^2}}{1 + x^2} = \frac{x}{\sqrt{1+x^2}} = LHS$ .

Therefore, the given function is a solution of the corresponding differential equation.

Question:5 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = Ax\qquad :\ xy' = y\;(x\neq 0)$

Answer:

Given,

$y = Ax$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(Ax) = A$

Substituting the values of y' in LHS,

$xy' = x(A) = Ax = y = RHS$ .

Therefore, the given function is a solution of the corresponding differential equation.

Question:6. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = x\sin x\qquad :\ xy' = y + x\sqrt{x^2 - y^2}\ (x\neq 0\ \textup{and} \ x > y\ or \ x < -y)$

Answer:

Given,

$y = x\sin x$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(xsinx) = sinx + xcosx$

Substituting the values of y' in LHS,

$xy' = x(sinx + xcosx)$

Substituting the values of y in RHS.

$\\xsinx + x\sqrt{x^2 - x^2sin^2x} = xsinx + x^2\sqrt{1-sinx^2} = x(sinx+xcosx) = LHS$

Therefore, the given function is a solution of the corresponding differential equation.

Question:7 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$xy = \log y + C\qquad :\ y' = \frac{y^2}{1 - xy}\ (xy\neq 1)$

Answer:

Given,

$xy = \log y + C$

Now, differentiating both sides w.r.t. x,

$\\ y + x\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(logy) = \frac{1}{y}\frac{\mathrm{d}y }{\mathrm{d} x}\\ \\ \implies y^2 + xyy' = y' \\ \\ \implies y^2 = y'(1-xy) \\ \\ \implies y' = \frac{y^2}{1-xy}$

Substituting the values of y' in LHS,

$y' = \frac{y^2}{1-xy} = RHS$

Therefore, the given function is a solution of the corresponding differential equation.

Question:8 In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y - cos y = x \qquad :(\ y\sin y + \cos y + x) y' = y$

Answer:

Given,

$y - cos y = x$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} +siny\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(x) = 1$

$\implies$ y' + siny.y' = 1

$\implies$ y'(1 + siny) = 1

$\implies y' = \frac{1}{1+siny}$

Substituting the values of y and y' in LHS,

$(\ (x+cosy)\sin y + \cos y + x) (\frac{1}{1+siny})$

$= [x(1+siny) + cosy(1+siny)]\frac{1}{1+siny}$

= (x + cosy) = y = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Question:9 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$x + y = \tan^{-1}y\qquad :\ y^2y' + y^2 + 1 = 0$

Answer:

Given,

$x + y = \tan^{-1}y$

Now, differentiating both sides w.r.t. x,

$\\ 1 + \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{1 + y^2}\frac{\mathrm{d} y}{\mathrm{d} x}\\ \\ \implies1+y^2 = y'(1-(1+y^2)) = -y^2y' \\ \implies y' = -\frac{1+y^2}{y^2}$

Substituting the values of y' in LHS,

$y^2(-\frac{1+y^2}{y^2}) + y^2 + 1 = -1- y^2+ y^2 +1 = 0 = RHS$

Therefore, the given function is a solution of the corresponding differential equation.

Question:10 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = \sqrt{a^2 - x^2}\ x\in (-a,a)\qquad : \ x + y \frac{dy}{dx} = 0\ (y\neq 0)$

Answer:

Given,

$y = \sqrt{a^2 - x^2}$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{\mathrm{d}}{\mathrm{d} x}(\sqrt{a^2 - x^2}) = \frac{-2x}{2\sqrt{a^2 - x^2}} = \frac{-x}{\sqrt{a^2 - x^2}}$

Substituting the values of y and y' in LHS,

$x + y \frac{dy}{dx} = x + (\sqrt{a^2 - x^2})(\frac{-x}{\sqrt{a^2 - x^2}}) = 0 = RHS$

Therefore, the given function is a solution of the corresponding differential equation.

Question:11 The number of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4

Answer:

(D) 4

The number of constants in the general solution of a differential equation of order n is equal to its order.

Question:12 The number of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3
(B) 2
(C) 1
(D) 0

Answer:

(D) 0

In a particular solution of a differential equation, there is no arbitrary constant.

Differential Equations Class 12 NCERT Solutions - Exercise: 9.3

Question:1 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

$\frac{x}{a} + \frac{y}{b} = 1$

Answer:

Given equation is

$\frac{x}{a} + \frac{y}{b} = 1$
Differentiate both the sides w.r.t x
$\frac{d\left ( \frac{x}{a}+\frac{y}{b} \right )}{dx}=\frac{d(1)}{dx}$
$\frac{1}{a}+\frac{1}{b}.\frac{dy}{dx} = 0\\ \frac{dy}{dx} = -\frac{b}{a}$
Now, again differentiate it w.r.t x
$\frac{d^2y}{dx^2} =0$
Therefore, the required differential equation is $\frac{d^2y}{dx^2} =0$ or $y^{''} =0$

Question:2 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

$y^2 = a(b^2 - x^2)$

Answer:

Given equation is
$y^2 = a(b^2 - x^2)$
Differentiate both the sides w.r.t x
$\frac{d\left ( y^2 \right )}{dx}=\frac{d(a(b^2-x^2))}{dx}$
$2y\frac{dy}{dx}= -2ax\\ \\ y.\frac{dy}{dx}= -ax\\ \\ y.y^{'}=-ax$ -(i)
Now, again differentiate it w.r.t x
$y^{'}.y^{'}+y.y^{''}= -a\\ (y^{'})^2+y.y^{''}=-a$ -(ii)
Now, divide equation (i) and (ii)
$\frac{(y^{'})^2+y.y^{''}}{y.y^{'}}= \frac{-a}{-ax}\\ \\ x(y^{'})^2+x.y.y^{''}=y.y^{'}\\ \\ x(y^{'})^2+x.y.y^{''}-y.y^{'}=0$
Therefore, the required differential equation is $x(y^{'})^2+x.y.y^{''}-y.y^{'}=0$

Question:3 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. $y = ae^{3x} + b e^{-2x}$

Answer:

Given equation is
$y = ae^{3x} + b e^{-2x}$ -(i)
Differentiate both the sides w.r.t x
$\frac{d\left ( y \right )}{dx}=\frac{d(ae^{3x}+be^{-2x})}{dx}$
$y^{'}=\frac{dy}{dx}= 3ae^{3x}-2be^{-2x}\\ \\$ -(ii)
Now, again differentiate w.r.t. x
$y^{''}= \frac{d^2y}{dx^2} = 9ae^{3x}+4be^{-2x}$ -(iii)
Now, multiply equation (i) with 2 and add equation (ii)
$2(ae^{3x}+be^{-2x})+(3a-2be^{-x}) = 2y+y^{'}\\ 5ae^{3x} = 2y+y^{'}\\ ae^{3x}= \frac{2y+y^{'}}{5}$ -(iv)
Now, multiply equation (i) with 3 and subtract from equation (ii)
$3(ae^{3x}+be^{-2x})-(3a-2be^{-x}) = 3y-y^{'}\\ 5be^{-2x} = 3y-y^{'}\\ be^{-2x}= \frac{3y-y^{'}}{5}$ -(v)
Now, put values from (iv) and (v) in equation (iii)
$y^{''}= 9.\frac{2y+y^{'}}{5}+4.\frac{3y-y^{'}}{5}\\ \\ y^{''}= \frac{18y+9y^{'}+12y-4y^{'}}{5}\\ \\ y^{''}= \frac{5(6y-y^{'})}{5}=6y-y^{'}\\ \\ y^{''}+y^{'}-6y=0$

Therefore, the required differential equation is $y^{''}+y^{'}-6y=0$

Question:4 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. $y = e^{2x}(a+bx)$

Answer:

Given equation is
$y = e^{2x}(a+bx)$ -(i)
Now, differentiate w.r.t x
$\frac{dy}{dx}= \frac{d(e^{2x}(a+bx))}{dx}= 2e^{2x}(a+bx)+e^{2x}.b$ -(ii)
Now, again differentiate w.r.t x
$y^{''}= \frac{d^2y}{dx^2}= \frac{d}{dx}\frac{dy}{dx} = 4e^{2x}(a+bx)+2be^{2x}+2be^{2x}= 4e^{2x}(a+bx)+4be^{2x}$ -(iii)
Now, multiply equation (ii) with 2 and subtract from equation (iii)
$4e^{2x}(a+bx)+4be^{2x}-2\left ( 2e^{2x}(a+bx)+be^{2x} \right )=y^{''}-2y^{'}\\ \\ 2be^{2x} = y^{''}-2y^{'}\\ \\ be^{2x}= \frac{y^{''}-2y^{'}}{2}$ -(iv)
Now,put the value in equation (iii)
$y^{''}=4y+4.\frac{y^{''}-2y^{'}}{2}\\ \\ y^{''}= 4y+2y^{''}-4y^{'}\\ \\ y^{''}-4y^{'}+4y=0$
Therefore, the required equation is $y^{''}-4y^{'}+4y=0$

Question:5 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

$y=e^x(a\cos x + b\sin x)$

Answer:

Given equation is
$y=e^x(a\cos x + b\sin x)$ -(i)
Now, differentiate w.r.t x
$\frac{dy}{dx}= \frac{d(e^{x}(a\cos x+b\sin x))}{dx}= e^{x}(a\cos x+b\sin x)+e^x(-a\sin x+b\cos x )$ -(ii)
Now, again differentiate w.r.t x
$y^{''}= \frac{d^2y}{dx^2}= \frac{d}{dx}\frac{dy}{dx} =e^{x}(a\cos x+b\sin x)+e^x(-a\sin x+b\cos x )$ $+e^x(-a\sin x+b\cos x )+e^x(-a\cos x-b\sin x)$
$=2e^x(-a\sin x+b\cos x )$ -(iii)
Now, multiply equation (i) with 2 and multiply equation (ii) with 2 and add and subtract from equation (iii) respectively
we will get

$y^{''}-2y^{'}+2y = 0$
Therefore, the required equation is $y^{''}-2y^{'}+2y = 0$

Question:6 Form the differential equation of the family of circles touching the y-axis at origin.

Answer:

1628484808301 If the circle touches y-axis at the origin then the centre of the circle lies at the x-axis
Let r be the radius of the circle
Then, the equation of a circle with centre at (r,0) is
$(x-r)^2+(y-0)^2 = r^2$
$x^2+r^2-2xr+y^2=r^2\\ x^2+y^2-2xr=0$ -(i)
Now, differentiate w.r.t x
$2x+2y\frac{dy}{dx}-2r=0\\ y\frac{dy}{dx}\Rightarrow yy^{'}+x-r=0$
$yy^{'}+x=r$ -(ii)
Put equation (ii) in equation (i)
$x^2+y^2=2x(yy^{'}+x)\\ y^2=2xyy^{'}+x^2$
Therefore, the required equation is $y^2=2xyy^{'}+x^2$

Question:7 Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

Answer:

Equation of perabola having vertex at origin and axis along positive y-axis is
$x^2= 4ay$ (i)
Now, differentiate w.r.t. c
$2x= 4a\frac{dy}{dx}\\ \\ \frac{dy}{dx} =y^{'}= \frac{x}{2a}$
$a=\frac{x}{2y^{'}}$ -(ii)
Put value from equation (ii) in (i)
$x^2= 4y.\frac{x}{2y^{'}}\\ xy^{'}-2y = 0$
Therefore, the required equation is $xy^{'}-2y = 0$

Question:8 Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

Answer:

Equation of ellipses having foci on y-axis and centre at origin is
$\frac{x^2}{b^2}+\frac{y^2}{a^2} = 1$ -
Now, differentiate w..r.t. x
$\frac{2x}{b^2}+\frac{2y}{a^2}.\frac{dy}{dx}=0\\$ -(i)
Now, again differentiate w.r.t. x
$\frac{2}{b^2}+\frac{2}{a^2}.y^{'}.y^{'}+\frac{2y}{a^2}.y^{''}=0\\ \\ \frac{1}{b^2}=-\frac{1}{a^2}\left ( (y^{'})^2+yy^{''} \right )$ -(ii)
Put value from equation (ii) in (i)
Our equation becomes
$\frac{2y}{a^2}y^{'}-\frac{2x}{a^2}\left ( (y^{'})^2+yy^{''} \right )=0\\ \\ 2yy^{'}-2(y^{'})^2x+2yy^{''}x=0\\ \\ xyy^{''}-x(y^{'})^2+yy^{'}= 0$
Therefore, the required equation is $xyy^{''}-x(y^{'})^2+yy^{'}= 0$

Question:9 Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Answer:

Equation of hyperbolas having foci on x-axis and centre at the origin
$\frac{x^2}{b^2}+\frac{y^2}{a^2} = 1$
Now, differentiate w..r.t. x
$\frac{2x}{b^2}+\frac{2y}{a^2}.\frac{dy}{dx}=0\\$ -(i)
Now, again differentiate w.r.t. x
$\frac{2}{b^2}+\frac{2}{a^2}.y^{'}.y^{'}+\frac{2y}{a^2}.y^{''}=0\\ \\ \frac{1}{b^2}=-\frac{1}{a^2}\left ( (y^{'})^2+yy^{''} \right )$ -(ii)
Put value from equation (ii) in (i)
Our equation becomes
$\frac{2y}{a^2}y^{'}-\frac{2x}{a^2}\left ( (y^{'})^2+yy^{''} \right )=0\\ \\ 2yy^{'}-2(y^{'})^2x+2yy^{''}x=0\\ \\ xyy^{''}-x(y^{'})^2+yy^{'}= 0$
Therefore, the required equation is $xyy^{''}-x(y^{'})^2+yy^{'}= 0$

Question:10 Form the differential equation of the family of circles having centre on y-axis and radius 3 units.

