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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

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.

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

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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 - 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

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>> 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 (y''')^2 + (y'')^3 + (y')^4 + y^5= 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 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

(C) 1

(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

(C) 0

(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

(C) \frac{d^2y}{dx^2} +1 = 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

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

(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,

1517900463071792

1517900463878580

1517900464704155

1517900465489721

Now comparing the coefficients

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

Solving these:

1517900467056911

Putting the values of A,B,C:

1517900467842360

Therefore,

1517900468626376

1517900469359944

1517900470082760

1517900470844580

151790047163329

151790047240963

1517900473174127

1517900473936125

1517900474702185

1517900475486768

Now, y= 1 when x = 0

1517900477940526

c = 1

Putting the value of c, we get:

1517900478708257

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

1517900483388423

1517900484190962

Now comparing the values of A,B,C

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

Solving these:

1517900484965476

Now putting the values of A,B,C

15179004857815100

1517900486545318

1517900487390967

1517900488152880

Given, y =0 when x =2

1517900489745379

1517900490509113

1517900492095338

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}}


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

(C) x = vy

(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

(C) (x^3 +2y^2)dx + 2xydy =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}

(C) \frac{1}{x}

(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}}

(C) \frac{1}{{1 - y^2 }}

(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,

1517901119530459

Now, 1517901120316962

1517901121031268

Let 151790112181538

151790112256298

\implies 1517901123344265

\implies 1517901124127226

Now, 1517901124894198

1517901125674574

Let v 2 = p

151790112647622

1517901127259961

1517901128042674

1517901128803851

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

1517901129585614

Thus,

1517901130366853

1517901131127676

1517901131975586

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

1517901143721318

1517901144865249

Integrating both sides,

15179011455756

1517901146296446

1517901147071246

1517901147830720

1517901148634163

1517901149451944

1517901150235484

1517901151028875

1517901151812195

Let 1517901152593257

1517901153375926

Let A = 1517901154163100 ,

1517901154943810

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}

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

1517901178627857

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 1517901191734385

p = 1517901192495126 and Q = 151790119325976

Now, I.F. = 1517901194040946

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

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

151790119561795

1517901196417819

151790119717945

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

(C) y = Cx

(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

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

(C) ye^x + x^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

More About NCERT Solutions for Class 12 Maths Chapter 9

This class 12 differential equations NCERT solutions has 5 marks weightage in 12th board final examination. Generally, one question is asked from this Chapter 9 Class 12 Maths that can be studied in detail from the NCERT Class 12 maths book in the 12th board final exam. You can score these 5 marks very easily with the help of these Ncert Solutions For Class 12 Maths Chapter 9 Differential Equations.

Class 12 Maths ch 9 is very important for the students aspiring for the 12th board exam. This NCERT Class 12 Maths Chapter 9 solutions holds good weightage in competitive exams like JEE Main, VITEEE, BITSAT. In this chapter, there are 6 exercises with 95 questions. All these questions are prepared and explained in this class 12 differential equations NCERT solutions article.

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

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:

  1. 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.

  2. 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.

  3. 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.

  4. 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.

  5. 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 Question (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 12th board 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.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hello,


Yes, you can give improvement exams for 2-3 subjects in CBSE (Central Board of Secondary Education). CBSE allows students who have appeared for their Class 12 board exams to improve their scores by re-appearing for exams in up to five subjects the following year.


Hope this helps,


Thank you

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  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

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Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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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