Let $x=x(y)$ be the solution of the differential equation $y^2 \mathrm{~d} x+\left(x-\frac{1}{y}\right) \mathrm{d} y=0$. If $x(1)=1$, then $x\left(\frac{1}{2}\right)$ is :

$\begin{aligned} & y^2 d x+\left(x-\frac{1}{y}\right) d y=0 \\ & y^2 d x=\left(\frac{1}{y}-x\right) d y \\ & \Rightarrow y^2 \frac{d x}{d y}=\frac{1}{y}-x \\ & \Rightarrow \frac{d x}{d y}+\frac{x}{y^2}=\frac{1}{y^3} \\ & \text { I.F. }=e^{\int \frac{1}{y^2} d y}=e^{\frac{-1}{y}}\end{aligned}$ $\therefore$ Solution is $$ x e^{\frac{-1}{y}}=\int e^{-\frac{1}{y}} \times \frac{1}{y^3} d y+C $$ Let $\frac{-1}{y}=t$ $$ \Rightarrow \frac{1}{y^2} d y=d t $$ $\begin{aligned} & \Rightarrow x e^{-\frac{1}{

Differential Equations — Page 16 | JEE Mathematics

Q151

Let for some function

\mathrm{y}=f(x), \int_0^x t f(t) d t=x^2 f(x), x>0

and

f(2)=3

. Then

f(6)

is equal to

A 1

B 6

C 2

D 3

Correct Answer

Option A

Solution

\begin{aligned} &\int_0^{\mathrm{x}} \mathrm{tf}(\mathrm{t}) \mathrm{dt}=\mathrm{x}^2+(\mathrm{x}), \mathrm{x}>0\\ &\text{ Diff. both side w.r. to } x\\ &\begin{aligned} & x f(x)=x^2 f^{\prime}(x)+2 x f(x) \\ & -x f(x)=x^2 f^{\prime}(x) \\ & \int \frac{f^{\prime}(x)}{f(x)} d x=\int \frac{-1}{x} d x \\ & \operatorname{logdf}(x)=-\log x+\log c \\ & f(x)=\frac{c}{x} \end{aligned}\\ &\mathrm{f}(2)=3 \Rightarrow 3=\frac{\mathrm{c}}{2} \Rightarrow \mathrm{c}=6\\ &f(x)=\frac{6}{x}\\ &f(6)=1\quad\therefore (1) \end{aligned}

Q152

Let

x=x(y)

be the solution of the differential equation

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

. If

x(1)=1

, then

x\left(\dfrac{1}{2}\right)

is :

A

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

B

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

C

3+e

D

3-e

Correct Answer

Option D

Solution

$\begin{aligned} & y^2 d x+\left(x-\dfrac{1}{y}\right) d y=0 \\ & y^2 d x=\left(\dfrac{1}{y}-x\right) d y \\ & \Rightarrow y^2 \dfrac{d x}{d y}=\dfrac{1}{y}-x \\ & \Rightarrow \dfrac{d x}{d y}+\dfrac{x}{y^2}=\dfrac{1}{y^3} \\ & \text{ I.F. }=e^{\int \dfrac{1}{y^2} d y}=e^{\dfrac{-1}{y}}\end{aligned}$ $\therefore$ Solution is

x e^{\frac{-1}{y}}=\int e^{-\frac{1}{y}} \times \frac{1}{y^3} d y+C

Let $\dfrac{-1}{y}=t$

\Rightarrow \frac{1}{y^2} d y=d t

$\begin{aligned} & \Rightarrow x e^{-\dfrac{1}{y}}=-\int e^t t d t+C \\ & \Rightarrow x e^{-\dfrac{1}{y}}=-e^t(t-1)+C \\ & \Rightarrow x e^{-\dfrac{1}{y}}=-e^{\dfrac{-1}{y}}\left(\dfrac{-1}{y}-1\right)+C\end{aligned}$ $\begin{aligned} & x(1)=1 \\ & \Rightarrow e^{-1}=-e^{-1}(-2)+C \\ & \Rightarrow C=-e^{-1} \\ & \Rightarrow x=\dfrac{1}{y}+1-e^{-1+\dfrac{1}{y}} \\ & x\left(\dfrac{1}{2}\right)=3-e\end{aligned}$

