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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
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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Also read :
>> Ordinary Differential Equations (ODEs): Ordinary Differential Equations involve derivatives of a function concerning a single independent variable. They are commonly used to model dynamic systems and phenomena.
>> Partial Differential Equations (PDEs): Partial Differential Equations involve derivatives of a function concerning multiple independent variables. They are frequently used in physics to describe phenomena like heat diffusion, wave propagation, and fluid dynamics.
>> Types of Differential Equations: Differential equations can be categorised based on their order, linearity, and specific properties. Common types include:
First-Order Differential Equations
Second-Order Differential Equations
Linear Differential Equations
Nonlinear Differential Equations
Homogeneous Differential Equations
Non-Homogeneous Differential Equations
>> Methods for Solving Differential Equations: Various techniques can be employed to solve differential equations, including:
Separation of Variables
Integrating Factors
Exact Differential Equations
Linear Differential Equations with Constant Coefficients
Method of Undetermined Coefficients
Variation of Parameters
Laplace Transforms
>> Applications of Differential Equations: Differential equations have widespread applications in science and engineering. Some examples include modelling population growth, describing electrical circuits, predicting radioactive decay, and simulating fluid flow.
Free download NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations for CBSE Exam.
NCERT differential equations class 12 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.
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
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'''}$
Therefore, order of given differential equation is 3 Now, the given differential equation is a polynomial equation in it's dervatives $y^{'''} , y^{''} \ and \ y^{'}$ and power raised to $y^{'''}$ is 2
Therefore, it's degree is 2
Question:7 Determine order and degree (if defined) of differential equation
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
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
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
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
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
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
$y = e^x + 1 \qquad :\ y'' -y'=0$
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.
$y = x^2 + 2x + C\qquad:\ y' -2x - 2 =0$
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.
$y = \cos x + C\qquad :\ y' + \sin x = 0$
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.
$y = \sqrt{1 + x^2}\qquad :\ y' = \frac{xy}{1 + x^2}$
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.
$y = Ax\qquad :\ xy' = y\;(x\neq 0)$
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.
$y = x\sin x\qquad :\ xy' = y + x\sqrt{x^2 - y^2}\ (x\neq 0\ \textup{and} \ x > y\ or \ x < -y)$
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.
$xy = \log y + C\qquad :\ y' = \frac{y^2}{1 - xy}\ (xy\neq 1)$
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.
$y - cos y = x \qquad :(\ y\sin y + \cos y + x) y' = y$
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.
$x + y = \tan^{-1}y\qquad :\ y^2y' + y^2 + 1 = 0$
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.
$y = \sqrt{a^2 - x^2}\ x\in (-a,a)\qquad : \ x + y \frac{dy}{dx} = 0\ (y\neq 0)$
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.
Answer:
(D) 4
The number of constants in the general solution of a differential equation of order n is equal to its order.
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
$\frac{x}{a} + \frac{y}{b} = 1$
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 $y^{''} =0$
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 $x(y^{'})^2+x.y.y^{''}-y.y^{'}=0$
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 $y^{''}+y^{'}-6y=0$
Answer:
Given equation is -(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 $y^{''}-4y^{'}+4y=0$
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:
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 $y^2=2xyy^{'}+x^2$
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 $xyy^{''}-x(y^{'})^2+yy^{'}= 0$
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:
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,
Now comparing the coefficients
A + B = 2; B + C = 1; A + C = 0
Solving these:
Putting the values of A,B,C:
Therefore,
Now, y= 1 when x = 0
c = 1
Putting the value of c, we get:
Question:12 Find a particular solution satisfying the given condition:
$x(x^2 -1)\frac{dy}{dx} =1;\ y = 0\ \textup{when} \ x = 2$
Answer:
Given, in the question
$x(x^2 -1)\frac{dy}{dx} =1$
$\\ \implies \int dy=\int \frac{dx}{x(x^2 -1)} \\ \implies \int dy=\int \frac{dx}{x(x -1)(x+1)}$
Let,
Now comparing the values of A,B,C
A + B + C = 0; B-C = 0; A = -1
Solving these:
Now putting the values of A,B,C
Given, y =0 when x =2
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:
$\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + \frac{1}{2} \\ \implies 2y -1 = e^x(\sin x - \cos x)$
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:
$\\ y - 2\log (y+2) = x + 2\log x + -2 \\ \implies y -x + 2 = 2\log x(y+2)$
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
$\\ \frac{y^2}{2} = \frac{x^2}{2} + 2 \\ \implies y^2 = x^2 + 4$
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
$\\ \implies y+3 = (x+4)^2$
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:
$\\ \frac{4}{3}\pi r ^3 = 84\pi t + 36\pi \\ \implies r ^3 = (21 t + 9)(3) = 62t + 27 \\ \implies r = \sqrt[3]{62t + 27}$
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%
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
Answer:
Let n be the number of bacteria at any time t.