Answer:

1628484856150 Equation of the family of circles having centre on y-axis and radius 3 units
Let suppose centre is at (0,b)
Now, equation of circle with center (0,b) an radius = 3 units
$(x-0)^2+(y-b)^2=3^2 \ \ \ \ \ \ \ \ \ \ \ -(i)\\ x^2+y^2+b^2-2yb = 9$
Now, differentiate w.r.t x
we get,
$2x+2yy^{'}-2by^{'}= 0\\ 2x+2y(y-b)= 0\\ (y-b)=\frac{-x}{y^{'}} \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Put value fro equation (ii) in (i)
$(x-0)^2+(\frac{-x}{y^{'}})^2=3^2 \\ x^2+\frac{x^2}{(y^{'})^2}=9\\ x^2(y^{'})^2+x^2=9(y^{'})^2\\ \\ (x^2-9)(y^{'})^2+x^2 = 0$
Therefore, the required differential equation is $(x^2-9)(y^{'})^2+x^2 = 0$

Question:11 Which of the following differential equations has $y = c_1e^x + c_2e^{-x}$ as the general solution?

(A) $\frac{d^2y}{dx^2} + y = 0$

(B) $\frac{d^2y}{dx^2} - y = 0$

(D) $\frac{d^2y}{dx^2} -1 = 0$

Answer:

Given general solution is
$y = c_1e^x + c_2e^{-x}$
Differentiate it w.r.t x
we will get
$\frac{dy}{dx} = c_1e^x-c_2e^{-x}$
Again, Differentiate it w.r.t x
$\frac{d^2y}{dx^2} = c_1e^x+c_2e^{x}=y\\ \frac{d^2y}{dx^2} - y = 0$
Therefore, (B) is the correct answer

Question:12 Which of the following differential equations has $y = x$ as one of its particular solution?

(A) $\frac{d^2y}{dx^2} - x^2\frac{dy}{dx} + xy =x$

(B) $\frac{d^2y}{dx^2} + x\frac{dy}{dx} + xy =x$

(D) $\frac{d^2y}{dx^2} + x\frac{dy}{dx} + xy =0$

Answer:

Given equation is
$y = x$
Now, on differentiating it w.r.t x
we get,
$\frac{dy}{dx} = 1$
and again on differentiating it w.r.t x
we get,

$\frac{d^2y}{dx^2} = 0$
Now, on substituting the values of $\frac{d^2y}{dx^2} , \frac{dy}{dx} \ and \ y$ in all the options we will find that only option c which is $\frac{d^2y}{dx^2} - x^2\frac{dy}{dx} + xy =0$ satisfies
Therefore, the correct answer is (C)

NCERT class 12 maths chapter 9 question answer - Exercise: 9.4

Question:1 Find the general solution: $\frac{dy}{dx} = \frac{1-\cos x}{1 + \cos x}$

Answer:

Given,

$\frac{dy}{dx} = \frac{1-\cos x}{1 + \cos x}$

$\\ \implies\frac{dy}{dx} = \frac{2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}} = tan^2\frac{x}{2} \\ \implies dy = (sec^2\frac{x}{2} - 1)dx$

$\\ \implies \int dy = \int sec^2\frac{x}{2}dx - \int dx \\ \implies y = 2tan^{-1}\frac{x}{2} - x + C$

Question:2 Find the general solution: $\frac{dy}{dx} = \sqrt{4-y^2}\ (-2 < y < 2)$

Answer:

Given, in the question

$\frac{dy}{dx} = \sqrt{4-y^2}$

$\\ \implies \frac{dy}{\sqrt{4-y^2}} = dx \\ \implies \int \frac{dy}{\sqrt{4-y^2}} = \int dx$

$\\ (\int \frac{dy}{\sqrt{a^2-y^2}} = sin^{-1}\frac{y}{a})\\$

The required general solution:

$\\ \implies sin^{-1}\frac{y}{2} = x + C$

Question:3 Find the general solution: $\frac{dy}{dx} + y = 1 (y\neq 1)$

Answer:

Given, in the question

$\frac{dy}{dx} + y = 1$

$\\ \implies \frac{dy}{dx} = 1- y \\ \implies \int\frac{dy}{1-y} = \int dx$

$(\int\frac{dx}{x} = lnx)$

$\\ \implies -log(1-y) = x + C\ \ (We\ can\ write\ C= log k) \\ \implies log k(1-y) = -x \\ \implies 1- y = \frac{1}{k}e^{-x} \\$

The required general equation

$\implies y = 1 -\frac{1}{k}e^{-x}$

Question:4 Find the general solution: $\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0$

Answer:

Given,

$\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0$

$\\ \implies \frac{sec^2 y}{tan y}dy = -\frac{sec^2 x}{tan x}dx \\ \implies \int \frac{sec^2 y}{tan y}dy = - \int \frac{sec^2 x}{tan x}dx$

Now, let tany = t and tanx = u

$sec^2 y dy = dt\ and\ sec^2 x dx = du$

$\\ \implies \int \frac{dt}{t} = -\int \frac{du}{u} \\ \implies log t = -log u +logk \\ \implies t = \frac{1}{ku} \\ \implies tany = \frac{1}{ktanx}$

Question:5 Find the general solution:

$(e^x + e^{-x})dy - (e^x - e^{-x})dx = 0$

Answer:

Given, in the question

$(e^x + e^{-x})dy - (e^x - e^{-x})dx = 0$

$\\ \implies dy = \frac{(e^x - e^{-x})}{(e^x + e^{-x})}dx$

Let,

$\\ (e^x + e^{-x}) = t \\ \implies (e^x - e^{-x})dx = dt$

$\\ \implies \int dy = \int \frac{dt}{t} \\ \implies y = log t + C \\ \implies y = log(e^x + e^{-x}) + C$

This is the general solution

Question:6 Find the general solution: $\frac{dy}{dx} = (1+x^2)(1+y^2)$

Answer:

Given, in the question

$\frac{dy}{dx} = (1+x^2)(1+y^2)$

$\\ \implies \int \frac{dy}{(1+y^2)} = \int (1+x^2)dx$

$(\int \frac{dx}{(1+x^2)} =tan^{-1}x +c)$

$\\ \implies tan^{-1}y = x+\frac{x^3}{3} + C$

Question:7 Find the general solution: $y\log y dx - x dy = 0$

Answer:

Given,

$y\log y dx - x dy = 0$

$\\ \implies \frac{1}{ylog y}dy = \frac{1}{x}dx$

let logy = t

=> 1/ydy = dt

$\\ \implies \int \frac{dt}{t} = \int \frac{1}{x}dx \\ \implies \log t = \log x + \log k \\ \implies t = kx \\ \implies \log y = kx$

This is the general solution

Question:8 Find the general solution: $x^5\frac{dy}{dx} = - y^5$

Answer:

Given, in the question

$x^5\frac{dy}{dx} = - y^5$

$\\ \implies \int \frac{dy}{y^5} = - \int \frac{dx}{x^5} \\ \implies \frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C \\ \implies \frac{1}{y^4} + \frac{1}{x^4} = C$

This is the required general equation.

Question:9 Find the general solution: $\frac{dy}{dx} = \sin^{-1}x$

Answer:

Given, in the question

$\frac{dy}{dx} = \sin^{-1}x$

$\implies \int dy = \int \sin^{-1}xdx$

Now,

$\int (u.v)dx = u\int vdx - \int(\frac{du}{dx}.\int vdx)dx$

Here, u = $\sin^{-1}x$ and v = 1

$\implies y = \sin^{-1}x .x - \int(\frac{1}{\sqrt{1-x^2}}.x)dx$

$\\ Let\ 1- x^2 = t \\ \implies -2xdx = dt \implies xdx = -dt/2$

$\\ \implies y = x\sin^{-1}x+ \int(\frac{dt}{2\sqrt{t}}) \\ \implies y = x\sin^{-1}x + \frac{1}{2}.2\sqrt{t} + C \\ \implies y = x\sin^{-1}x + \sqrt{1-x^2} + C$

Question:10 Find the general solution $e^x\tan y dx + (1-e^x)\sec^2 y dy = 0$

Answer:

Given,

$e^x\tan y dx + (1-e^x)\sec^2 y dy = 0$

$\\ \implies e^x\tan y dx = - (1-e^x)\sec^2 y dy \\ \implies \int \frac{\sec^2 y }{\tan y}dy = -\int \frac{e^x }{(1-e^x)}dx$

$\\ let\ tany = t \ and \ 1-e^x = u \\ \implies \sec^2 ydy = dt\ and \ -e^xdx = du$

$\\ \therefore \int \frac{dt }{t} = \int \frac{du }{u} \\ \implies \log t = \log u + \log k \\ \implies t = ku \\ \implies \tan y= k (1-e^x)$

Question:11 Find a particular solution satisfying the given condition:

$(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x; \ y = 1\ \textup{when}\ x = 0$

Answer:

Given, in the question

$(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x$

$\\ \implies \int dy = \int\frac{2x^2 + x}{(x^3 + x^2 + x + 1)}dx$

$(x^3 + x^2 + x + 1) = (x +1)(x^2+1)$

Now,

Now comparing the coefficients

A + B = 2; B + C = 1; A + C = 0

Solving these:

Putting the values of A,B,C:

Therefore,

Now, y= 1 when x = 0

c = 1

Putting the value of c, we get:

Question:12 Find a particular solution satisfying the given condition:

$x(x^2 -1)\frac{dy}{dx} =1;\ y = 0\ \textup{when} \ x = 2$

Answer:

Given, in the question

$x(x^2 -1)\frac{dy}{dx} =1$

$\\ \implies \int dy=\int \frac{dx}{x(x^2 -1)} \\ \implies \int dy=\int \frac{dx}{x(x -1)(x+1)}$

Let,

1628485409903

Now comparing the values of A,B,C

A + B + C = 0; B-C = 0; A = -1

Solving these:

Now putting the values of A,B,C

1517900488152880

Given, y =0 when x =2

1517900489745379

Therefore,

$\\ \implies y = \frac{1}{2}\log[\frac{4(x-1)(x+1)}{3x^2}]$

$\\ \implies y = \frac{1}{2}\log[\frac{4(x^2-1)}{3x^2}]$

Question:13 Find a particular solution satisfying the given condition:

$\cos\left(\frac{dy}{dx} \right ) = a\ (a\in R);\ y = 1\ \textup{when}\ x = 0$

Answer:

Given,

$\cos\left(\frac{dy}{dx} \right ) = a$

$\\ \implies \frac{dy}{dx} = \cos^{-1}a \\ \implies \int dy = \int\cos^{-1}a\ dx \\ \implies y = x\cos^{-1}a + c$

Now, y =1 when x =0

1 = 0 + c

Therefore, c = 1

Putting the value of c:

$\implies y = x\cos^{-1}a + 1$

Question:14 Find a particular solution satisfying the given condition:

$\frac{dy}{dx} = y\tan x; \ y =1\ \textup{when}\ x = 0$

Answer:

Given,

$\frac{dy}{dx} = y\tan x$

$\\ \implies \int \frac{dy}{y} = \int \tan x\ dx \\ \implies \log y = \log \sec x + \log k \\ \implies y = k\sec x$

Now, y=1 when x =0

1 = ksec0

$\implies$ k = 1

Putting the vlue of k:

y = sec x

Question:15 Find the equation of a curve passing through the point (0, 0) and whose differential equation is $y' = e^x\sin x$ .

Answer:

We first find the general solution of the given differential equation

Given,

$y' = e^x\sin x$

$\\ \implies \int dy = \int e^x\sin xdx$

$\\ Let I = \int e^x\sin xdx \\ \implies I = \sin x.e^x - \int(\cos x. e^x)dx \\ \implies I = e^x\sin x - [e^x\cos x - \int(-\sin x.e^x)dx] \\ \implies 2I = e^x(\sin x - \cos x) \\ \implies I = \frac{1}{2}e^x(\sin x - \cos x)$

$\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + c$

Now, Since the curve passes through (0,0)

y = 0 when x =0

$\\ \therefore 0 = \frac{1}{2}e^0(\sin 0 - \cos 0) + c \\ \implies c = \frac{1}{2}$

Putting the value of c, we get:

$\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + \frac{1}{2} \\ \implies 2y -1 = e^x(\sin x - \cos x)$

Question:16 For the differential equation $xy\frac{dy}{dx} = (x+2)(y+2)$ , find the solution curve passing through the point (1, –1).