Q153

Let

f(x)

be a real differentiable function such that

f(0)=1

and

f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y)

for all

x, y \in \mathbf{R}

. Then

\sum\limits_{n=1}^{100} \log _e f(n)

is equal to :

A 2406

B 5220

C 2525

D 2384

Correct Answer

Option C

Solution

$\begin{aligned} & f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(x) \\ & \text{ Put }=x=y=0 \\ & f(0)=f(0) f^{\prime}(0)+f^{\prime}(0) f(0) \\ & f^{\prime}(0)=\dfrac{1}{2} \\ & \text{ Put } y=0 \\ & f(x)=f(x) f^{\prime}(0)+f^{\prime}(x) f(0) \\ & f(x)=\dfrac{1}{2} f(x)+f^{\prime}(x) \\ & f^{\prime}(x)=\dfrac{f(x)}{2}\end{aligned}$

\frac{d y}{d x}=\frac{y}{2} \Rightarrow \int \frac{d y}{y}=\int \frac{d x}{2}

\Rightarrow \ln y=\frac{x}{2}+c

$\begin{aligned} & \because \mathrm{f}(0)=1 \Rightarrow \mathrm{C}=0 \\ & \ell \mathrm{ny}=\dfrac{\pi}{2} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x} / 2} \\ & \ln \mathrm{f}(\mathrm{n})=\dfrac{\mathrm{n}}{2} \\ & \sum_{\mathrm{n}=1}^{100} \ell \mathrm{f}(\mathrm{n})=\dfrac{1}{2} \sum_{\mathrm{n}=1}^{100} \mathrm{n}=\dfrac{5050}{2} \\ & =2525\end{aligned}$

Q154

Let

f: \mathbf{R} \rightarrow \mathbf{R}

be a twice differentiable function such that

f(x+y)=f(x) f(y)

for all

x, y \in \mathbf{R}

. If

f^{\prime}(0)=4 \mathrm{a}

and

f

satisfies

f^{\prime \prime}(x)-3 \mathrm{a} f^{\prime}(x)-f(x)=0, \mathrm{a}>0

, then the area of the region

\mathrm{R}=\{(x, y) \mid 0 \leq y \leq f(a x), 0 \leq x \leq 2\}

is :

A

\mathrm{e}^2-1

B

e^4+1

C

\mathrm{e}^2+1

D

e^4-1

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y}) \\ & \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\lambda \mathrm{x}} \mathrm{f}^{\prime}(0)=4 \mathrm{a} \\ & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\lambda \mathrm{e}^{\lambda \mathrm{x}} \Rightarrow \lambda=4 \mathrm{a} \end{aligned}

So, $\mathrm{f}(\mathrm{x})=\mathrm{e}^{4 \mathrm{ax}}$

\begin{aligned} & \mathrm{f}^{\prime \prime}(\mathrm{x})-3 \mathrm{af}^{\prime}(\mathrm{x})-\mathrm{f}(\mathrm{x})=0 \\ & \Rightarrow \lambda^2-3 \mathrm{a} \lambda-1=0 \\ & \Rightarrow 16 \mathrm{a}^2-12 \mathrm{a}^2-1=0 \Rightarrow 4 \mathrm{a}^2=1 \Rightarrow \mathrm{a}=\frac{1}{2} \end{aligned}

$\begin{aligned} & \mathrm{F}(\mathrm{x})=\mathrm{e}^{2 \mathrm{x}} \\ & \text{ Area }=\int_0^2 \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{e}^2-1\end{aligned}$

Q155

If

x=f(y)

is the solution of the differential equation

\left(1+y^2\right)+\left(x-2 \mathrm{e}^{\tan ^{-1} y}\right) \dfrac{\mathrm{d} y}{\mathrm{~d} x}=0, y \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)

with

f(0)=1

, then

f\left(\dfrac{1}{\sqrt{3}}\right)

is equal to :

A

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

B

e^{\pi / 12}

C

\mathrm{e}^{\pi / 6}

D

e^{\pi / 3}

Correct Answer

Option C

Solution

\begin{aligned} & \frac{d x}{d y}+\frac{x}{1+y^2}=\frac{2 e^{\tan ^{-1} y}}{1+y^2} \\ & \text{ I.F. }=e^{\tan ^{-1} y} \\ & x^{\tan ^{-1} y}=\int \frac{2\left(e^{\tan ^{-1} y}\right)^2 d y}{1+y^2} \\ & \text{ Put } \tan ^{-1} y=t, \frac{d y}{1+y^2}=d t \\ & x e^{\tan ^{-1} y}=\int 2 e^{2 t} d t \\ & x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+c \\ & x=e^{\tan ^{-1} y}+c e^{-\tan ^{-1} y} \\ & \because y=0, x=1 \\ & 1=1+c \Rightarrow c=0 \\ & y=\frac{1}{\sqrt{3}}, x=e^{\pi / 6} \end{aligned}

