Let $y=y(x)$ be the solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=2 x(x+y)^3-x(x+y)-1, y(0)=1$. Then, $\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2$ e

$\begin{aligned} & \frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1 \\\\ & \text { Put } x+y=t \\\\ & \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1 \\\\ & \frac{d t}{d x}-1=2 x(t)^3-x t\end{aligned}$ $\begin{aligned} \Rightarrow & \frac{d t}{2 t^3-t}=x d x \\\\ & \int \frac{1}{2 t^3-t} d t=\int x d x \\\\ \Rightarrow & \int \frac{t}{2 t^4-t^2} d t=\int x d x\end{aligned}$ $\begin{aligned} & t^2=z \\\\ & 2 t d t=d z \\\\ & \frac{1}{2} \int \frac{d z}{2 z^2-z}=\int x d x \\\\ & \ln \left|\frac{z-\frac{1}{2}}{z

Differential Equations — Page 14 | JEE Mathematics

Q131

Let

y=y(x)

be the solution of the differential equation

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

. Then,

\left(\dfrac{1}{\sqrt{2}}+y\left(\dfrac{1}{\sqrt{2}}\right)\right)^2

equals :

A

\dfrac{4}{4+\sqrt{\mathrm{e}}}

B

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

C

\dfrac{2}{1+\sqrt{\mathrm{e}}}

D

\dfrac{1}{2-\sqrt{\mathrm{e}}}

Correct Answer

Option D

Solution

$\begin{aligned} & \dfrac{d y}{d x}=2 x(x+y)^3-x(x+y)-1 \\\\ & \text{ Put } x+y=t \\\\ & \Rightarrow \dfrac{d y}{d x}=\dfrac{d t}{d x}-1 \\\\ & \dfrac{d t}{d x}-1=2 x(t)^3-x t\end{aligned}$ $\begin{aligned} \Rightarrow & \dfrac{d t}{2 t^3-t}=x d x \\\\ & \int \dfrac{1}{2 t^3-t} d t=\int x d x \\\\ \Rightarrow & \int \dfrac{t}{2 t^4-t^2} d t=\int x d x\end{aligned}$ $\begin{aligned} & t^2=z \\\\ & 2 t d t=d z \\\\ & \dfrac{1}{2} \int \dfrac{d z}{2 z^2-z}=\int x d x \\\\ & \ln \left|\dfrac{z-\dfrac{1}{2}}{z}\right|=x^2+c\end{aligned}$ $\ln \left|\dfrac{(x+y)^2-\dfrac{1}{2}}{(x+y)^2}\right|=x^2+c$ $y(0)=1 \Rightarrow c=\ln \left(\dfrac{1}{2}\right)$ $\Rightarrow \dfrac{(x+y)^2-\dfrac{1}{2}}{(x+y)^2}=e^{x^2} \times \dfrac{1}{2}$ $\dfrac{(x+y)^2-\dfrac{1}{2}}{(x+y)^2}=\sqrt{e} \times \dfrac{1}{2}$ $\Rightarrow(x+y)^2=\dfrac{1}{2-\sqrt{e}}$

Q132

Let

x=x(\mathrm{t})

and

y=y(\mathrm{t})

be solutions of the differential equations

\dfrac{\mathrm{d} x}{\mathrm{dt}}+\mathrm{a} x=0

and

\dfrac{\mathrm{d} y}{\mathrm{dt}}+\mathrm{by}=0

respectively,

\mathrm{a}, \mathrm{b} \in \mathbf{R}

. Given that

x(0)=2 ; y(0)=1

and

3 y(1)=2 x(1)

, the value of

\mathrm{t}

, for which

x(\mathrm{t})=y(\mathrm{t})

, is :