Answer:

We first find the general solution of the given differential equation

Given,

$xy\frac{dy}{dx} = (x+2)(y+2)$

$\\ \implies \int \frac{y}{y+2}dy = \int \frac{x+2}{x}dx \\ \implies \int \frac{(y+2)-2}{y+2}dy = \int (1 + \frac{2}{x})dx \\ \implies \int (1 - \frac{2}{y+2})dy = \int (1 + \frac{2}{x})dx \\ \implies y - 2\log (y+2) = x + 2\log x + C$

Now, Since the curve passes through (1,-1)

y = -1 when x = 1

$\\ \therefore -1 - 2\log (-1+2) = 1 + 2\log 1 + C \\ \implies -1 -0 = 1 + 0 +C \\ \implies C = -2$

Putting the value of C:

$\\ y - 2\log (y+2) = x + 2\log x + -2 \\ \implies y -x + 2 = 2\log x(y+2)$

Question:17 Find the equation of a curve passing through the point $( 0 ,-2)$ given that at any point $(x,y)$ on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Answer:

According to the question,

$y\frac{dy}{dx} =x$

$\\ \implies \int ydy =\int xdx \\ \implies \frac{y^2}{2} = \frac{x^2}{2} + c$

Now, Since the curve passes through (0,-2).

x =0 and y = -2

$\\ \implies \frac{(-2)^2}{2} = \frac{0^2}{2} + c \\ \implies c = 2$

Putting the value of c, we get

$\\ \frac{y^2}{2} = \frac{x^2}{2} + 2 \\ \implies y^2 = x^2 + 4$

Question:18 At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

Answer:

Slope m of line joining (x,y) and (-4,-3) is $\frac{y+3}{x+4}$

According to the question,

$\\ \frac{dy}{dx} = 2(\frac{y+3}{x+4}) \\ \implies \int \frac{dy}{y+3} = 2\int \frac{dx}{x+4} \\ \implies \log (y+3) = 2\log (x+4) + \log k \\ \implies (y+3) = k(x+4)^2$

Now, Since the curve passes through (-2,1)

x = -2 , y =1

$\\ \implies (1+3) = k(-2+4)^2 \\ \implies k =1$

Putting the value of k, we get

$\\ \implies y+3 = (x+4)^2$

Question:19 The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Answer:

Volume of a sphere, $V = \frac{4}{3}\pi r ^3$

Given, Rate of change is constant.

$\\ \therefore \frac{dV}{dt} = c \\ \implies \frac{d}{dt} (\frac{4}{3}\pi r ^3) = c \\ \implies \int d(\frac{4}{3}\pi r ^3) = c\int dt \\ \implies \frac{4}{3}\pi r ^3 = ct + k$

Now, at t=0, r=3 and at t=3 , r =6

Putting these value:

$\frac{4}{3}\pi (3) ^3 = c(0) + k \\ \implies k = 36\pi$

Also,

$\frac{4}{3}\pi (6) ^3 = c(3) + 36\pi \\ \implies 3c = 252\pi \\ \implies c = 84\pi$

Putting the value of c and k:

$\\ \frac{4}{3}\pi r ^3 = 84\pi t + 36\pi \\ \implies r ^3 = (21 t + 9)(3) = 62t + 27 \\ \implies r = \sqrt[3]{62t + 27}$

Question:20 In a bank, principal increases continuously at the rate of r % per year. Find the value of r if Rs 100 double itself in 10 years (log _e 2 = 0.6931).

Answer:

Let p be the principal amount and t be the time.

According to question,

$\frac{dp}{dt} = (\frac{r}{100})p$

$\\ \implies \int\frac{dp}{p} = \int (\frac{r}{100})dt \\ \implies \log p = \frac{r}{100}t + C$

$\\ \implies p = e^{\frac{rt}{100} + C}$

Now, at t =0 , p = 100

and at t =10, p = 200

Putting these values,

$\\ \implies 100 = e^{\frac{r(0)}{100} + C} = e^C$

Also,

, $\\ \implies 200 = e^{\frac{r(10)}{100} + C} = e^{\frac{r}{10}}.e^C = e^{\frac{r}{10}}.100 \\ \implies e^{\frac{r}{10}} = 2 \\ \implies \frac{r}{10} = \ln 2 = 0.6931 \\ \implies r = 6.93$

So value of r = 6.93%

Question:21 In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e ^0.5 = 1.648).

Answer:

Let p be the principal amount and t be the time.

According to question,

$\frac{dp}{dt} = (\frac{5}{100})p$

$\\ \implies \int\frac{dp}{p} = \int (\frac{1}{20})dt \\ \implies \log p = \frac{1}{20}t + C$

$\\ \implies p = e^{\frac{t}{20} + C}$

Now, at t =0 , p = 1000

Putting these values,

$\\ \implies 1000 = e^{\frac{(0)}{20} + C} = e^C$

Also, At t=10

, $\\ \implies p = e^{\frac{(10)}{20} + C} = e^{\frac{1}{2}}.e^C = e^{\frac{1}{2}}.1000 \\ \implies p =(1.648)(1000) = 1648$

After 10 years, the total amount would be Rs.1648

Question:22 In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Answer:

Let n be the number of bacteria at any time t.

According to question,

$\frac{dn}{dt} = kn\ \ (k\ is\ a\ constant)$

$\\ \implies \int \frac{dn}{n} = \int kdt \\ \implies \log n = kt + C$

Now, at t=0, n = 100000

$\\ \implies \log (100000) = k(0) + C \\ \implies C = 5$

Again, at t=2, n= 110000

$\\ \implies \log (110000) = k(2) + 5 \\ \implies \log 11 + 4 = 2k + 5 \\ \implies 2k = \log 11 -1 =\log \frac{11}{10} \\ \implies k = \frac{1}{2}\log \frac{11}{10}$

Using these values, for n= 200000

$\\ \implies \log (200000) = kt + C \\ \implies \log 2 +5 = kt + 5 \\ \implies (\frac{1}{2}\log \frac{11}{10})t = \log 2 \\ \implies t = \frac{2\log 2}{ \log \frac{11}{10}}$

Question:23 The general solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ is

(A) $e^x + e^{-y} = C$

(B) $e^{x }+ e^{y} = C$

(D) $e^{-x }+ e^{-y} = C$

Answer:

Given,

$\frac{dy}{dx} = e^{x+y}$

$\\ \implies\frac{dy}{dx} = e^x.e^y \\ \implies\int \frac{dy}{e^y} = \int e^x.dx \\ \implies -e^{-y} = e^x + C \\ \implies e^x + e^{-y} = K\ \ \ \ (Option A)$

NCERT class 12 maths chapter 9 question answer - Exercise: 9.5

Question:1 Show that the given differential equation is homogeneous and solve each of them. $(x^2 + xy)dy = (x^2 + y^2)dx$

Answer:

The given diffrential eq can be written as
$\frac{dy}{dx}=\frac{x^{2}+y^{2}}{x^{2}+xy}$
Let $F(x,y)=\frac{x^{2}+y^{2}}{x^{2}+xy}$
Now, $F(\lambda x,\lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{(\lambda x)^{2}+(\lambda x)(\lambda y)}$
$=\frac{x^{2}+y^{2}}{x^{2}+xy} = \lambda ^{0}F(x,y)$ Hence, it is a homogeneous equation.

To solve it put y = vx
Diff erentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\v +x\frac{dv}{dx} = \frac{x^{2}+(vx)^{2}}{x^{2}+x(vx)}\\ v +x\frac{dv}{dx} = \frac{1+v^{2}}{1+v}$

$x\frac{dv}{dx} = \frac{(1+v^{2})-v(1+v)}{1+v} = \frac{1-v}{1+v}$

$( \frac{1+v}{1-v})dv = \frac{dx}{x}$

$( \frac{2}{1-v}-1)dv = \frac{dx}{x}$
Integrating on both side, we get;
$\\-2\log(1-v)-v=\log x -\log k\\ v= -2\log (1-v)-\log x+\log k\\ v= \log\frac{k}{x(1-v)^{2}}\\$
Again substitute the value $y = \frac{v}{x}$ ,we get;

$\\\frac{y}{x}= \log\frac{kx}{(x-y)^{2}}\\ \frac{kx}{(x-y)^{2}}=e^{y/x}\\ (x-y)^{2}=kxe^{-y/x}$
This is the required solution of given diff. equation

Question:2 Show that the given differential equation is homogeneousand solve each of them. $y' = \frac{x+y}{x}$

Answer:

the above differential eq can be written as,

$\frac{dy}{dx} = F(x,y)=\frac{x+y}{x}$ ............................(i)

Now, $F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x} = \lambda ^{0}F(x,y)$
Thus the given differential eq is a homogeneous equaion
Now, to solve substitute y = vx
Diff erentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx}= \frac{x+vx}{x} = 1+v$
$\\x\frac{dv}{dx}= 1\\ dv = \frac{dx}{x}$
Integrating on both sides, we get; (and substitute the value of $v =\frac{y}{x}$ )

$\\v =\log x+C\\ \frac{y}{x}=\log x+C\\ y = x\log x +Cx$
this is the required solution

Question:3 Show that the given differential equation is homogeneous and solve each of them.

$(x-y)dy - (x+y)dx = 0$

Answer:

The given differential eq can be written as;

$\frac{dy}{dx}=\frac{x+y}{x-y} = F(x,y)(let\ say)$ ....................................(i)

$F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}= \lambda ^{0}F(x,y)$
Hence it is a homogeneous equation.

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\v+x\frac{dv}{dx}= \frac{1+v}{1-v}\\ x\frac{dv}{dx} = \frac{1+v}{1-v}-v =\frac{1+v^{2}}{1-v}$

$\frac{1-v}{1+v^{2}}dv = (\frac{1}{1+v^{2}}-\frac{v}{1-v^{2}})dv=\frac{dx}{x}$
Integrating on both sides, we get;

$\tan^{-1}v-1/2 \log(1+v^{2})=\log x+C$
again substitute the value of $v=y/x$
$\\\tan^{-1}(y/x)-1/2 \log(1+(y/x)^{2})=\log x+C\\ \tan^{-1}(y/x)-1/2 [\log(x^{2}+y^{2})-\log x^{2}]=\log x+C\\ tan^{-1}(y/x) = 1/2[\log (x^{2}+y^{2})]+C$

This is the required solution.

Question:4 Show that the given differential equation is homogeneous and solve each of them.

$(x^2 - y^2)dx + 2xydy = 0$

Answer:

we can write it as;

$\frac{dy}{dx}= -\frac{(x^{2}-y^{2})}{2xy} = F(x,y)\ (let\ say)$ ...................................(i)

$F(\lambda x,\lambda y) = \frac{(\lambda x)^{2}-(\lambda y)^{2}}{2(\lambda x)(\lambda y)} = \lambda ^{0}.F(x,y)$
Hence it is a homogeneous equation

Now, to solve substitute y = vx
Diff erentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx} = \frac{ x^{2}-(vx)^{2}}{2x(vx)} =\frac{v^{2}-1}{2v}$
$\\x\frac{dv}{dx} =\frac{v^{2}+1}{2v}\\ \frac{2v}{1+v^{2}}dv=\frac{dx}{x}$
integrating on both sides, we get

$\log (1+v^{2})= -\log x +\log C = \log C/x$
$\\= 1+v^{2} = C/x\\ = x^2+y^{2}=Cx$ .............[ $v =y/x$ ]
This is the required solution.

Question:5 Show that the given differential equation is homogeneous and solve it.

$x^2\frac{dy}{dx} = x^2 - 2y^2 +xy$

Answer:

$\frac{dy}{dx}= \frac{x^{2}-2y^{2}+xy}{x^{2}} = F(x,y)\ (let\ say)$

$F(\lambda x,\lambda y)= \frac{(\lambda x)^{2}-2(\lambda y)^{2}+(\lambda .\lambda )xy}{(\lambda x)^{2}} = \lambda ^{0}.F(x,y)$ ............(i)
Hence it is a homogeneous eq

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\v+x\frac{dv}{dx}= 1-2v^{2}+v\\ x\frac{dv}{dx} = 1-2v^{2}\\ \frac{dv}{1-2v^{2}}=\frac{dx}{x}$

$1/2[\frac{dv}{(1/\sqrt{2})^{2}-v^{2}}] = \frac{dx}{x}$

On integrating both sides, we get;

$\frac{1}{2\sqrt{2}}\log (\frac{1/\sqrt{2}+v}{1/\sqrt{2}-v}) = \log x +C$
after substituting the value of $v= y/x$

$\frac{1}{2\sqrt{2}}\log (\frac{x+\sqrt{2}y}{x-\sqrt{2}y}) = \log \left | x \right | +C$

This is the required solution

Question:6 Show that the given differential equation is homogeneous and solve it.