Q156

Let a curve

y=f(x)

pass through the points

(0,5)

and

\left(\log _e 2, k\right)

. If the curve satisfies the differential equation

2(3+y) e^{2 x} d x-\left(7+e^{2 x}\right) d y=0

, then

k

is equal to

A 32

B 8

C 4

D 16

Correct Answer

Option B

Solution

\begin{aligned} & \frac{d y}{d x}=\frac{2(3+y) \cdot e^{2 x}}{7+e^{2 x}} \\ & \frac{d y}{d x}-\frac{2 y \cdot e^{2 x}}{7+e^{2 x}}=\frac{6 \cdot e^{2 x}}{7+e^{2 x}} \\ & \text{ I.F. }=e^{-\int \frac{2 e^{2 x}}{7+e^{2 x}}}=\frac{1}{7+e^{2 x}} \end{aligned}

\begin{aligned} & \therefore y \cdot \frac{1}{7+\mathrm{e}^{2 \mathrm{x}}}=\int \frac{6 \mathrm{e}^{2 \mathrm{x}}}{\left(7+3^{2 x}\right)^2} \mathrm{dx} \\ & \quad \frac{\mathrm{y}}{7+\mathrm{e}^{2 \mathrm{x}}}=\frac{-3}{7+\mathrm{e}^{2 \mathrm{x}}}+C \\ & (0,5) \Rightarrow \frac{5}{8}=\frac{-3}{8}+C \Rightarrow C=1 \\ & \therefore y=-3+7+\mathrm{e}^{2 \mathrm{x}} \\ & y=\mathrm{e}^{2 \mathrm{x}}+4 \\ & \therefore \mathrm{k}=8 \end{aligned}

Q157

Let

x=x(y)

be the solution of the differential equation

y=\left(x-y \dfrac{\mathrm{~d} x}{\mathrm{~d} y}\right) \sin \left(\dfrac{x}{y}\right), y>0

and

x(1)=\dfrac{\pi}{2}

. Then

\cos (x(2))

is equal to :

A

2\left(\log _e 2\right)-1

B

1-2\left(\log _e 2\right)^2

C

1-2\left(\log _{\mathrm{e}} 2\right)

D

2\left(\log _e 2\right)^2-1

Correct Answer

Option D

Solution

\begin{aligned} \quad y d y & =(x d y-y d x) \sin \left(\frac{x}{y}\right) \\ \frac{d y}{y} & =\left(\frac{x d y-y d x}{y^2}\right) \sin \left(\frac{x}{y}\right) \\ \frac{d y}{y} & =\sin \left(\frac{x}{y}\right) d\left(-\frac{x}{y}\right) \\ \Rightarrow \quad \ell n y & =\cos \frac{x}{y}+C \end{aligned}

\begin{aligned} & x(1)=\frac{\pi}{2} \Rightarrow 0=\cos \frac{\pi}{2}+C \Rightarrow C=0 \\ & \text{ थny }=\cos \frac{x}{y} \\ & \text{ but } y=2 \Rightarrow \cos \frac{x}{2}=\ln 2 \\ & \qquad \begin{aligned} \cos x & =2 \cos ^2 \frac{x}{2}-1 \\ & =2(\ln 2)^2-1 \end{aligned} \end{aligned}

Q158

Let

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

be the solution of the differential equation

\left(x y-5 x^2 \sqrt{1+x^2}\right) d x+\left(1+x^2\right) d y=0, y(0)=0

. Then

y(\sqrt{3})

