Let $$y=y(x)$$ be the solution of the differential equation $$\left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}$$, $$y(1)=0$$. Then $$y(0)$$ is

To determine $ y(0) $, we start by solving the differential equation given: $ (1 + x^2) \frac{dy}{dx} + y = e^{\tan^{-1} x} $ First, we rewrite it in the standard form for a linear differential equation: $ \frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{e^{\tan^{-1} x}}{1 + x^2} $ Next, we find the integrating factor (I.F.): $ \text{I.F.} = e^{\int \frac{1}{1 + x^2} dx} = e^{\tan^{-1} x} $ Multiply through by the integrating factor: $ y \cdot e^{\tan^{-1} x} = \int e^{\tan^{-1} x} \cdot \frac{e^{\tan^

Differential Equations — Page 15 | JEE Mathematics

Q141

Let

y=y(x)

be the solution curve of the differential equation

\sec y \dfrac{\mathrm{d} y}{\mathrm{~d} x}+2 x \sin y=x^3 \cos y, y(1)=0

. Then

y(\sqrt{3})

is equal to:

A

\dfrac{\pi}{6}

B

\dfrac{\pi}{12}

C

\dfrac{\pi}{3}

D

\dfrac{\pi}{4}

Correct Answer

Option D

Solution

\begin{aligned} & \sec y \frac{d y}{d x}+2 x \sin y=x^3 \cos y \\ & \Rightarrow \sec ^2 y \frac{d y}{d x}+2 x \tan y=x^3 \end{aligned}

Let

z=\tan y

\begin{aligned} & \frac{d z}{d x}=\sec ^2 y \frac{d y}{d x} \\ & \Rightarrow \frac{d z}{d x}+2 x z=x^3 \end{aligned}

\begin{aligned} & \text{ I.F. }=e^{x^2} \\ & \Rightarrow z . e^{x^2}=\int e^{x^2} \cdot x^3 d x+c \end{aligned}

\Rightarrow \tan y \cdot e^{x^2}=\frac{1}{2}\left(x^2 e^{x^2}-e^{x^2}\right)+c

\Rightarrow \tan (0) \cdot e=\frac{1}{2}(1 \cdot e-e)+c

\Rightarrow c=0

\Rightarrow \tan y=\frac{x^2-1}{2}

f(x)=\tan ^{-1}\left(\frac{x^2-1}{2}\right) \Rightarrow f(\sqrt{3})=\frac{\pi}{4}

Q142

Let

f(x)

be a positive function such that the area bounded by

y=f(x), y=0

from

x=0

to

x=a>0

is

e^{-a}+4 a^2+a-1

. Then the differential equation, whose general solution is

y=c_1 f(x)+c_2

, where

c_1

and

c_2

are arbitrary constants, is

A

\left(8 e^x+1\right) \dfrac{d^2 y}{d x^2}-\dfrac{d y}{d x}=0

B

\left(8 e^x+1\right) \dfrac{d^2 y}{d x^2}+\dfrac{d y}{d x}=0

C

\left(8 e^x-1\right) \dfrac{d^2 y}{d x^2}-\dfrac{d y}{d x}=0

D

\left(8 e^x-1\right) \dfrac{d^2 y}{d x^2}+\dfrac{d y}{d x}=0

Correct Answer

Option B

Solution

\int\limits_0^a f(x) d x=e^{-a}+4 a^2+a-1

Differentiate equation w.r.t. 'a'

\begin{aligned} & f(a)=-e^{-a}+8 a+1 \\ & \Rightarrow f(x)=-e^{-x}+8 x+1 \end{aligned}

And

y=c_1 f(x)+c_2

\begin{aligned} & y=c_1\left(-e^{-x}+8 x+1\right)+c_2 \\ & y^{\prime}=c_1\left(e^{-x}+8\right) \Rightarrow c_1=\frac{y^{\prime}}{e^{-x}+8} \end{aligned}

y^{\prime \prime}=-c_1 e^{-x}

put value of

c_1

\begin{aligned} & \frac{d^2 y}{d x^2}=\frac{-\frac{d y}{d x} \cdot e^{-x}}{\left(e^{-x}+8\right)}=\frac{\frac{d y}{d x}}{\left(1+8 e^x\right)} \\ & \Rightarrow\left(1+8 e^x\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=1 \end{aligned}