$xdy - ydx = \sqrt{x^2 + y^2}dx$

Answer:

$\frac{dy}{dx}=\frac{y+\sqrt{x^{2}+y^{2}}}{x} = F(x,y)$ .................................(i)

$F(\mu x,\mu y)=\frac{\mu y+\sqrt{(\mu x)^{2}+(\mu y)^{2}}}{\mu x} =\mu^{0}.F(x,y)$
henxe it is a homogeneous equation

Now, to solve substitute y = vx

Diff erentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx}= v+\sqrt{1+v^{2}}=\sqrt{1+v^{2}}$

$=\frac{dv}{\sqrt{1+v^{2}}} =\frac{dx}{x}$

On integrating both sides,

$\Rightarrow \log \left | v+\sqrt{1+v^{2}} \right | = \log \left | x \right |+\log C$
Substitute the value of v=y/x , we get

$\\\Rightarrow \log \left | \frac{y+\sqrt{x^{2}+y^{2}}}{x} \right | = \log \left | Cx \right |\\ y+\sqrt{x^{2}+y^{2}} = Cx^{2}$

Required solution

Question:7 Solve.

$\left\{x\cos\left(\frac{y}{x} \right ) + y\sin\left(\frac{y}{x} \right ) \right \}ydx = \left\{y\sin\left(\frac{y}{x} \right ) - x\cos\left(\frac{y}{x} \right ) \right \}xdy$

Answer:

$\frac{dy}{dx} =\frac{x \cos(y/x)+y\sin(y/x)}{y\sin(y/x)-x\cos(y/x)}.\frac{y}{x} = F(x,y)$ ......................(i)
By looking at the equation we can directly say that it is a homogenous equation.

Now, to solve substitute y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx} =\frac{v \cos v+v^{2}\sin v}{v\sin v-\cos v}\\ =x\frac{dv}{dx} = \frac{2v\cos v}{v\sin v-\cos v}\\ =(\tan v-1/v)dv = \frac{2dx}{x}$

integrating on both sides, we get

$\\=\log(\frac{\sec v}{v})= \log (Cx^{2})\\=\sec v/v =Cx^{2}$
substitute the value of v= y/x , we get

$\\\sec(y/x) =Cxy \\ xy \cos (y/x) = k$

Required solution

Question:8 Solve.

$x\frac{dy}{dx} - y + x\sin\left(\frac{y}{x}\right ) = 0$

Answer:

$\frac{dy}{dx}=\frac{y-x \sin(y/x)}{x} = F(x,y)$ ...............................(i)

$F(\mu x, \mu y)=\frac{\mu y-\mu x \sin(\mu y/\mu x)}{\mu x} = \mu^{0}.F(x,y)$
it is a homogeneous equation

Now, to solve substitute y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx}= v- \sin v = -\sin v$
$\Rightarrow -\frac{dv}{\sin v} = -(cosec\ v)dv=\frac{dx}{x}$

On integrating both sides we get;

$\\\Rightarrow \log \left | cosec\ v-\cot v \right |=-\log x+ \log C\\ \Rightarrow cosec (y/x) - \cot (y/x) = C/x$

$= x[1-\cos (y/x)] = C \sin (y/x)$ Required solution

Question:9 Solve.

$ydx + x\log\left(\frac{y}{x} \right ) -2xdy = 0$

Answer:

$\frac{dy}{dx}= \frac{y}{2x-x \log(y/x)} = F(x,y)$ ..................(i)

$\frac{\mu y}{2\mu x-\mu x \log(\mu y/\mu x)} = F(\mu x,\mu y) = \mu^{0}.F(x,y)$

hence it is a homogeneous eq

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx}= \frac{v}{2-\log v}\\ =x\frac{dv}{dx} = \frac{v\log v-v}{2-\log v}\\ =[\frac{1}{v(\log v-1)}-\frac{1}{v}]dv=\frac{dx}{x}$
integrating on both sides, we get; ( substituting v =y/x)

$\\\Rightarrow \log[\log(y/x)-1]-\log(y/x)=\log(Cx)\\\Rightarrow \frac{x}{y}[\log(y/x)-1]=Cx\\ \Rightarrow \log (y/x)-1=Cy$

This is the required solution of the given differential eq

Question:10 Solve.

$\left(1 + e^{\frac{x}{y}} \right )dx + e^\frac{x}{y}\left(1-\frac{x}{y}\right )dy = 0$

Answer:

$\frac{dx}{dy}=\frac{-e^{x/y}(1-x/y)}{1+e^{x/y}} = F(x,y)$ .......................................(i)

$= F(\mu x,\mu y)=\frac{-e^{\mu x/\mu y}(1-\mu x/\mu y)}{1+e^{\mu x/\mu y}} =\mu^{0}.F(x,y)$
Hence it is a homogeneous equation.

Now, to solve substitute x = yv

Diff erentiating on both sides wrt $x$
$\frac{dx}{dy}= v +y\frac{dv}{dy}$

Substitute this value in equation (i)

$\\=v+y\frac{dv}{dy} = \frac{-e^{v}(1-v)}{1+e^{v}} \\ =y\frac{dv}{dy} = -\frac{v+e^{v}}{1+e^{v}}\\ =\frac{1+e^{v}}{v+e^{v}}dv=-\frac{dy}{y}$

Integrating on both sides, we get;

$\dpi{100} \log(v+e^{v})=-\log y+ \log c =\log (c/y)\\ =[\frac{x}{y}+e^{x/y}]= \frac{c}{y}\\\Rightarrow x+ye^{x/y}=c$
This is the required solution of the diff equation.

Question:11 Solve for particular solution.

$(x + y)dy + (x -y)dx = 0;\ y =1\ when \ x =1$

Answer:

$\frac{dy}{dx}=\frac{-(x-y)}{x+y} =F(x,y)$ ..........................(i)

We can clearly say that it is a homogeneous equation.

Now, to solve substitute y = vx

Diff erentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\v+x\frac{dv}{dx}=\frac{v-1}{v+1}\\ \Rightarrow x\frac{dv}{dx} = -\frac{(1+v^{2})}{1+v}$

$\frac{1+v}{1+v^{2}}dv = [\frac{v}{1+v^{2}}+\frac{1}{1+v^{2}}]dv=-\frac{dx}{x}$

On integrating both sides

$\\=\frac{1}{2}[\log (1+v^{2})]+\tan^{-1}v = -\log x +k\\ =\log(1+v^{2})+2\tan^{-1}v=-2\log x +2k\\ =\log[(1+(y/x)^{2}).x^{2}]+2\tan^{-1}(y/x)=2k\\ =\log(x^{2}+y^{2})+2\tan^{-1}(y/x) = 2k$ ......................(ii)

Now, y=1 and x= 1

$\\=\log 2 +2\tan^{-1}1=2k\\ =\pi/2+\log 2 = 2k\\$

After substituting the value of 2k in eq. (ii)

$\log(x^{2}+y^2)+2\tan^{-1}(y/x)=\pi/2+\log 2$

This is the required solution.

Question:12 Solve for particular solution.

$x^2dy + (xy + y^2)dx = 0; y =1\ \textup{when}\ x = 1$

Answer:

$\frac{dy}{dx}= \frac{-(xy+y^{2})}{x^{2}} = F(x,y)$ ...............................(i)

$F(\mu x, \mu y)=\frac{-\mu^{2}(xy+(\mu y)^{2})}{(\mu x)^{2}} =\mu ^{0}. F(x,y)$
Hence it is a homogeneous equation

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i), we get

$\\=v+\frac{xdv}{dx}= -v- v^{2}\\ =\frac{xdv}{dx}=-v(v+2)\\ =\frac{dv}{v+2}=-\frac{dx}{x}\\ =1/2[\frac{1}{v}-\frac{1}{v+2}]dv=-\frac{dx}{x}$

Integrating on both sides, we get;

$\\=\frac{1}{2}[\log v -\log(v+2)]= -\log x+\log C\\ =\frac{v}{v+2}=(C/x)^{2}$

replace the value of v=y/x

$\frac{x^{2}y}{y+2x}=C^{2}$ .............................(ii)

Now y =1 and x = 1

$C = 1/\sqrt{3}$
therefore,

$\frac{x^{2}y}{y+2x}=1/3$

Required solution

Question:13 Solve for particular solution.

$\left [x\sin^2\left(\frac{y}{x} \right ) - y \right ]dx + xdy = 0;\ y =\frac{\pi}{4}\ when \ x = 1$

Answer:

$\frac{dy}{dx}=\frac{-[x\sin^{2}(y/x)-y]}{x} = F(x,y)$ ..................(i)

$F(\mu x,\mu y)=\frac{-[\mu x\sin^{2}(\mu y/\mu x)-\mu y]}{\mu x}=\mu ^{0}.F(x,y)$

Hence it is a homogeneous eq

Now, to solve substitute y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

on integrating both sides, we get;

$\\-\cot v =\log\left | x \right | -C\\ =\cot v = \log\left | x \right |+\log C$

On substituting v =y/x

$=\cot (y/x) = \log\left | Cx \right |$ ............................(ii)

Now, $y = \pi/4\ @ x=1$

$\\\cot (\pi/4) = \log C \\ =C=e^{1}$

put this value of C in eq (ii)

$\cot (y/x)=\log\left | ex \right |$

Required solution.

Question:14 Solve for particular solution.

$\frac{dy}{dx} - \frac{y}{x} + \textup{cosec}\left (\frac{y}{x} \right ) = 0;\ y = 0 \ \textup{when}\ x = 1$

Answer:

$\frac{dy}{dx} = \frac{y}{x} -cosec(y/x) =F(x,y)$ ....................................(i)

the above eq is homogeneous. So,
Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx}=v- cosec\ v\\ =x\frac{dv}{dx} = -cosec\ v\\ =-\frac{dv}{cosec\ v}= \frac{dx}{x}\\ =-\sin v dv = \frac{dx}{x}$

on integrating both sides, we get;

$\\=cos\ v = \log x +\log C =\log Cx\\ =\cos(y/x)= \log Cx$ .................................(ii)

now y = 0 and x =1 , we get

$C =e^{1}$

put the value of C in eq 2

$\cos(y/x)=\log \left | ex \right |$

Question:15 Solve for particular solution.

$2xy + y^2 - 2x^2\frac{dy}{dx} = 0 ;\ y = 2\ \textup{when}\ x = 1$

Answer:

The above eq can be written as;

$\frac{dy}{dx}=\frac{2xy+y^{2}}{2x^{2}} = F(x,y)$
By looking, we can say that it is a homogeneous equation.

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx}= \frac{2v+v^{2}}{2}\\ =x\frac{dv}{dx} = v^{2}/2\\ = \frac{2dv}{v^{2}}=\frac{dx}{x}$

integrating on both sides, we get;

$\\=-2/v=\log \left | x \right |+C\\ =-\frac{2x}{y}=\log \left | x \right |+C$ .............................(ii)

Now, y = 2 and x =1, we get

C =-1
put this value in equation(ii)

$\\=-\frac{2x}{y}=\log \left | x \right |-1\\ \Rightarrow y = \frac{2x}{1- \log x}$

Question:16 A homogeneous differential equation of the from $\frac{dx}{dy}= h\left(\frac{x}{y} \right )$ can be solved by making the substitution.

(A) $y = vx$

(B) $v = yx$

(D) $x =v$

Answer:

$\frac{dx}{dy}= h\left(\frac{x}{y} \right )$
for solving this type of equation put x/y = v
x = vy

option C is correct

Question:17 Which of the following is a homogeneous differential equation?

(A) $(4x + 6x +5)dy - (3y + 2x +4)dx = 0$

(B) $(xy)dx - (x^3 + y^3)dy = 0$

(D) $y^2dx + (x^2 -xy -y^2)dy = 0$

Answer:

Option D is the right answer.