is equal to

A

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

B

\sqrt{\dfrac{15}{2}}

C

\sqrt{\dfrac{14}{3}}

D

2 \sqrt{2}

Correct Answer

Option A

Solution

\begin{aligned} & \left(1+x^2\right) \frac{d y}{d x}+x y=5 x^1 \sqrt{1+x^2} \\ & \frac{d y}{d x}+\frac{x y}{1+x^2}=\frac{5 x^2}{\sqrt{1+x^2}} \\ & \therefore \text{ I.F. }=e^{\int \frac{x}{1+x^2} d x}=e^{\frac{\ln \left(1+x^2\right)}{2}}=\sqrt{1+x^2} \\ & \therefore y \sqrt{1+x^2}=\int \frac{5 x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} d x \\ & \therefore y \sqrt{1+x^2}=\int \frac{5 x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} d x \\ & y \sqrt{1+x^2}=\frac{5 x^3}{3}+C \\ & \because y(0)=0 \Rightarrow 0=0+C \Rightarrow C=0 \\ & \therefore y=\frac{5 x^3}{3 \sqrt{1+x^2}} \\ & y(\sqrt{3})=\frac{15 \sqrt{3}}{32}=\frac{5 \sqrt{3}}{2} \end{aligned}

Q159

Let

f(x) = x - 1

and

g(x) = e^x

for

x \in \mathbb{R}

. If

\dfrac{dy}{dx} = \left( e^{-2\sqrt{x}} g\left(f(f(x))\right) - \dfrac{y}{\sqrt{x}} \right)

,

y(0) = 0

, then

y(1)

is

A

\dfrac{1 - e^3}{e^4}

B

\dfrac{e-1}{e^4}

C

\dfrac{1 - e^2}{e^4}

D

\dfrac{2e - 1}{e^3}

Correct Answer

Option B

Solution

\begin{aligned} & f(x)=x-1 \\ & f(f(x))=f(x)-1=x-1-1=x-2 \\ & g(f(f(x)))=e^{x-2} \\ & \therefore \frac{d y}{d x}=e^{-2 \sqrt{x}} \times e^{x-2}-\frac{1}{\sqrt{x}} y \\ & \frac{d y}{d x}+\frac{1}{\sqrt{x}} y=e^{x-2 \sqrt{x}-2} \text{ which is L.D.E } \\ & \text{ I.F. }=e^{\int \frac{d y}{\sqrt{x}}}=e^{2 \sqrt{x}} \end{aligned}

Its solution is

\begin{aligned} & y \times e^{2 \sqrt{x}}=\int e^{2 \sqrt{x}} \times e^{x-2 \sqrt{x}-2} d x+c \\ & y \times e^{2 \sqrt{x}}=\int e^{x-2} d x+c \\ & y \times e^{2 \sqrt{x}}=e^{x-2}+c \end{aligned}

Given $\mathrm{x}=0, \mathrm{y}=0 \Rightarrow 0=\mathrm{e}^{-2}+\mathrm{c} \quad ; \mathrm{c}=-\mathrm{e}^{-2}$

\therefore \mathrm{y} \times \mathrm{e}^{2 \sqrt{x}}=\mathrm{e}^{\mathrm{x}-2}-\mathrm{e}^{-2}

when $\mathrm{x}=1, \mathrm{y} \times \mathrm{e}^2=\mathrm{e}^{-1}-\mathrm{e}^{-2}$

y=\frac{e^{-1}-e^{-2}}{e^2}=\frac{\frac{1}{e}-\frac{1}{e^2}}{e^2}=\frac{e^2-e}{e^5}=\frac{e-1}{e^4}

Option (1) is correct

Q160

If

{{dy} \over {dx}} + {3 \over {{{\cos }^2}x}}y = {1 \over {{{\cos }^2}x}},\,\,x \in \left( {{{ - \pi } \over 3},{\pi \over 3}} \right)

and

y\left( {{\pi \over 4}} \right) = {4 \over 3},

then

y\left( { - {\pi \over 4}} \right)

equals -

A

{1 \over 3} + {e^6}

B

{1 \over 3}

C

{1 \over 3}

+ e3

D

-

{4 \over 3}

Correct Answer

Option A

Solution

{{dy} \over {dx}} + 3{\sec ^2}x.y = {\sec ^2}x

I.F. =

{e^{3\int {{{\sec }^2}xdx} }} = {e^{3\tan x}}

or

y.e{}^{3\tan x} = \int {{{\sec }^2}x.{e^{3\tan x}}}

or

y.{e^{3\tan x}} = {1 \over 3}{e^{3\tan x}} + C

Given

y\left( {{\pi \over 4}} \right) = {4 \over 3}

$\therefore$

{4 \over 3}.{e^3} = {1 \over 3}{e^3} + C

$\therefore$ C = e3