Q143

Let

y=y(x)

be the solution of the differential equation

(1+y^2) e^{\tan x} d x+\cos ^2 x(1+e^{2 \tan x}) d y=0, y(0)=1

. Then

y\left(\dfrac{\pi}{4}\right)

is equal to

A

\dfrac{1}{e^2}

B

\dfrac{2}{e^2}

C

\dfrac{2}{e}

D

\dfrac{1}{e}

Correct Answer

Option D

Solution

\begin{aligned} & \left(1+y^2\right) e^{\tan x} d x+\cos ^2 x\left(1+e^{2 \tan x}\right) d y=0 \\ & \frac{d y}{1+y^2}=-\frac{e^{\tan x} \cdot \sec ^2 x d x}{1+e^{2 \tan x}} \\ & \int \frac{d y}{1+y^2}=-\int \frac{e^{\tan x} \cdot \sec ^2 x d x}{1+e^{2 \tan x}} \end{aligned}

Let

e^{\tan x}=t

\begin{aligned} & e^{\tan x} \cdot \sec ^2 x d x=d t \\ & \int \frac{d y}{1+y^2}=-\int \frac{d t}{1+t^2} \end{aligned}

\begin{aligned} & \tan ^{-1} y=-\tan ^{-1} t+c \\ & \tan ^{-1} y=-\tan ^{-1}\left(e^{\tan x}\right)+c \\ & \text{ at } x=0, y=1 \\ & \tan ^{-1}(1)=-\tan ^{-1}(1)+c \\ & \frac{\pi}{4}=-\frac{\pi}{4}+c \\ & c=\frac{\pi}{2} \\ & \tan ^{-1} y=-\tan ^{-1}\left(e^{\tan x}\right)+\frac{\pi}{2} \end{aligned}

Now, at

x=\frac{\pi}{4}

\begin{aligned} & \tan ^{-1} y=-\tan ^{-1}(e)+\frac{\pi}{2} \\ & \tan ^{-1} y=\cot ^{-1} e=\tan ^{-1} \frac{1}{e} \\ & \Rightarrow y=\frac{1}{e} \end{aligned}

Q144

If

y=y(x)

is the solution of the differential equation

\dfrac{\mathrm{d} y}{\mathrm{~d} x}+2 y=\sin (2 x), y(0)=\dfrac{3}{4}

, then

y\left(\dfrac{\pi}{8}\right)

is equal to :

A

\mathrm{e}^{-\pi / 8}

B

\mathrm{e}^{\pi / 4}

C

\mathrm{e}^{-\pi / 4}

D

\mathrm{e}^{\pi / 8}

Correct Answer

Option C

Solution

\begin{aligned} & \frac{d y}{d x}+2 y=\sin 2 x \\ & \text{ IF }=e^{2 d x}=e^{2 x} \\ & y \cdot e^{2 x}=\int e^{2 x} \sin 2 x d x+c \\ & =\frac{e^{2 x}}{8}(2 \sin 2 x-2 \cos 2 x)+c \\ & y(0)=\frac{3}{4} \\ & \frac{3}{4}=\frac{1}{8}(-2)+c \Rightarrow c=1 \\ & \text{ Put } x=\frac{\pi}{8} \\ & y=\frac{1}{8} \times 2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)+e^{-\pi / 4} \\ & y=e^{-\pi / 4} \end{aligned}

Q145

The differential equation of the family of circles passing through the origin and having centre at the line

y=x

is :

A

\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2+2 x y\right) \mathrm{d} y

B

\left(x^2+y^2-2 x y\right) \mathrm{d} x=\left(x^2+y^2+2 x y\right) \mathrm{d} y

C

\left(x^2+y^2+2 x y\right) \mathrm{d} x=\left(x^2+y^2-2 x y\right) \mathrm{d} y

D

\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2-2 x y\right) \mathrm{d} y

Correct Answer

Option D

Solution

Equation of circle passing through origin & having centre at the line

y=x

is

\begin{aligned} & (x-t)^2+(y-t)^2=2 t^2 \\ & x^2+y^2+t^2+t^2-2 t x-2 t y=2 t^2 \\ & x^2+y^2=2 t(x+y) \end{aligned}

Now differentiate

\begin{aligned} & 2 x+2 y y^{\prime}=2 t\left(1+y^{\prime}\right) \\ & t=\frac{x+y y^{\prime}}{1+y^{\prime}} \end{aligned}