$y^2dx + (x^2 -xy -y^2)dy = 0$
$\frac{dy}{dx}=\frac{y^{2}}{x^{2}-xy-y^{2}} = F(x,y)$
we can take out lambda as a common factor and it can be cancelled out

NCERT class 12 maths chapter 9 question answer - Exercise: 9.6

Question:1 Find the general solution:

$\frac{dy}{dx} + 2y = \sin x$

Answer:

Given equation is
$\frac{dy}{dx} + 2y = \sin x$
This is $\frac{dy}{dx} + py = Q$ type where p = 2 and Q = sin x
Now,
$I.F. = e^{\int pdx}= e^{\int 2dx}= e^{2x}$
Now, the solution of given differential equation is given by relation
$Y(I.F.) =\int (Q\times I.F.)dx +C$
$Y(e^{\int 2x }) =\int (\sin x\times e^{\int 2x })dx +C$
Let $I =\int (\sin x\times e^{\int 2x })$
$I = \sin x \int e^{2x}dx- \int \left ( \frac{d(\sin x)}{dx}.\int e^{2x}dx \right )dx\\ \\ I = \sin x.\frac{e^{2x}}{2}- \int \left ( \cos x.\frac{e^{2x}}{2} \right )\\ \\ I = \sin x. \frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x\int e^{2x}dx- \left ( \frac{d(\cos x)}{dx}.\int e^{2x}dx \right ) \right )dx\\ \\ I = \sin x\frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x.\frac{e^{2x}}{2}+ \int \left ( \sin x.\frac{e^{2x}}{2} \right ) \right )\\ \\ I = \sin x\frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x.\frac{e^{2x}}{2}+\frac{I}{2} \right ) \ \ \ \ \ \ \ \ \ \ \ (\because I = \int \sin xe^{2x})\\ \\ \frac{5I}{4}= \frac{e^{2x}}{4}\left ( 2\sin x-\cos x \right )\\ \\ I = \frac{e^{2x}}{5}\left ( 2\sin x-\cos x \right )$
Put the value of I in our equation
Now, our equation become
$Y.e^{x^2 }= \frac{e^{2x}}{5}\left (2 \sin x-\cos x \right )+C$
$Y= \frac{1}{5}\left (2 \sin x-\cos x \right )+C.e^{-2x}$
Therefore, the general solution is $Y= \frac{1}{5}\left (2 \sin x-\cos x \right )+C.e^{-2x}$

Question:2 Solve for general solution:

$\frac{dy}{dx} + 3y = e^{-2x}$

Answer:

Given equation is
$\frac{dy}{dx} + 3y = e^{-2x}$
This is $\frac{dy}{dx} + py = Q$ type where p = 3 and $Q = e^{-2x}$
Now,
$I.F. = e^{\int pdx}= e^{\int 3dx}= e^{3x}$
Now, the solution of given differential equation is given by the relation
$Y(I.F.) =\int (Q\times I.F.)dx +C$
$Y(e^{ 3x }) =\int (e^{-2x}\times e^{ 3x })dx +C$
$Y(e^{ 3x }) =\int (e^{x})dx +C\\ Y(e^{3x})= e^x+C\\ Y = e^{-2x}+Ce^{-3x}$
Therefore, the general solution is $Y = e^{-2x}+Ce^{-3x}$

Question:3 Find the general solution

$\frac{dy}{dx} + \frac{y}{x} = x^2$

Answer:

Given equation is
$\frac{dy}{dx} + \frac{y}{x} = x^2$
This is $\frac{dy}{dx} + py = Q$ type where $p = \frac{1}{x}$ and $Q = x^2$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{1}{x}dx}= e^{\log x}= x$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(x) =\int (x^2\times x)dx +C$
$y(x) =\int (x^3)dx +C\\ y.x= \frac{x^4}{4}+C\\$
Therefore, the general solution is $yx =\frac{x^4}{4}+C$

Question:4 Solve for General Solution.

$\frac{dy}{dx} + (\sec x)y = \tan x \ \left(0\leq x < \frac{\pi}{2} \right )$

Answer:

Given equation is
$\frac{dy}{dx} + (\sec x)y = \tan x \ \left(0\leq x < \frac{\pi}{2} \right )$
This is $\frac{dy}{dx} + py = Q$ type where $p = \sec x$ and $Q = \tan x$
Now,
$I.F. = e^{\int pdx}= e^{\int \sec xdx}= e^{\log |\sec x+ \tan x|}= \sec x+\tan x$ $(\because 0\leq x\leq \frac{\pi}{2} \sec x > 0,\tan x > 0)$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\sec x+\tan x) =\int ((\sec x+\tan x)\times \tan x)dx +C$
$y(\sec x+ \tan x) =\int (\sec x\tan x+\tan^2 x)dx +C\\y(\sec x+ \tan x) =\sec x+\int (\sec^2x-1)dx +C\\ y(\sec x+ \tan x) = \sec x +\tan x - x+C$
Therefore, the general solution is $y(\sec x+ \tan x) = \sec x +\tan x - x+C$

Question:5 Find the general solution.

$\cos^2 x\frac{dy}{dx} + y = \tan x\left(0\leq x < \frac{\pi}{2} \right )$

Answer:

Given equation is
$\cos^2 x\frac{dy}{dx} + y = \tan x\left(0\leq x < \frac{\pi}{2} \right )$
we can rewrite it as
$\frac{dy}{dx}+\sec^2x y= \sec^2x\tan x$
This is $\frac{dy}{dx} + py = Q$ where $p = \sec ^2x$ and $Q =\sec^2x \tan x$
Now,
$I.F. = e^{\int pdx}= e^{\int \sec^2 xdx}= e^{\tan x}$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(e^{\tan x}) =\int ((\sec^2 x\tan x)\times e^{\tan x})dx +C$
$ye^{\tan x} =\int \sec^2 x\tan xe^{\tan x}dx+C\\$
take
$e^{\tan x } = t\\ \Rightarrow \sec^2x.e^{\tan x}dx = dt$
$\int t.\log t dt = \log t.\int tdt-\int \left ( \frac{d(\log t)}{dt}.\int tdt \right )dt \\ \\ \int t.\log t dt = \log t . \frac{t^2}{2}- \int (\frac{1}{t}.\frac{t^2}{2})dt\\ \\ \int t.\log t dt = \log t.\frac{t^2}{2}- \int \frac{t}{2}dt\\ \\ \int t.\log t dt = \log t.\frac{t^2}{2}- \frac{t^2}{4}\\ \\ \int t.\log t dt = \frac{t^2}{4}(2\log t -1)$
Now put again $t = e^{\tan x}$
$\int \sec^2x\tan xe^{\tan x}dx = \frac{e^{2\tan x}}{4}(2\tan x-1)$
Put this value in our equation

$ye^{\tan x} =\frac{e^{2\tan x}}{4}(2\tan x-1)+C\\ \\$
Therefore, the general solution is $y =\frac{e^{\tan x}}{4}(2\tan x-1)+Ce^{-\tan x }\\$

Question:6 Solve for General Solution.

$x\frac{dy}{dx} + 2y = x^2\log x$

Answer:

Given equation is
$x\frac{dy}{dx} + 2y = x^2\log x$
Wr can rewrite it as
$\frac{dy}{dx} +2.\frac{y}{x}= x\log x$
This is $\frac{dy}{dx} + py = Q$ type where $p = \frac{2}{x}$ and $Q = x\log x$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{2}{x}dx}= e^{2\log x}=e^{\log x^2} = x^2$ $(\because 0\leq x\leq \frac{\pi}{2} \sec x > 0,\tan x > 0)$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(x^2) =\int (x\log x\times x^2)dx +C$
$x^2y = \int x^3\log x+ C$
Let
$I = \int x^3\log x\\ \\ I = \log x\int x^3dx-\int \left ( \frac{d(\log x)}{dx}.\int x^3dx \right )dx\\ \\ I = \log x.\frac{x^4}{4}- \int \left ( \frac{1}{x}.\frac{x^4}{4} \right )dx\\ \\ I = \log x.\frac{x^4}{4}- \int \left ( \frac{x^3}{4} \right )dx\\ \\ I = \log x.\frac{x^4}{4}-\frac{x^4}{16}$
Put this value in our equation
$x^2y =\log x.\frac{x^4}{4}-\frac{x^4}{16}+ C\\ \\ y = \frac{x^2}{16}(4\log x-1)+C.x^{-2}$
Therefore, the general solution is $y = \frac{x^2}{16}(4\log x-1)+C.x^{-2}$

Question:7 Solve for general solutions.

$x\log x\frac{dy}{dx} + y = \frac{2}{x}\log x$

Answer:

Given equation is
$x\log x\frac{dy}{dx} + y = \frac{2}{x}\log x$
we can rewrite it as
$\frac{dy}{dx}+\frac{y}{x\log x}= \frac{2}{x^2}$
This is $\frac{dy}{dx} + py = Q$ type where $p = \frac{1}{x\log x}$ and $Q =\frac{2}{x^2}$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{1}{x\log x} dx}= e^{\log(\log x)} = \log x$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\log x) =\int ((\frac{2}{x^2})\times \log x)dx +C$

take
$I=\int ((\frac{2}{x^2})\times \log x)dx$
$I = \log x.\int \frac{2}{x^2}dx-\int \left ( \frac{d(\log x)}{dt}.\int \frac{x^2}{2}dx \right )dx \\ \\ I= -\log x . \frac{2}{x}+ \int (\frac{1}{x}.\frac{2}{x})dx\\ \\ I = -\log x.\frac{2}{x}+ \int \frac{2}{x^2}dx\\ \\I = -\log x.\frac{2}{x}- \frac{2}{x}\\ \\$
Put this value in our equation

$y\log x =-\frac{2}{x}(\log x+1)+C\\ \\$
Therefore, the general solution is $y\log x =-\frac{2}{x}(\log x+1)+C\\ \\$

Question:8 Find the general solution.

$(1 + x^2)dy + 2xydx = \cot x dx\ (x\neq 0)$

Answer:

Given equation is
$(1 + x^2)dy + 2xydx = \cot x dx\ (x\neq 0)$
we can rewrite it as
$\frac{dy}{dx}+\frac{2xy}{(1+x^2)}= \frac{\cot x}{1+x^2}$
This is $\frac{dy}{dx} + py = Q$ type where $p = \frac{2x}{1+ x^2}$ and $Q =\frac{\cot x}{1+x^2}$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{2x}{1+ x^2} dx}= e^{\log(1+ x^2)} = 1+x^2$
Now, the solution of the given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(1+x^2) =\int ((\frac{\cot x}{1+x^2})\times (1+ x^2))dx +C$
$y(1+x^2) =\int \cot x dx+C\\ \\ y(1+x^2)= \log |\sin x|+ C\\ \\ y = (1+x^2)^{-1}\log |\sin x|+C(1+x^2)^{-1}$
Therefore, the general solution is $y = (1+x^2)^{-1}\log |\sin x|+C(1+x^2)^{-1}$

Question:9 Solve for general solution.

$x\frac{dy}{dx} + y -x +xy \cot x = 0\ (x \neq 0)$

Answer:

Given equation is
$x\frac{dy}{dx} + y -x +xy \cot x = 0\ (x \neq 0)$
we can rewrite it as
$\frac{dy}{dx}+y.\left ( \frac{1}{x}+\cot x \right )= 1$
This is $\frac{dy}{dx} + py = Q$ type where $p =\left ( \frac{1}{x}+\cot x \right )$ and $Q =1$
Now,
$I.F. = e^{\int pdx}= e^{\int \left ( \frac{1}{x}+\cot x \right ) dx}= e^{\log x +\log |\sin x|} = x.\sin x$
Now, the solution of the given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(x.\sin x) =\int 1\times x\sin xdx +C$
$y(x.\sin x) =\int x\sin xdx +C$
Lets take
$I=\int x\sin xdx \\ \\ I = x .\int \sin xdx-\int \left ( \frac{d(x)}{dx}.\int \sin xdx \right )dx\\ \\ I =- x.\cos x+ \int (\cos x)dx\\ \\ I = -x\cos x+\sin x$
Put this value in our equation
$y(x.\sin x)= -x\cos x+\sin x + C\\ y = -\cot x+\frac{1}{x}+\frac{C}{x\sin x}$
Therefore, the general solution is $y = -\cot x+\frac{1}{x}+\frac{C}{x\sin x}$

Question:10 Find the general solution.

$(x+y)\frac{dy}{dx} = 1$

Answer:

Given equation is
$(x+y)\frac{dy}{dx} = 1$
we can rewrite it as
$\frac{dy}{dx} = \frac{1}{x+y}\\ \\ x+ y =\frac{dx}{dy}\\ \\ \frac{dx}{dy}-x=y$
This is $\frac{dx}{dy} + px = Q$ type where $p =-1$ and $Q =y$
Now,
$I.F. = e^{\int pdy}= e^{\int -1 dy}= e^{-y}$
Now, the solution of given differential equation is given by relation
$x(I.F.) =\int (Q\times I.F.)dy +C$
$x(e^{-y}) =\int y\times e^{-y}dy +C$
$xe^{-y}= \int y.e^{-y}dy + C$
Lets take
$I=\int ye^{-y}dy \\ \\ I = y .\int e^{-y}dy-\int \left ( \frac{d(y)}{dy}.\int e^{-y}dy \right )dy\\ \\ I =- y.e^{-y}+ \int e^{-y}dy\\ \\ I = - ye^{-y}-e^{-y}$
Put this value in our equation
$x.e^{-y} = -e^{-y}(y+1)+C\\ x = -(y+1)+Ce^{y}\\ x+y+1=Ce^y$
Therefore, the general solution is $x+y+1=Ce^y$

Question:11 Solve for general solution.