A

\log _{\dfrac{2}{3}} 2

B

\log _{\dfrac{4}{3}} 2

C

\log _4 3

D

\log _3 4

Correct Answer

Option B

Solution

\begin{aligned} & \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0 \\ & \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{adt} \\ & \int \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{a} \int \mathrm{dt} \\ & \ln |\mathrm{x}|=-\mathrm{at}+\mathrm{c} \\ & \mathrm{at} \mathrm{t}=0, \mathrm{x}=2 \\ & \ln 2=0+\mathrm{c} \\ & \ln \mathrm{x}=-\mathrm{at}+\ln 2 \\ & \frac{\mathrm{x}}{2}=\mathrm{e}^{-\mathrm{at}} \\ & \mathrm{x}=2 \mathrm{e}^{-\mathrm{at}} \quad \text{.... (i)} \end{aligned}

\begin{aligned} & \frac{d y}{d t}+b y=0 \\ & \frac{d y}{y}=-b d t \\ & \ln |y|=-b t+\lambda \\ & t=0, y=1 \\ & 0=0+\lambda \\ & y=e^{-b t} \quad \text{..... (ii)} \end{aligned}

According to question

\begin{aligned} & 3 \mathrm{y}(1)=2 \mathrm{x}(1) \\ & 3 \mathrm{e}^{-\mathrm{b}}=2\left(2 \mathrm{e}^{-\mathrm{a}}\right) \\ & \mathrm{e}^{\mathrm{a}-\mathrm{b}}=\frac{4}{3} \end{aligned}

\begin{aligned} & \text{ For } x(t)=y(t) \\ & \begin{array}{r} 2 \mathrm{e}^{-a t}=e^{-b t} \\ 2=e^{(a-b) t} \\ 2=\left(\frac{4}{3}\right)^t \\ \log _{\frac{4}{3}} 2=t \end{array} \end{aligned}

Q133

The solution of the differential equation

(x^2+y^2) \mathrm{d} x-5 x y \mathrm{~d} y=0, y(1)=0

, is :

A

\left|x^2-4 y^2\right|^5=x^2

B

\left|x^2-2 y^2\right|^6=x

C

\left|x^2-2 y^2\right|^5=x^2

D

\left|x^2-4 y^2\right|^6=x

Correct Answer

Option A

Solution

\begin{aligned} & \left(x^2+y^2\right) d x-5 x y d y=0 \\ & \frac{d y}{d x}=\frac{x^2+y^2}{5 x y} \\ & \text{ Let } y=v x \\ & \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & V+x \frac{d v}{d x}=\frac{1+v^2}{5 v} \\ & x \frac{d v}{d x}=\frac{1+v^2-5 v^2}{5 v} \end{aligned}

\begin{aligned} & \frac{1}{8} \int \frac{8 \times 5 v d v}{1-4 v^2}=\int \frac{d x}{x} \\ & \frac{-5}{8} \ln \left|1-4 v^2\right|=\ln |x|+\ln c \\ & \Rightarrow \frac{5}{8} \ln \frac{\left|x^2-4 y^2\right|}{x^2}+\ln |x|=\ln c \\ & \left|\frac{x^2-4 y^2}{x^2}\right|^{5 / 8}|x|=c \\ & \frac{\left|x^2-4 y^2\right|^{5 / 8}}{|x|^{\frac{1}{4}}}=c \because y(1)=0 \Rightarrow c=1 \\ \end{aligned}

\begin{aligned} & \Rightarrow\left|x^2-4 y^2\right|^{5 / 8}=|x|^{\frac{1}{4}} \\ & \left|x^2-4 y^2\right|^5=x^2 \end{aligned}

Q134

Let

y=y(x)

be the solution of the differential equation

(x^2+4)^2 d y+(2 x^3 y+8 x y-2) d x=0

. If

y(0)=0

, then

y(2)

is equal to

A

2 \pi

B

\dfrac{\pi}{8}

C

\dfrac{\pi}{16}

D

\dfrac{\pi}{32}

Correct Answer

Option D

Solution

\frac{d y}{d x}+\frac{y\left(2 x^3+8 x\right)}{\left(x^2+4\right)^2}=\frac{2}{\left(x^2+4\right)^2}