Now,

x^2+y^2=2\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)(x+y)

\begin{aligned} & x^2+y^2+x^2 \frac{d y}{d x}+y^2 \frac{d y}{d x} \\ & =2\left(x^2+x y+x y \frac{d y}{d x}+y^2 \frac{d y}{d x}\right) \\ & d x\left(x^2+y^2-2 x^2-2 x y\right)=d y\left(2 x y+2 y^2-x^2-y^2\right) \\ & d x\left(x^2-y^2+2 x y\right)=d y\left(x^2-y^2-2 x y\right) \end{aligned}

Q146

Suppose the solution of the differential equation

\dfrac{d y}{d x}=\dfrac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)}

represents a circle passing through origin. Then the radius of this circle is :

A

\sqrt{17}

B 2

C

\dfrac{\sqrt{17}}{2}

D

\dfrac{1}{2}

Correct Answer

Option C

Solution

\begin{aligned} & \frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)} \\ & \beta x d y-2 \alpha y d y-(\beta \gamma-4 \alpha) d y \\ & =2 x d x+\alpha x d x-\beta y d x+2 d x \\ & \beta \int(x d y+y d y)-\alpha y^2-(\beta \gamma-4 x) y=x^2+\frac{\alpha x^2}{2}+2 x \\ & \beta x y-\alpha y^2-(\beta \gamma-4 \alpha) y=x^2+\frac{\alpha x^2}{2}+2 x \\ & \left(1+\frac{\alpha}{2}\right) x^2+\alpha y^2-\beta x y+2 x+(\beta \gamma-4 \alpha) y=0 \\ & \because \text{ this represents circle passing through origin } \\ & \Rightarrow \beta=0 \text{ and } 1+\frac{\alpha}{2}=\alpha \\ & \Rightarrow \alpha=2 \end{aligned}

\begin{aligned} & \therefore C: 2 x^2+2 y^2+2 x-8 y=0 \\ & x^2+y^2+x-4 y=0 \\ & \text{ Radius }=\sqrt{\frac{1}{4}+4-0} \\ & \quad=\frac{\sqrt{17}}{2} \end{aligned}

Q147

Let

y=y(x)

be the solution of the differential equation

\left(2 x \log _e x\right) \dfrac{d y}{d x}+2 y=\dfrac{3}{x} \log _e x, x>0

and

y\left(e^{-1}\right)=0

. Then,

y(e)

is equal to

A

-\dfrac{3}{\mathrm{e}}

B

-\dfrac{3}{2 \mathrm{e}}

C

-\dfrac{2}{3 \mathrm{e}}

D

-\dfrac{2}{\mathrm{e}}

Correct Answer

Option A

Solution

\begin{aligned} & (2 x \log x) \frac{d y}{d x}+2 y=\frac{3}{x} \log x \\ & \frac{d y}{d x}+\frac{y}{x \log x}=\frac{3}{2 x^2} \\ & I F=e^{\int \frac{1}{x \log x}}=e^{\log (\log x)}=\log x \\ & y \times \log x=\int \log x \times \frac{3}{2 x^2} d x+C \\ & y \log x=\frac{3}{2}\left[\frac{-\log x}{x}-\frac{1}{x}\right]+C \\ & y\left(e^{-1}\right)=0 \\ & \Rightarrow 0=\frac{3}{2}[e-e]+C \\ & C=0 \\ & y \log x=\frac{-3}{2}\left[\frac{\log x+1}{x}\right] \\ & y(e) \rightarrow y(e) \times 1=\frac{-3}{2}\left[\frac{1+1}{e}\right]=-\frac{3}{e} \end{aligned}

Q148

Let

y=y(x)

be the solution of the differential equation

\left(1+x^2\right) \dfrac{d y}{d x}+y=e^{\tan ^{-1} x}

,

y(1)=0

. Then

y(0)

is

A

\dfrac{1}{4}\left(e^{\pi / 2}-1\right)

B

\dfrac{1}{2}\left(1-e^{\pi / 2}\right)

C

\dfrac{1}{4}\left(1-e^{\pi / 2}\right)

D

\dfrac{1}{2}\left(e^{\pi / 2}-1\right)