$y dx + (x - y^2)dy = 0$

Answer:

Given equation is
$y dx + (x - y^2)dy = 0$
we can rewrite it as
$\frac{dx}{dy}+\frac{x}{y}=y$
This is $\frac{dx}{dy} + px = Q$ type where $p =\frac{1}{y}$ and $Q =y$
Now,
$I.F. = e^{\int pdy}= e^{\int \frac{1}{y} dy}= e^{\log y } = y$
Now, the solution of given differential equation is given by relation
$x(I.F.) =\int (Q\times I.F.)dy +C$
$x(y) =\int y\times ydy +C$
$xy= \int y^2dy + C$
$xy = \frac{y^3}{3}+C$
$x = \frac{y^2}{3}+\frac{C}{y}$
Therefore, the general solution is $x = \frac{y^2}{3}+\frac{C}{y}$

Question:12 Find the general solution.

$(x+3y^2)\frac{dy}{dx} = y\ (y > 0)$

Answer:

Given equation is
$(x+3y^2)\frac{dy}{dx} = y\ (y > 0)$
we can rewrite it as
$\frac{dx}{dy}-\frac{x}{y}= 3y$
This is $\frac{dx}{dy} + px = Q$ type where $p =\frac{-1}{y}$ and $Q =3y$
Now,
$I.F. = e^{\int pdy}= e^{\int \frac{-1}{y} dy}= e^{-\log y } =y^{-1}= \frac{1}{y}$
Now, the solution of given differential equation is given by relation
$x(I.F.) =\int (Q\times I.F.)dy +C$
$x(\frac{1}{y}) =\int 3y\times \frac{1}{y}dy +C$
$\frac{x}{y}= \int 3dy + C$
$\frac{x}{y}= 3y+ C$
$x = 3y^2+Cy$
Therefore, the general solution is $x = 3y^2+Cy$

Question:13 Solve for particular solution.

$\frac{dy}{dx} + 2y \tan x = \sin x; \ y = 0 \ when \ x =\frac{\pi}{3}$

Answer:

Given equation is
$\frac{dy}{dx} + 2y \tan x = \sin x; \ y = 0 \ when \ x =\frac{\pi}{3}$
This is $\frac{dy}{dx} + py = Q$ type where $p = 2\tan x$ and $Q = \sin x$
Now,
$I.F. = e^{\int pdx}= e^{\int 2\tan xdx}= e^{2\log |\sec x|}= \sec^2 x$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\sec^2 x) =\int ((\sin x)\times \sec^2 x)dx +C$
$y(\sec^2 x) =\int (\sin \times \frac{1}{\cos x}\times \sec x)dx +C\\ \\ y(\sec^2 x) = \int \tan x\sec xdx+ C\\ \\ y.\sec^2 x= \sec x+C$
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when $x= \frac{\pi}{3}$
at $x= \frac{\pi}{3}$
$0.\sec \frac{\pi}{3} = \sec \frac{\pi}{3}+C\\ \\ C = - 2$
Now,

$y.\sec^2 x= \sec x - 2\\ \frac{y}{\cos ^2x}= \frac{1}{\cos x}- 2\\ y = \cos x- 2\cos ^2 x$
Therefore, the particular solution is $y = \cos x- 2\cos ^2 x$

Question:14 Solve for particular solution.

$(1 + x^2)\frac{dy}{dx} + 2xy =\frac{1}{1 + x^2}; \ y = 0 \ when \ x = 1$

Answer:

Given equation is
$(1 + x^2)\frac{dy}{dx} + 2xy =\frac{1}{1 + x^2}; \ y = 0 \ when \ x = 1$
we can rewrite it as
$\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{(1+x^2)^2}$
This is $\frac{dy}{dx} + py = Q$ type where $p =\frac{2x}{1+x^2}$ and $Q = \frac{1}{(1+x^2)^2}$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{2x}{1+x^2}dx}= e^{\log |1+x^2|}= 1+x^2$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(1+ x^2) =\int (\frac{1}{(1+x^2)^2}\times (1+x^2))dx +C$
$y(1+x^2) =\int \frac{1}{(1+x^2)}dx +C\\ \\ y(1+x^2) = \tan^{-1}x+ C\\ \\$
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 1
at x = 1
$0.(1+1^2) = \tan^{-1}1+ C\\ \\ C =- \frac{\pi}{4}$
Now,
$y(1+x^2)= \tan^{-1}x- \frac{\pi}{4}$
Therefore, the particular solution is $y(1+x^2)= \tan^{-1}x- \frac{\pi}{4}$

Question:15 Find the particular solution.

$\frac{dy}{dx} - 3y \cot x = \sin 2x;\ y = 2\ when \ x = \frac{\pi}{2}$

Answer:

Given equation is
$\frac{dy}{dx} - 3y \cot x = \sin 2x;\ y = 2\ when \ x = \frac{\pi}{2}$
This is $\frac{dy}{dx} + py = Q$ type where $p =-3\cot x$ and $Q =\sin 2x$
Now,
$I.F. = e^{\int pdx}= e^{-3\cot xdx}= e^{-3\log|\sin x|}= \sin ^{-3}x= \frac{1}{\sin^3x}$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\frac{1}{\sin^3 x}) =\int (\sin 2x\times\frac{1}{\sin^3 x})dx +C$
$\frac{y}{\sin^3 x} =\int (2\sin x\cos x\times\frac{1}{\sin^3 x})dx +C$
$\frac{y}{\sin^3 x} =\int (2\times \frac{\cos x}{\sin x}\times\frac{1}{\sin x})dx +C$
$\frac{y}{\sin^3 x} =\int (2\times\cot x\times cosec x)dx +C$
$\frac{y}{\sin^3 x} =-2cosec x +C$
Now, by using boundary conditions we will find the value of C
It is given that y = 2 when $x= \frac{\pi}{2}$
at $x= \frac{\pi}{2}$
$\frac{2}{\sin^3\frac{\pi}{2}} = -2cosec \frac{\pi}{2}+C\\ \\ 2 = -2 +C\\ C = 4$
Now,
$y= 4\sin^3 x-2\sin^2x\\$
Therefore, the particular solution is $y= 4\sin^3 x-2\sin^2x\\$

Question:16 Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Answer:

Let f(x , y) is the curve passing through origin
Then, the slope of tangent to the curve at point (x , y) is given by $\frac{dy}{dx}$
Now, it is given that
$\frac{dy}{dx} = y + x\\ \\ \frac{dy}{dx}-y=x$
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = -1 \ and \ Q = x$
Now,
$I.F. = e^{\int pdx}= e^{\int -1dx }= e^{-x}$
Now,
$y(I.F.)= \int (Q \times I.F. )dx+ C$
$y(e^{-x})= \int (x \times e^{-x} )dx+ C$
Now, Let
$I= \int (x \times e^{-x} )dx \\ \\ I = x.\int e^{-x}dx-\int \left ( \frac{d(x)}{dx}.\int e^{-x}dx \right )dx\\ \\ I = -xe^{-x}+\int e^{-x}dx\\ \\ I = -xe^{-x}-e^{-x}\\ \\ I = -e^{-x}(x+1)$
Put this value in our equation
$ye^{-x}= -e^{-x}(x+1)+C$
Now, by using boundary conditions we will find the value of C
It is given that curve passing through origin i.e. (x , y) = (0 , 0)
$0.e^{-0}= -e^{-0}(0+1)+C\\ \\ C = 1$
Our final equation becomes
$ye^{-x}= -e^{-x}(x+1)+1\\ y+x+1=e^x$
Therefore, the required equation of the curve is $y+x+1=e^x$

Question:17 Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Answer:

Let f(x , y) is the curve passing through point (0 , 2)
Then, the slope of tangent to the curve at point (x , y) is given by $\frac{dy}{dx}$
Now, it is given that
$\frac{dy}{dx} +5= y + x \\ \\ \frac{dy}{dx}-y=x-5$
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = -1 \ and \ Q = x- 5$
Now,
$I.F. = e^{\int pdx}= e^{\int -1dx }= e^{-x}$
Now,
$y(I.F.)= \int (Q \times I.F. )dx+ C$
$y(e^{-x})= \int ((x-5) \times e^{-x} )dx+ C$
Now, Let
$I= \int ((x-5) \times e^{-x} )dx \\ \\ I = (x-5).\int e^{-x}dx-\int \left ( \frac{d(x-5)}{dx}.\int e^{-x}dx \right )dx\\ \\ I = -(x-5)e^{-x}+\int e^{-x}dx\\ \\ I = -xe^{-x}-e^{-x}+5e^{-x}\\ \\ I = -e^{-x}(x-4)$
Put this value in our equation
$ye^{-x}= -e^{-x}(x-4)+C$
Now, by using boundary conditions we will find the value of C
It is given that curve passing through point (0 , 2)
$2.e^{-0}= -e^{-0}(0-4)+C\\ \\ C = -2$
Our final equation becomes
$ye^{-x}= -e^{-x}(x-4)-2\\ y=4-x-2e^x$
Therefore, the required equation of curve is $y=4-x-2e^x$

Question:18 The Integrating Factor of the differential equation $x\frac{dy}{dx} - y = 2x^2$ is

(A) $e^{-x}$

(B) $e^{-y}$

(D) $x$

Answer:

Given equation is
$x\frac{dy}{dx} - y = 2x^2$
we can rewrite it as
$\frac{dy}{dx}-\frac{y}{x}= 2x$
Now,
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = \frac{-1}{x} \ and \ Q = 2x$
Now,
$I.F. = e^{\int pdx} = e^{\int \frac{-1}{x}dx}= e^{\int -\log x }= x^{-1}= \frac{1}{x}$
Therefore, the correct answer is (C)

Question:19 The Integrating Factor of the differential equation $(1 - y^2)\frac{dx}{dy} + yx = ay \ \ (-1<y<1)$ is

(A) $\frac{1}{{y^2 -1}}$

(B) $\frac{1}{\sqrt{y^2 -1}}$

(D) $\frac{1}{\sqrt{1 - y^2 }}$

Answer:

Given equation is
$(1 - y^2)\frac{dx}{dy} + yx = ay \ \ (-1<y<1)$
we can rewrite it as
$\frac{dx}{dy}+\frac{yx}{1-y^2}= \frac{ay}{1-y^2}$
It is $\frac{dx}{dy}+px= Q$ type of equation where $p = \frac{y}{1-y^2}\ and \ Q = \frac{ay}{1-y^2}$
Now,
$I.F. = e^{\int pdy}= e^{\int \frac{y}{1-y^2}dy}= e^{\frac{\log |1 - y^2|}{-2}}= (1-y^2)^{\frac{-1}{2}}= \frac{1}{\sqrt{1-y^2}}$
Therefore, the correct answer is (D)

Class 12 Maths Chapter 9 NCERT solutions - Miscellaneous Exercise

Question:1 Indicate Order and Degree.

(i) $\frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x$

Answer:

Given function is
$\frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x$
We can rewrite it as
$y''+5x(y')^2-6y = \log x$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y''$

Therefore, the order of the given differential equation $\frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x$ is 2
Now, the given differential equation is a polynomial equation in its derivative y '' and y 'and power raised to y '' is 1
Therefore, it's degree is 1

Question:1 Indicate Order and Degree.

(ii) $\left(\frac{dy}{dx} \right )^3 - 4\left(\frac{dy}{dx} \right )^2 + 7y = \sin x$

Answer:

Given function is
$\left(\frac{dy}{dx} \right )^3 - 4\left(\frac{dy}{dx} \right )^2 + 7y = \sin x$
We can rewrite it as
$(y')^3-4(y')^2+7y=\sin x$
Now, it is clear from the above that, the highest order derivative present in differential equation is y'

Therefore, order of given differential equation is 1
Now, the given differential equation is a polynomial equation in it's dervatives y 'and power raised to y ' is 3
Therefore, it's degree is 3

Question:1 Indicate Order and Degree.

(iii) $\frac{d^4 y}{dx^4} - \sin\left(\frac{d^3y}{dx^3} \right ) = 0$

Answer:

Given function is
$\frac{d^4 y}{dx^4} - \sin\left(\frac{d^3y}{dx^3} \right ) = 0$
We can rewrite it as
$y''''-\sin y''' = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is y''''

Therefore, order of given differential equation is 4
Now, the given differential equation is not a polynomial equation in it's dervatives
Therefore, it's degree is not defined

Question:2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i) $xy = ae^x + be^{-x} + x^2\qquad :\ x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy +x^2 -2 =0$

Answer:

Given,

$xy = ae^x + be^{-x} + x^2$

Now, differentiating both sides w.r.t. x,

$x\frac{dy}{dx} + y = ae^x - be^{-x} + 2x$

Again, differentiating both sides w.r.t. x,

$\\ (x\frac{d^2y}{dx^2} + \frac{dy}{dx}) + \frac{dy}{dx} = ae^x + be^{-x} + 2 \\ \implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = ae^x + be^{-x} + 2 \\ \implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = xy -x^2 + 2 \\ \implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 + 2$

Therefore, the given function is the solution of the corresponding differential equation.