According to question,
$\frac{dn}{dt} = kn\ \ (k\ is\ a\ constant)$
$\\ \implies \int \frac{dn}{n} = \int kdt \\ \implies \log n = kt + C$
Now, at t=0, n = 100000
$\\ \implies \log (100000) = k(0) + C \\ \implies C = 5$
Again, at t=2, n= 110000
$\\ \implies \log (110000) = k(2) + 5 \\ \implies \log 11 + 4 = 2k + 5 \\ \implies 2k = \log 11 -1 =\log \frac{11}{10} \\ \implies k = \frac{1}{2}\log \frac{11}{10}$
Using these values, for n= 200000
$\\ \implies \log (200000) = kt + C \\ \implies \log 2 +5 = kt + 5 \\ \implies (\frac{1}{2}\log \frac{11}{10})t = \log 2 \\ \implies t = \frac{2\log 2}{ \log \frac{11}{10}}$
Question:23 The general solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ is
Answer:
Given,
$\frac{dy}{dx} = e^{x+y}$
$\\ \implies\frac{dy}{dx} = e^x.e^y \\ \implies\int \frac{dy}{e^y} = \int e^x.dx \\ \implies -e^{-y} = e^x + C \\ \implies e^x + e^{-y} = K\ \ \ \ (Option A)$
NCERT class 12 maths chapter 9 question answer - Exercise: 9.5
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.
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.
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.
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.
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)
$\\=-\frac{2x}{y}=\log \left | x \right |-1\\ \Rightarrow y = \frac{2x}{1- \log x}$
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:
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.
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.
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 $y= 4\sin^3 x-2\sin^2x\\$
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 $y+x+1=e^x$
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
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
(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
(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.
(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.
(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.
(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.
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 $y' = \frac{2y^2-x^2}{4xy}$
Answer:
Given,
$(x^3 - 3x y^2 ) dx = (y^3 - 3x^2 y) dy$
$\implies \frac{ dy}{dx} = \frac{(x^3 - 3x y^2 )}{(y^3 - 3x^2 y)}$
Now, let y = vx
$\implies \frac{ dy}{dx} = \frac{ d(vx)}{dx} = v + x\frac{dv}{dx}$
Substituting the values of y and y' in the equation,
$v + x\frac{dv}{dx} = \frac{(x^3 - 3x (vx)^2 )}{((vx)^3 - 3x^2 (vx))}$
$\\\implies v + x\frac{dv}{dx} = \frac{1 - 3v^2 }{v^3 - 3v}\\ \implies x\frac{dv}{dx} = \frac{1 - 3v^2 }{v^3 - 3v} -v = \frac{1 - v^4 }{v^3 - 3v}$
$\implies (\frac{v^3 - 3v }{1 - v^4})dv = \frac{dx}{x}$
Integrating both sides we get,
Now,
Let
$\implies$
$\implies$
Now,
Let v 2 = p
Now, substituting the values of I 1 and I 2 in the above equation, we get,
Thus,
$\\ (x^2 - y^2)^2 = C'^4(x^2 + y^2 )^4 \\ \implies (x^2 - y^2) = C'^2(x^2 + y^2 )^2 \\ \implies (x^2 - y^2) = K(x^2 + y^2 )^2, where\ K = C'^2$
Answer:
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 $(x-y)^2\left ( (y')^2+1 \right )=(x+yy')^2$
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 $\sin^{-1}y+\sin^{-1}x= C$
Answer:
Given,
$\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$
Integrating both sides,
Let
Let A = ,
Hence proved.
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 $\cos y = \frac{\sec x}{\sqrt 2}$
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,
$\\ \frac{dt}{dy} = 1 \\ \implies \int dt = \int dy \\ \implies t = y + C$
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 $\log|x-y|= x+y+1$
Answer:
Given,
$\left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x} \right ]\frac{dx}{dy} = 1$
This is equation is in the form of
p = and Q =
Now, I.F. =
We know that the solution of the given differential equation is:
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 $y\sin x= 2x^2-\frac{\pi^2}{2}, (sinx\neq0)$
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 $y = \log \frac{2x-1}{1+x}, x\neq-1$
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
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
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
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.
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
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-
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-
Differential equations class 12 ncert solutions are designed to help students understand the various concepts and techniques involved in solving differential equations. Some of the key features of these solutions are:
Comprehensive coverage: The class 12 maths ch 9 question answer cover all the topics included in the Class 12 Maths syllabus, ensuring that students are well-prepared for their exams.
Simple language: The class 12 maths ch 9 question answer are written in simple language, making it easy for students to understand the concepts and techniques involved in solving differential equations.
Step-by-step approach: The class 12 differential equations solutions follow a step-by-step approach, which helps students to understand the solution process in a structured way.
Well-illustrated solutions: The maths chapter 9 class 12 solutions are accompanied by diagrams and illustrations, which help students to visualize the solution process and understand the concepts better.
Conceptual clarity: The maths chapter 9 class 12 solutions aim to develop the conceptual clarity of students, rather than just providing them with the final answers. This helps students to build a strong foundation in the subject.
NCERT solutions for class 12 maths chapter 9 differential equations are very helpful for the preparation of this chapter. Here are some tips to get command on it.
Happy learning !!!
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.
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.
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.
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.
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
Hello there! Thanks for reaching out to us at Careers360.
Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.
Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!
Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.
If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.
Let me know if you need any other tips for your math prep. Good luck with your studies!
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I hope this information helps you.
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If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
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