Correct Answer

Option B

Solution

To determine $y(0)$ , we start by solving the differential equation given: $(1 + x^2) \dfrac{dy}{dx} + y = e^{\tan^{-1} x}$ First, we rewrite it in the standard form for a linear differential equation: $\dfrac{dy}{dx} + \dfrac{y}{1 + x^2} = \dfrac{e^{\tan^{-1} x}}{1 + x^2}$ Next, we find the integrating factor (I.F.): $\text{I.F.} = e^{\int \dfrac{1}{1 + x^2} dx} = e^{\tan^{-1} x}$ Multiply through by the integrating factor: $y \cdot e^{\tan^{-1} x} = \int e^{\tan^{-1} x} \cdot \dfrac{e^{\tan^{-1} x}}{1 + x^2} dx$ This simplifies to: $y \cdot e^{\tan^{-1} x} = \int \dfrac{e^{2 \tan^{-1} x}}{1 + x^2} dx$ Make the substitution $\tan^{-1} x = t$ , then $\dfrac{1}{1 + x^2} dx = dt$ : $\int e^{2t} dt = \dfrac{e^{2t}}{2} + \dfrac{C}{2}$ Rewrite in terms of $x$ : $y \cdot e^{\tan^{-1} x} = \dfrac{e^{2 \tan^{-1} x}}{2} + \dfrac{C}{2}$ Use the initial condition $y(1) = 0$ : $0 = \dfrac{e^{2 \cdot \tan^{-1} 1}}{2} + \dfrac{C}{2}$ Since $\tan^{-1} 1 = \dfrac{\pi}{4}$ : $0 = \dfrac{e^{\pi/2}}{2} + \dfrac{C}{2} \implies C = -e^{\pi/2}$ Thus, the solution is: $y \cdot e^{\tan^{-1} x} = \dfrac{e^{2 \tan^{-1} x} - e^{\pi/2}}{2}$ Evaluating $y(0)$ : $y(0) = y \cdot 1 = \dfrac{1 - e^{\pi/2}}{2}$ Therefore, $y(0) = \dfrac{1}{2} (1 - e^{\pi/2})$ Hence, the correct answer is: Option B $\dfrac{1}{2}\left(1-e^{\pi / 2}\right)$

Q149

Let y = y(x) be the solution of the differential equation :

\cos x\left(\log _e(\cos x)\right)^2 d y+\left(\sin x-3 y \sin x \log _e(\cos x)\right) d x=0

, x ∈ (0,

\dfrac{\pi}{2}

). If

y(\dfrac{\pi}{4})

=

-\dfrac{1}{\log_{e}2}

, then

y(\dfrac{\pi}{6})

is equal to :

A

\dfrac{2}{\log_{e}(3)−\log_{e}(4)}

B

-\dfrac{1}{\log_{e}(4)}

C

\dfrac{1}{\log_{e}(4)−\log_{e}(3)}

D

\dfrac{1}{\log_{e}(3)−\log_{e}(4)}

Correct Answer

Option D

Solution

\begin{aligned} & \cos x(\ln (\cos x))^2 d y+(\sin x-3 y(\sin x) \ln (\cos x)) d x=0 \\ & \cos x(\ln (\cos x))^2 \frac{d y}{d x}-3 \sin x \cdot \ln (\cos x) y=-\sin x \\ & \frac{d y}{d x}-\frac{3 \tan x}{\ln (\cos x)} y=\frac{-\tan x}{(\ln (\cos x))^2} \\ & \frac{d y}{d x}+\frac{3 \tan x}{\ln (\sec x)} y=\frac{-\tan x}{(\ln (\sec x))^2} \\ & \text{ I.F. }=e^{\int \frac{3 \tan x}{\ln (\sec x)} d x}=(\ln (\sec x))^3 \end{aligned}

\begin{aligned} & y \times(\ln (\sec x))^3=-\int \frac{\tan x}{(\ln (\sec x))^2}(\ln (\sec x))^3 d x \\ & y \times(\ln (\sec x))^3=-\frac{1}{2}(\ln (\sec x))^2+C \\ & \text{ Given }: x=\frac{\pi}{4}, y=-\frac{1}{\ln 2} \\ & \frac{-1}{\ln 2} \times(\ln \sqrt{2})^3=-\frac{1}{2} \times(\ln \sqrt{2})^2+C \\ & \Rightarrow \frac{-1}{8 \ln 2} \times(\ln 2)^3=\frac{-1}{2} \times \frac{1}{4}(\ln 2)^2+C \\ & -\frac{1}{8}(\ln 2)^2=\frac{-1}{8}(\ln 2)^2+C \\ & \Rightarrow C=0 \\ & \therefore y(\ln (\sec x))^3=\frac{-1}{2}(\ln (\sec x))^2+0 \\ & y=\frac{-1}{2 \ln (\sec x)} \\ & y=\frac{1}{2 \ln (\cos x)} \end{aligned}