Question:2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(ii) $y = e^x(a\cos x + b \sin x )\qquad : \ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$

Answer:

Given,

$y = e^x(a\cos x + b \sin x )$

Now, differentiating both sides w.r.t. x,

$\frac{dy}{dx} = e^x(-a\sin x + b \cos x ) + e^x(a\cos x + b \sin x ) =e^x(-a\sin x + b \cos x ) +y$

Again, differentiating both sides w.r.t. x,

$\\ \frac{d^2y}{dx^2} = e^x(-a\cos x - b \sin x ) + e^x(-a\sin x + b \cos x ) + \frac{dy}{dx} \\ = -y + (\frac{dy}{dx} -y) + \frac{dy}{dx} \\ \implies \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$

Therefore, the given function is the solution of the corresponding differential equation.

Question:2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iii) $y= x\sin 3x \qquad : \ \frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0$

Answer:

Given,

$y= x\sin 3x$

Now, differentiating both sides w.r.t. x,

$y= x\sin 3x \frac{dy}{dx} = x(3\cos 3x) + \sin 3x$

Again, differentiating both sides w.r.t. x,

$\\ \frac{d^2y}{dx^2} = 3x(-3\sin 3x) + 3\cos 3x + 3\cos 3x \\ = -9y + 6\cos 3x \\ \implies \frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0$

Therefore, the given function is the solution of the corresponding differential equation.

Question:2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iv) $x^2 = 2y^2\log y\qquad : \ (x^2 + y^2)\frac{dy}{dx} - xy = 0$

Answer:

Given,

$x^2 = 2y^2\log y$

Now, differentiating both sides w.r.t. x,

$\\ 2x = (2y^2.\frac{1}{y} + 2(2y)\log y)\frac{dy}{dx} = 2(y + 2y\log y)\frac{dy}{dx} \\ \implies \frac{dy}{dx} = \frac{x}{y(1+ 2\log y)}$

Putting $\frac{dy}{dx}\ and \ x^2$ values in LHS

$\\ (2y^2\log y + y^2)\frac{dy}{dx} - xy = y^2(2\log y + 1)\frac{x}{y(1+ 2\log y)} -xy \\ = xy - xy = 0 = RHS$

Therefore, the given function is the solution of the corresponding differential equation.

Question:3 Form the differential equation representing the family of curves given by $(x-a)^2 + 2y^2 = a^2$ , where a is an arbitrary constant.

Answer:

Given equation is
$(x-a)^2 + 2y^2 = a^2$
we can rewrite it as
$2y^2 = a^2-(x-a)^2$ -(i)
Differentiate both the sides w.r.t x
$\frac{d\left ( 2y^2 \right )}{dx}=\frac{d(a^2-(x-a)^2)}{dx}$
$4yy^{'}=4y\frac{dy}{dx}=-2(x-a)\\ \\$
$(x-a)= -2yy'\Rightarrow a = x+2yy'$ -(ii)
Put value from equation (ii) in (i)
$(-2yy')^2+2y^2= (x+2yy')^2\\ 4y^2(y')^2+2y^2= x^2+4y^2(y')^2+4xyy'\\ y' = \frac{2y^2-x^2}{4xy}$
Therefore, the required differential equation is $y' = \frac{2y^2-x^2}{4xy}$

Question:4 Prove that $x^2 - y^2 = c (x^2 + y^2 )^2$ is the general solution of differential equation $(x^3 - 3x y^2 ) dx = (y^3 - 3x^2 y) dy$ , where c is a parameter.

Answer:

Given,

$(x^3 - 3x y^2 ) dx = (y^3 - 3x^2 y) dy$

$\implies \frac{ dy}{dx} = \frac{(x^3 - 3x y^2 )}{(y^3 - 3x^2 y)}$

Now, let y = vx

$\implies \frac{ dy}{dx} = \frac{ d(vx)}{dx} = v + x\frac{dv}{dx}$

Substituting the values of y and y' in the equation,

$v + x\frac{dv}{dx} = \frac{(x^3 - 3x (vx)^2 )}{((vx)^3 - 3x^2 (vx))}$

$\\\implies v + x\frac{dv}{dx} = \frac{1 - 3v^2 }{v^3 - 3v}\\ \implies x\frac{dv}{dx} = \frac{1 - 3v^2 }{v^3 - 3v} -v = \frac{1 - v^4 }{v^3 - 3v}$

$\implies (\frac{v^3 - 3v }{1 - v^4})dv = \frac{dx}{x}$

Integrating both sides we get,

Now,

Let

$\implies$

Now,

Let v ² = p

1517901128803851

Now, substituting the values of I ₁ and I ₂ in the above equation, we get,

1517901129585614

Thus,

1517901130366853

1517901131127676

1517901132764415

$\\ (x^2 - y^2)^2 = C'^4(x^2 + y^2 )^4 \\ \implies (x^2 - y^2) = C'^2(x^2 + y^2 )^2 \\ \implies (x^2 - y^2) = K(x^2 + y^2 )^2, where\ K = C'^2$

Question:5 Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Answer:

1646973511355 Now, equation of the circle with center at (x,y) and radius r is
$(x-a)^2+(y-b)^2 = r^2$
Since, it touch the coordinate axes in first quadrant
Therefore, x = y = r
$(x-a)^2+(y-a)^2 = a^2$ -(i)
Differentiate it w.r.t x
we will get
$2(x-a)+2(y-a)y'= 0\\ \\ 2x-2a+2yy'-2ay' = 0\\ a=\frac{x+yy'}{1+y'}$ -(ii)
Put value from equation (ii) in equation (i)
$(x-\frac{x+yy'}{1+y'})^2+(y-\frac{x+yy'}{1+y'})^2=\left ( \frac{x+yy'}{1+y'} \right )^2\\ \\ (x+xy'-x-yy')^2+(y+yy'-x-yy')^2=(x+yy')^2\\ \\ (y')^2(x-y)^2+(x-y)^2=(x+yy')^2\\ \\ (x-y)^2\left ( (y')^2+1 \right )=(x+yy')^2$
Therefore, the differential equation of the family of circles in the first quadrant which touches the coordinate axes is $(x-y)^2\left ( (y')^2+1 \right )=(x+yy')^2$

Question:6 Find the general solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0$

Answer:

Given equation is
$\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0$
we can rewrite it as
$\frac{dy}{dx } =- \sqrt{\frac{1-y^2}{1-x^2}}\\ \\ \frac{dy}{\sqrt{1-y^2}}= \frac{-dx}{\sqrt{1-x^2}}$
Now, integrate on both the sides
$\sin^{-1}y + C =- \sin ^{-1}x + C'\\ \\ \sin^{-1}y+\sin^{-1}x= C$
Therefore, the general solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0$ is $\sin^{-1}y+\sin^{-1}x= C$

Question:7 Show that the general solution of the differential equation $\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$ is given by $(x + y + 1) = A (1 - x - y - 2xy)$ , where A is parameter.

Answer:

Given,

$\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$

1517901142938688

Integrating both sides,

1517901146296446

1517901147071246

1517901148634163

1517901149451944

1517901150235484

1517901151028875

1517901151812195

Let

Let A = ,

Hence proved.

Question:8 Find the equation of the curve passing through the point $\left(0,\frac{\pi}{4} \right )$ whose differential equation is $\sin x \cos y dx + \cos x \sin y dy = 0.$

Answer:

Given equation is
$\sin x \cos y dx + \cos x \sin y dy = 0.$
we can rewrite it as
$\frac{dy}{dx}= -\tan x\cot y\\ \\ \frac{dy}{\cot y}= -\tan xdx\\ \\ \tan y dy =- \tan x dx$
Integrate both the sides
$\log |\sec y|+C' = -\log|sec x|- C''\\ \\ \log|\sec y | +\log|\sec x| = C\\ \\ \sec y .\sec x = e^{C}$
Now by using boundary conditiond, we will find the value of C
It is given that the curve passing through the point $\left(0,\frac{\pi}{4} \right )$
So,
$\sec \frac{\pi}{4} .\sec 0 = e^{C}\\ \\ \sqrt2.1= e^C\\ \\ C = \log \sqrt2$
Now,
$\sec y.\sec x= e^{\log \sqrt 2}\\ \\ \frac{\sec x}{\cos y} = \sqrt 2\\ \\ \cos y = \frac{\sec x}{\sqrt 2}$
Therefore, the equation of the curve passing through the point $\left(0,\frac{\pi}{4} \right )$ whose differential equation is $\sin x \cos y dx + \cos x \sin y dy = 0.$ is $\cos y = \frac{\sec x}{\sqrt 2}$

Question:9 Find the particular solution of the differential equation $(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$ , given that $y = 1$ when $x = 0$ .

Answer:

Given equation is
$(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$
we can rewrite it as
$\frac{dy}{dx}= -\frac{(1+y^2)e^x}{(1+e^{2x})}\\ \\ \frac{dy}{1+y^2}= \frac{-e^xdx}{1+e^{2x}}$
Now, integrate both the sides
$\tan^{-1}y + C' =\int \frac{-e^{x}dx}{1+e^{2x}}$
$\int \frac{-e^{x}dx}{1+e^{2x}}\\$
Put
$e^x = t \\ e^xdx = dt$
$\int \frac{dt}{1+t^2}= \tan^{-1}t + C''$
Put $t = e^x$ again
$\int \frac{-e^{x}dx}{1+e^{2x}} = -\tan ^{-1}e^x+C''$
Put this in our equation
$\tan^{-1}y = -\tan ^{-1}e^x+C\\ \tan^{-1}y +\tan ^{-1}e^x=C$
Now, by using boundary conditions we will find the value of C
It is given that
y = 1 when x = 0
$\\ \tan^{-1}1 +\tan ^{-1}e^0=C\\ \\ \frac{\pi}{4}+\frac{\pi}{4}= C\\ C = \frac{\pi}{2}$
Now, put the value of C

$\tan^{-1}y +\tan ^{-1}e^x=\frac{\pi}{2}$
Therefore, the particular solution of the differential equation $(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$ is $\tan^{-1}y +\tan ^{-1}e^x=\frac{\pi}{2}$

Question:10 Solve the differential equation $ye^\frac{x}{y}dx = \left(xe^\frac{x}{y} + y^2 \right )dy\ (y \neq 0)$

Answer:

Given,

$ye^\frac{x}{y}dx = (xe^\frac{x}{y} + y^2)dy$

$\\ ye^\frac{x}{y}\frac{dx}{dy} = xe^\frac{x}{y} + y^2 \\ \implies e^\frac{x}{y}[y\frac{dx}{dy} -x] = y^2 \\ \implies \frac{e^\frac{x}{y}[y\frac{dx}{dy} -x]}{y^2} = 1$

Let $\large e^\frac{x}{y} = t$

Differentiating it w.r.t. y, we get,

$\\ \frac{d}{dy}e^\frac{x}{y} = \frac{dt}{dy} \\ \implies e^\frac{x}{y}.\frac{d}{dy}(\frac{x}{y}) = \frac{dt}{dy} \\ \implies \frac{e^\frac{x}{y}[y\frac{dx}{dy} -x]}{y^2} =\frac{dt}{dy}$

Thus from these two equations,we get,

$\\ \frac{dt}{dy} = 1 \\ \implies \int dt = \int dy \\ \implies t = y + C$

Question:11 Find a particular solution of the differential equation $(x - y) (dx + dy) = dx - dy,$ , given that $y = -1$ , when $x = 0$ . (Hint: put $x - y = t$ )

Answer:

Given equation is
$(x - y) (dx + dy) = dx - dy,$
Now, integrate both the sides
Put
$(x-y ) = t\\ dx - dy = dt$
Now, given equation become
$dx+dy= \frac{dt}{t}$
Now, integrate both the sides
$x+ y + C '= \log t + C''$
Put $t = x- y$ again
$x+y = \log (x-y)+ C$
Now, by using boundary conditions we will find the value of C
It is given that
y = -1 when x = 0
$0+(-1) = \log (0-(-1))+ C\\ C = -1$
Now, put the value of C

$x+y = \log |x-y|-1\\ \log|x-y|= x+y+1$
Therefore, the particular solution of the differential equation $(x - y) (dx + dy) = dx - dy,$ is $\log|x-y|= x+y+1$

Question:12 Solve the differential equation $\left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x} \right ]\frac{dx}{dy} = 1\; \ (x\neq 0)$ .