\mathrm{IF}=e^{\int \frac{2 x^3+8 x}{\left(x^2+4\right)^2} d x}

\text{ Let }\left(x^2+4\right)^2=t \quad \Rightarrow 2\left(x^2+4\right)(2 x) d x=d t

\begin{gathered} =e^{\int \frac{d t}{2 t}}=e^{\log \sqrt{t}}=\sqrt{t}=\left(x^2+4\right) \\ \therefore \quad y\left(x^2+4\right)=\int \frac{2}{x^2+4}+c \\ \Rightarrow y\left(x^2+4\right)=\tan ^{-1}\left(\frac{x}{2}\right)+c \\ y(0)=0 \end{gathered}

\begin{aligned} & \Rightarrow 0=0+c \quad \Rightarrow c=0 \\ & \text{ put } x=2 \\ & y(8)=\frac{\pi}{4} \quad \Rightarrow \quad y=\frac{\pi}{32} \\ \end{aligned}

Q135

If

y=y(x)

is the solution curve of the differential equation

\left(x^2-4\right) \mathrm{d} y-\left(y^2-3 y\right) \mathrm{d} x=0, x>2, y(4)=\dfrac{3}{2}

and the slope of the curve is never zero, then the value of

y(10)

equals :

A

\dfrac{3}{1+(8)^{1 / 4}}

B

\dfrac{3}{1-(8)^{1 / 4}}

C

\dfrac{3}{1-2 \sqrt{2}}

D

\dfrac{3}{1+2 \sqrt{2}}

Correct Answer

Option A

Solution

\begin{aligned} & \left(x^2-4\right) d y-\left(y^2-3 y\right) d x=0 \\ & \Rightarrow \int \frac{d y}{y^2-3 y}=\int \frac{d x}{x^2-4} \\ & \Rightarrow \frac{1}{3} \int \frac{y-(y-3)}{y(y-3)} d y=\int \frac{d x}{x^2-4} \\ & \Rightarrow \frac{1}{3}(\ln |y-3|-\ln |y|)=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \\ & \Rightarrow \frac{1}{3} \ln \left|\frac{y-3}{y}\right|=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \end{aligned}

$$\begin{aligned} & \text { At } \mathrm{x}=4, \mathrm{y}=\frac{3}{2} \\ & \therefore \mathrm{C}=\frac{1}{4} \ln 3 \\ & \therefore \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{\mathrm{x}-2}{\mathrm{x}+2}\right|+\frac{1}{4} \ln (3) \\ & \text { At } \mathrm{x}=10 \\ & \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{2}{3}\right|+\frac{1}{4} \ln (3) \\ & \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\ln 2^{3 / 4}, \forall \mathrm{x}>2, \frac{\mathrm{dy}}{\mathrm{dx}}

Q136

The temperature

T(t)

of a body at time

t=0

is

160^{\circ} \mathrm{F}

and it decreases continuously as per the differential equation

\dfrac{d T}{d t}=-K(T-80)

, where

K

is a positive constant. If

T(15)=120^{\circ} \mathrm{F}

, then

T(45)

is equal to

A 90

^\circ

F

B 85

^\circ

F

C 80

^\circ

F

D 95

^\circ

F

Correct Answer

Option A

Solution

\begin{aligned} & \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-80) \\ & \int\limits_{160}^{\mathrm{T}} \frac{\mathrm{dT}}{(\mathrm{T}-80)}=\int\limits_0^{\mathrm{t}}-\mathrm{Kdt} \\ & {[\ln |\mathrm{T}-80|]_{160}^{\mathrm{T}}=-\mathrm{kt}} \\ & \ln |\mathrm{T}-80|-\ln 80=-\mathrm{kt} \\ & \ln \left|\frac{\mathrm{T}-80}{80}\right|=-\mathrm{kt} \\ & \mathrm{T}=80+80 \mathrm{e}^{-\mathrm{kt}} \\ & 120=80+80 \mathrm{e}^{-\mathrm{k} .15} \\ & \frac{40}{80}=\mathrm{e}^{-\mathrm{k} 15}=\frac{1}{2} \\ & \therefore \mathrm{T}(45)=80+80 \mathrm{e}^{-\mathrm{k} .45} \\ & =80+80\left(\mathrm{e}^{-\mathrm{k} .15}\right)^3 \\ & =80+80 \times \frac{1}{8} \\ & =90 \end{aligned}