\begin{aligned} & \therefore y\left(\frac{\pi}{6}\right)=\frac{1}{2 \ln \left(\cos \frac{\pi}{6}\right)} \\ & =\frac{1}{2 \ln \left(\frac{\sqrt{3}}{2}\right)} \\ & =\frac{1}{2\left(\frac{1}{2} \ln 3-\ln 2\right)} \\ & =\frac{1}{\ln 3-\ln 4} \end{aligned}

Q150

If for the solution curve

y=f(x)

of the differential equation

\dfrac{d y}{d x}+(\tan x) y=\dfrac{2+\sec x}{(1+2 \sec x)^2}

,

x \in\left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right), f\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{10}

, then

f\left(\dfrac{\pi}{4}\right)

is equal to:

A

\dfrac{5-\sqrt{3}}{2 \sqrt{2}}

B

\dfrac{4 - \sqrt{2}}{14}

C

\dfrac{9\sqrt{3} + 3}{10(4 + \sqrt{3})}

D

\dfrac{\sqrt{3} + 1}{10(4 + \sqrt{3})}

Correct Answer

Option B

Solution

\begin{aligned} & \text{ If } \mathrm{e}^{\int \tan x d x}=\mathrm{e}^{\ln (\sec x)}=\sec x \\ & \therefore y \cdot \sec x=\int\left\{\frac{2+\sec x}{(1+2 \sec x)^2}\right\} \sec x d x \end{aligned}

\begin{aligned} &\begin{aligned} & =\int \frac{2 \cos x+1}{(\cos x+2)^2} d x \text{ Let } \cos x=\frac{1-t^2}{1+t^2} \\ & =\int \frac{2\left(\frac{1-t^2}{1+t^2}\right)+1}{\left(\frac{1-t^2}{1+t^2}+2\right)^2} 2 d t \\ & =\int \frac{2-2 t^2+1+t^2}{\left(1-t^2+2+2 t^2\right)^2} \times 2 d t \\ & =2 \int \frac{3-t^2}{\left(t^2+3\right)^2} d t \end{aligned}\\ &\text{ Let } \mathrm{t}+\frac{3}{\mathrm{t}}=\mathrm{u}\\ &\left(1-\frac{3}{\mathrm{t}^2}\right) \mathrm{dt}=\mathrm{du} \end{aligned}

$=-2 \int \dfrac{\mathrm{du}}{\mathrm{u}^2}$

\begin{aligned} & y \cdot(\sec x)=\frac{2}{u}+c \\ & y \cdot \sec x=\frac{2}{t+\frac{3}{t}}+c\quad\text{..... (I)} \end{aligned}

\begin{aligned} & \text{ At } \mathrm{x}=\frac{\pi}{3}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\frac{1}{\sqrt{3}} \\ & \text{ 2. } \frac{\sqrt{3}}{10}=\frac{2}{\frac{1}{\sqrt{3}}+3 \sqrt{3}}+\mathrm{c} \\ & \text{ 2. } \frac{\sqrt{3}}{10}=\frac{2 \sqrt{3}}{10}+\mathrm{c} \Rightarrow \mathrm{C}=0 \\ & \text{ At } \mathrm{x}=\frac{\pi}{4}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\sqrt{2}-1 \end{aligned}

\begin{aligned} & \therefore y \cdot \sqrt{2}=\frac{2}{\sqrt{2}-1+\frac{3}{\sqrt{2}-1}} \\ & y \cdot \sqrt{2}=\frac{2(\sqrt{2}-1)}{6-2 \sqrt{2}} \\ & y=\frac{\sqrt{2}(\sqrt{2}-1)}{2(3-\sqrt{2})}=\frac{1}{\sqrt{2}} \times \frac{2 \sqrt{2}-1}{7} \\ & =\frac{4-\sqrt{2}}{14} \end{aligned}