Answer:

Given,

$\left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x} \right ]\frac{dx}{dy} = 1$

1517901190189935

1517901190951594

This is equation is in the form of

p = and Q =

Now, I.F. =

We know that the solution of the given differential equation is:

$y(I.F.) = \int(Q\timesI.F.)dx + C$

151790119561795

Question:13 Find a particular solution of the differential equation $\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)$ , given that $y = 0 \ \textup{when}\ x = \frac{\pi}{2}$ .

Answer:

Given equation is
$\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)$
This is $\frac{dy}{dx} + py = Q$ type where $p =\cot x$ and $Q = 4xcosec x$ $Q = 4x \ cosec x$
Now,
$I.F. = e^{\int pdx}= e^{\int \cot xdx}= e^{\log |\sin x|}= \sin x$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\sin x ) =\int (\sin x\times 4x \ cosec x)dx +C$
$y(\sin x) =\int(\sin x\times \frac{4x}{\sin x}) +C\\ \\ y(\sin x) = \int 4x + C\\ y\sin x= 2x^2+C$
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when $x= \frac{\pi}{2}$
at $x= \frac{\pi}{2}$
$0.\sin \frac{\pi}{2 } = 2.\left ( \frac{\pi}{2} \right )^2+C\\ \\ C = - \frac{\pi^2}{2}$
Now, put the value of C
$y\sin x= 2x^2-\frac{\pi^2}{2}$
Therefore, the particular solution is $y\sin x= 2x^2-\frac{\pi^2}{2}, (sinx\neq0)$

Question:14 Find a particular solution of the differential equation $(x+1)\frac{dy}{dx} = 2e^{-y} -1$ , given that $y = 0$ when $x = 0$

Answer:

Given equation is
$(x+1)\frac{dy}{dx} = 2e^{-y} -1$
we can rewrite it as
$\frac{e^ydy}{2-e^y}= \frac{dx}{x+1}\\$
Integrate both the sides
$\int \frac{e^ydy}{2-e^y}= \log |x+1|\\$
$\int \frac{e^ydy}{2-e^y}$
Put
$2-e^y = t\\ -e^y dy = dt$
$\int \frac{-dt}{t}=- \log |t|$
put $t = 2- e^y$ again
$\int \frac{e^ydy}{2-e^y} =- \log |2-e^y|$
Put this in our equation
$\log |2-e^y| + C'= \log|1+x| + C''\\ \log (2-e^y)^{-1}= \log (1+x)+\log C\\ \frac{1}{2-e^y}= C(1+x)$

Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 0
at x = 0
$\frac{1}{2-e^0}= C(1+0)\\ C = \frac{1}{2}$
Now, put the value of C
$\frac{1}{2-e^y} = \frac{1}{2}(1+x)\\ \\ \frac{2}{1+x}= 2-e^y\\ \frac{2}{1+x}-2= -e^y\\ -\frac{2x-1}{1+x} = -e^y\\ y = \log \frac{2x-1}{1+x}$
Therefore, the particular solution is $y = \log \frac{2x-1}{1+x}, x\neq-1$

Question:15 The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Answer:

Let n be the population of the village at any time t.

According to question,

$\frac{dn}{dt} = kn\ \ (k\ is\ a\ constant)$

$\\ \implies \int \frac{dn}{n} = \int kdt \\ \implies \log n = kt + C$

Now, at t=0, n = 20000 (Year 1999)

$\\ \implies \log (20000) = k(0) + C \\ \implies C = \log2 + 4$

Again, at t=5, n= 25000 (Year 2004)

$\\ \implies \log (25000) = k(5) + \log2 + 4 \\ \implies \log 25 + 3 = 5k + \log2 +4 \\ \implies 5k = \log 25 - \log2 -1 =\log \frac{25}{20} \\ \implies k = \frac{1}{5}\log \frac{5}{4}$

Using these values, at t =10 (Year 2009)

$\\ \implies \log n = k(10)+ C \\ \implies \log n = \frac{1}{5}\log \frac{5}{4}(10) + \log2 + 4 \\ \implies \log n = \log(\frac{25.2.10000}{16}) = \log(31250) \\ \therefore n = 31250$

Therefore, the population of the village in 2009 will be 31250.

Question:16 The general solution of the differential equation $\frac{ydx - xdy}{y} = 0$ is

(A) $xy = C$

(B) $x = Cy^2$

(D) $y = Cx^2$

Answer:

Given equation is
$\frac{ydx - xdy}{y} = 0$
we can rewrite it as
$dx = \frac{x}{y}dy\\ \frac{dy}{y}=\frac{dx}{x}$
Integrate both the sides
we will get
$\log |y| = \log |x| + C\\ \log \frac{y}{x} = C \\ \frac{y}{x} = e^C\\ \frac{y}{x} = C\\ y = Cx$
Therefore, answer is (C)

Question:17 The general solution of a differential equation of the type $\frac{dx}{dy} + P_1 x = Q_1$ is

(A) $ye^{\int P_1 dy} = \int \left(Q_1 e^{\int P_1 dy} \right )dy +C$

(B) $ye^{\int P_1 dx} = \int \left(Q_1 e^{\int P_1 dx} \right )dx +C$

(D) $xe^{\int P_1 dx} = \int \left(Q_1 e^{\int P_1 dx} \right )dx +C$

Answer:

Given equation is
$\frac{dx}{dy} + P_1 x = Q_1$
and we know that the general equation of such type of differential equation is

$xe^{\int p_1dy} = \int (Q_1e^{\int p_1dy})dy+ C$
Therefore, the correct answer is (C)

Question:18 The general solution of the differential equation $e^x dy + (y e^x + 2x) dx = 0$ is

(A) $xe^y + x^2 = C$

(B) $xe^y + y^2 = C$

(D) $ye^y + x^2 = C$

Answer:

Given equation is
$e^x dy + (y e^x + 2x) dx = 0$
we can rewrite it as
$\frac{dy}{dx}+y=-2xe^{-x}$
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = 1 \ and \ Q = -2xe^{-x}$
Now,
$I.F. = e^{\int p dx }= e^{\int 1dx}= e^x$
Now, the general solution is
$y(I.F.) = \int (Q\times I.F.)dx+C$
$y(e^x) = \int (-2xe^{-x}\times e^x)dx+C\\ ye^x= \int -2xdx + C\\ ye^x=- x^2 + C\\ ye^x+x^2 = C$
Therefore, (C) is the correct answer

If you want to get command on concepts then differential equations solutions of NCERT exercise are listed below

Differential Equations Class 12 - Topics

9.1 Introduction

9.2 Basic Concepts

9.2.1. Order of a differential equation

9.2.2 Degree of a differential equation

9.3. General and Particular Solutions of a Differential Equation

9.4 Formation of a Differential Equation whose General Solution is given

9.4.1 Procedure to form a differential equation that will represent a given family of curves

9.5. Methods of Solving First Order, First Degree Differential Equations

9.5.1 Differential equations with variables separable

9.5.2 Homogeneous differential equations

9.5.3 Linear differential equations

NCERT Exemplar Class 12 Solutions - Subject Wise

So, what is basically a differential equation? A differential equation is an equation in which derivatives of the dependent variable with respect to independent variables involved. Let's understand it with an example from NCERT chapter 9 differential equations-

$\\x^2-3x+3=0\;\:\:\:\:\:\:\:\:\:\:\:\:.....(1)\\ Sin\:x+cos\:x=0\:\:\:\:\:\:\:\:\:\:\:\:....(2)\\x\:+\:y\:=0\;\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....(3)\\x\frac{dy}{dx}+y=0\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.....(4)$

From the above equations, we notice that equations (1), (2) and (3) involve dependent variable(variables) and/or independent only but equation (4) involves variables as well as derivative of the dependent variable (y) with respect to the independent variable (x). That type of equation is known as the differential equation.

Important terms used in class 12 chapter 9 differential equations-

Order of a differential equation - It is the order of the highest order derivative present in the equation.
Degree of a differential equation - It is the power of the highest order derivative in the differential equation.
Homogeneous differential equation - A differential equation that can be expressed in the form $\frac{dy}{dx}=f(x,\:y)$ where $f(x,\:y)$ is a homogeneous function of degree zero.
First order linear differential equation - A differential equation of the form $\dpi{80} \frac{dy}{dx}\:+\:Py=Q$ where P and Q are constants or functions of x only.

NCERT solutions for class 12 maths - Chapter Wise

chapter 1	NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
chapter 2	NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions
chapter 3	NCERT solutions for class 12 maths chapter 3 Matrices
chapter 4	NCERT solutions for class 12 maths chapter 4 Determinants
chapter 5	NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability
chapter 6	NCERT solutions for class 12 maths chapter 6 Application of Derivatives
chapter 7	NCERT solutions for class 12 maths chapter 7 Integrals
chapter 8	NCERT solutions for class 12 maths chapter 8 Application of Integrals
chapter 9	NCERT solutions for class 12 maths chapter 9 Differential Equations
chapter 10	NCERT solutions for class 12 maths chapter 10 Vector Algebra
chapter 11	NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry
chapter 12	NCERT solutions for class 12 maths chapter 12 Linear Programming
chapter 13	NCERT solutions for class 12 maths chapter 13 Probability

Key Features of Class 12 Differential Equations NCERT Solutions – Differential Equations

Differential equations class 12 ncert solutions are designed to help students understand the various concepts and techniques involved in solving differential equations. Some of the key features of these solutions are:

Comprehensive coverage: The class 12 maths ch 9 question answer cover all the topics included in the Class 12 Maths syllabus, ensuring that students are well-prepared for their exams.
Simple language: The class 12 maths ch 9 question answer are written in simple language, making it easy for students to understand the concepts and techniques involved in solving differential equations.
Step-by-step approach: The class 12 differential equations solutions follow a step-by-step approach, which helps students to understand the solution process in a structured way.
Well-illustrated solutions: The maths chapter 9 class 12 solutions are accompanied by diagrams and illustrations, which help students to visualize the solution process and understand the concepts better.
Conceptual clarity: The maths chapter 9 class 12 solutions aim to develop the conceptual clarity of students, rather than just providing them with the final answers. This helps students to build a strong foundation in the subject.

NCERT solutions for class 12 subject wise

NCERT Solutions class wise

Tips to use NCERT Class 12 Maths Chapter 9 Solutions

NCERT solutions for class 12 maths chapter 9 differential equations are very helpful for the preparation of this chapter. Here are some tips to get command on it.

Differential equations are the easiest part of the class 12 calculus. If your concepts of integration are clear then it won't take much effort to get command on this chapter.
First, solve all NCERT problems including examples and miscellaneous exercise on your own. If you are not able to solve you can take the help of NCERT solutions for ch 9 maths class 12 differential equations which are provided here
If you have solved NCERT problems, you can solve previous years paper. It gives you an idea about the type of questions and difficulty levels of questions that have been asked in previous years

NCERT Books and NCERT Syllabus

Happy learning !!!

Frequently Asked Questions (FAQs)

1. How the NCERT solutions are helpful in the board exam ?

As CBSE board exam paper is designed entirely based on NCERT textbooks and most of the questions in CBSE board exam are directly asked from NCERT textbook, students must know the NCERT very well to perform well in the exam. Only knowing the answer it not enough to perform well in the exam. In the NCERT solutions students will get know how best to write answer in the board exam in order to get good marks.

2. What is the weightage of the chapter Differential Equations for CBSE board exam ?

Generally, one question of 5 marks is asked from this chapter in the 12^thboard final exam. you should refer NCER syllabus for it. NCERT textbook and NCERT Notes are recommended if you want to obtain meritious marks in the Board exam.

3. What are the important topics in chapter Differential Equations ?

Basic concepts of differential equation, order and degree of the differential equation, general and particular solutions of a differential equation, formation of a differential equation, methods of solving first order,first degree differential equations, homogeneous differential equations and linear differential equations are the important topics of this chapter.

4. Are the answers to all the textbook questions available in the differential equations ncert solutions maths class 12 chapter 9?

The NCERT class 12 maths differential equations are available in PDF format and have been created by subject experts in line with the textbook questions. These solutions for ch 9 maths class 12 adhere to the latest CBSE Syllabus for 2023 and encompass all the significant concepts for the board exam. The problems in the textbook are solved step by step in accordance with the marks weightage in the CBSE Board exams. Careers360 website offers both chapter-wise and exercise-wise PDF links that can be used by students to instantly clarify their doubts.

Also Read

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

NCERT Class 12 Physics Chapter 8 Notes Electromagnetic Waves - Download PDF

NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

NCERT Class 12 Physics Chapter 6 Notes Electromagnetic Induction - Download PDF

NCERT Class 12 Physics Chapter 5 Notes Magnetism and Matter - Download PDF

NCERT Class 12 Physics Chapter 4 Notes Moving Charges and Magnetism - Download PDF

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer

kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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