Q137

Let

y=y(x)

be the solution of the differential equation

\dfrac{d y}{d x}=\dfrac{(\tan x)+y}{\sin x(\sec x-\sin x \tan x)}, x \in\left(0, \dfrac{\pi}{2}\right)

satisfying the condition

y\left(\dfrac{\pi}{4}\right)=2

. Then,

y\left(\dfrac{\pi}{3}\right)

is

A

\sqrt{3}\left(2+\log _e 3\right)

B

\sqrt{3}\left(1+2 \log _e 3\right)

C

\sqrt{3}\left(2+\log _e \sqrt{3}\right)

D

\dfrac{\sqrt{3}}{2}\left(2+\log _e 3\right)

Correct Answer

Option C

Solution

\begin{aligned} & \frac{d y}{d x}=\frac{\sin x+y \cos x}{\sin x \cdot \cos x\left(\frac{1}{\cos x}-\sin x \cdot \frac{\sin x}{\cos x}\right)} \\ & =\frac{\sin x+y \cos x}{\sin x\left(1-\sin ^2 x\right)} \\ & \frac{d y}{d x}=\sec ^2 x+y \cdot 2(\operatorname{cosec} 2 x) \\ & \frac{d y}{d x}-2 \operatorname{cosec}(2 x) \cdot y=\sec ^2 x \\ & \frac{d y}{d x}+p \cdot y=Q \end{aligned}

\text{ I.F. }=\mathrm{e}^{\int p \mathrm{dx}}=\mathrm{e}^{\int-2 \operatorname{cosec}(2 \mathrm{x}) \mathrm{dx}}

Let

2 \mathrm{x}=\mathrm{t}

2 \frac{\mathrm{dx}}{\mathrm{dt}}=1

\mathrm{dx}=\frac{\mathrm{dt}}{2}

=\mathrm{e}^{-\int \operatorname{cosec}(\mathrm{t}) \mathrm{dt}}

=\mathrm{e}^{-\ln \left|\tan \frac{t}{2}\right|}

=\mathrm{e}^{-\ln |\tan x|}=\frac{1}{|\tan x|}

\begin{aligned} & y(\text{ IF })=\int Q(I F) d x+c \\ & \Rightarrow y \frac{1}{|\tan x|}=\int \sec ^2 x \cdot \frac{1}{|\tan x|}+c \\ & y \cdot \frac{1}{|\tan x|}=\int \frac{d t}{|t|}+c \quad \text{ for } \tan x=t \\ & y \cdot \frac{1}{|\tan x|}=\ln |t|+c \\ & y=|\tan x|(\ln |\tan x|+c) \\ & \text{ Put } x=\frac{\pi}{4}, y=2 \\ & 2=\ln 1+c \Rightarrow c=2 \\ & y=|\tan x|(\ln |\tan x|+2) \\ & y\left(\frac{\pi}{3}\right)=\sqrt{3}(\ln \sqrt{3}+2) \end{aligned}

Q138

The solution curve of the differential equation

y \dfrac{d x}{d y}=x\left(\log _e x-\log _e y+1\right), x>0, y>0

passing through the point

(e, 1)

is

A

\left|\log _e \dfrac{y}{x}\right|=y^2

B

\left|\log _e \dfrac{y}{x}\right|=x

C

\left|\log _e \dfrac{x}{y}\right|=y

D

2\left|\log _e \dfrac{x}{y}\right|=y+1

Correct Answer

Option C

Solution

\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{x}}{\mathrm{y}}\left(\ln \left(\frac{\mathrm{x}}{\mathrm{y}}\right)+1\right)

Let

\frac{x}{y}=t \Rightarrow x=t y

\begin{aligned} & \frac{d x}{d y}=t+y \frac{d t}{d y} \\ & t+y \frac{d t}{d y}=t(\ln (t)+1) \end{aligned}

\mathrm{y} \frac{\mathrm{dt}}{\mathrm{dy}}=\mathrm{t} \ln (\mathrm{t}) \Rightarrow \frac{\mathrm{dt}}{\mathrm{t} \ln (\mathrm{t})}=\frac{\mathrm{dy}}{\mathrm{y}}

\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t} \cdot \ln (\mathrm{t})}=\int \frac{\mathrm{dy}}{\mathrm{y}}

\Rightarrow \int \frac{d p}{p}=\int \frac{d y}{y} \quad

let

\ln t=p

\frac{1}{\mathrm{t}} \mathrm{dt}=\mathrm{dp}

\begin{aligned} & \Rightarrow \ln p=\ln y+c \\ & \ln (\ln t)=\ln y+c \\ & \ln \left(\ln \left(\frac{x}{y}\right)\right)=\ln y+c \\ & \text{ at } x=e, y=1 \\ & \ln \left(\ln \left(\frac{e}{1}\right)\right)=\ln (1)+c \Rightarrow c=0 \end{aligned}

\begin{aligned} & \ln \left|\ln \left(\frac{x}{y}\right)\right|=\ln y \\ & \left|\ln \left(\frac{x}{y}\right)\right|=e^{\ln y} \\ & \left|\ln \left(\frac{x}{y}\right)\right|=y \end{aligned}

Q139

A function

y=f(x)

satisfies

f(x) \sin 2 x+\sin x-\left(1+\cos ^2 x\right) f^{\prime}(x)=0

with condition

f(0)=0

. Then,

f\left(\dfrac{\pi}{2}\right)

is equal to

A 2

B 1

C

-

1

D 0

Correct Answer

Option B

Solution

\begin{aligned} & \frac{d y}{d x}-\left(\frac{\sin 2 x}{1+\cos ^2 x}\right) y=\sin x \\ & \text{ I.F. }=1+\cos ^2 x \\ & y \cdot\left(1+\cos ^2 x\right)=\int(\sin x) d x \\ & =-\cos x+C \\ & x=0, C=1 \\ & y\left(\frac{\pi}{2}\right)=1 \end{aligned}

Q140

If

\sin \left(\dfrac{y}{x}\right)=\log _e|x|+\dfrac{\alpha}{2}

is the solution of the differential equation

x \cos \left(\dfrac{y}{x}\right) \dfrac{d y}{d x}=y \cos \left(\dfrac{y}{x}\right)+x

and

y(1)=\dfrac{\pi}{3}

, then

\alpha^2

is equal to

A 12

B 9

C 4

D 3

Correct Answer

Option D

Solution

Differential equation :-

\begin{aligned} & x \cos \frac{y}{x} \frac{d y}{d x}=y \cos \frac{y}{x}+x \\ & \cos \frac{y}{x}\left[x \frac{d y}{d x}-y\right]=x \end{aligned}

Divide both sides by

\mathrm{x}^2

\cos \frac{y}{x}\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)=\frac{1}{x}

Let

\frac{y}{x}=t

\begin{aligned} & \cos \mathrm{t}\left(\frac{\mathrm{dt}}{\mathrm{dx}}\right)=\frac{1}{\mathrm{x}} \\ & \cos \mathrm{t~dt}=\frac{1}{\mathrm{x}} \mathrm{dx} \end{aligned}

Integrating both sides

\begin{aligned} & \sin \mathrm{t}=\ln |\mathrm{x}|+\mathrm{c} \\ & \sin \frac{\mathrm{y}}{\mathrm{x}}=\ln |\mathrm{x}|+\mathrm{c} \end{aligned}

Using

\mathrm{y}(1)=\frac{\pi}{3}

, we get

\mathrm{c}=\frac{\sqrt{3}}{2}

So,

\alpha=\sqrt{3} \Rightarrow \alpha^2=3