If the solution curve of the differential equation $$\frac{d y}{d x}=\frac{x+y-2}{x-y}$$ passes through the points $$(2,1)$$ and $$(\mathrm{k}+1,2), \mathrm{k}>0$$, then

$\frac{d y}{d x}=\frac{x+y-2}{x-y}=\frac{(x-1)+(y-1)}{(x-1)-(y-1)}$ Let $x-1=X, y-1=Y$ $$ \frac{d Y}{d X}=\frac{X+Y}{X-Y} $$ Let $Y=t X \Rightarrow \frac{d Y}{d X}=t+X \frac{d t}{d X}$ $t+X \frac{d t}{d X}=\frac{1+t}{1-t}$ $X \frac{d t}{d X}=\frac{1+t}{1-t}-t=\frac{1+t^{2}}{1-t}$ $\int \frac{1-t}{1+t^{2}} d t=\int \frac{d X}{X}$ $$ \begin{aligned} &\tan ^{-1} t-\frac{1}{2} \ln \left(1+t^{2}\right)=\ln |X|+c \\\\ &\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1

Differential Equations — Page 11 | JEE Mathematics

Q101

Let

y=y(x)

be the solution of the differential equation

\sec x \mathrm{~d} y+\{2(1-x) \tan x+x(2-x)\} \mathrm{d} x=0

such that

y(0)=2

. Then

y(2)

is equal to:

A

2\{\sin (2)+1\}

B 2

C 1

D

2\{1-\sin (2)\}

Correct Answer

Option B

Solution

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

Now both side integrate

\begin{aligned} & y(x)=\int 2(x-1) \sin x d x+\left[\left(x^2-2 x\right)(\sin x)-\int(2 x-2) \sin x d x\right] \\ & y(x)=\left(x^2-2 x\right) \sin x+\lambda \\ & y(0)=0+\lambda \Rightarrow 2=\lambda \\ & y(x)=\left(x^2-2 x\right) \sin x+2 \\ & y(2)=2 \end{aligned}

Q102

If

y=y(x), x \in(0, \pi / 2)

be the solution curve of the differential equation

\left(\sin ^{2} 2 x\right) \dfrac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)

, with

y(\pi / 4)=\mathrm{e}^{-\pi}

, then

y(\pi / 6)

is equal to :

A

\dfrac{2}{\sqrt{3}} e^{-2 \pi / 3}

B

\dfrac{2}{\sqrt{3}} \mathrm{e}^{2 \pi / 3}

C

\dfrac{1}{\sqrt{3}} e^{-2 \pi / 3}

D

\dfrac{1}{\sqrt{3}} e^{2 \pi / 3}

Correct Answer

Option A

Solution

({\sin ^2}2x){{dy} \over {dx}} + (8{\sin ^2}2x + 2\sin 4x)y

= 2{e^{ - 4x}}(2\sin 2x + \cos 2x)

{{dy} \over {dx}} + (8 + 4\cot 2x)y = 2{e^{ - 4x}}\left( {{{2\sin 2x + \cos 2x} \over {{{\sin }^2}2x}}} \right)

Integrating factor

(I.F.) = {e^{\int {(8 + 4\cot 2x)dx} }}

= {e^{8x + 2\ln \sin 2x}}

Solution of differential equation

y.\,{e^{8x + 2\ln \sin 2x}}

= \int {2{e^{(4x + 2\ln \sin 2x)}}{{(2\sin 2x + \cos 2x)} \over {{{\sin }^2}2x}}dx}

= 2\int {{e^{4x}}(2\sin 2x + \cos 2x)dx}

y.\,{e^{8x + 2\ln \sin 2x}} = {e^{4x}}\sin 2x + c

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

{e^{ - \pi }}\,.\,{e^{2\pi }} = {e^\pi } + c \Rightarrow c = 0

y\left( {{\pi \over 6}} \right) = {{{e^{{{2\pi } \over 3}}}{{\sqrt 3 } \over 2}} \over {{e^{\left( {{{4\pi } \over 3} + 2\ln {{\sqrt 3 } \over 2}} \right)}}}}

= {e^{{{ - 2\pi } \over 3}}}\,.\,{2 \over {\sqrt 3 }}

Q103

Let

y=y(x)

be the solution curve of the differential equation

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

,

x >1

passing through the point

\left(2, \sqrt{\dfrac{1}{3}}\right)

. Then

\sqrt{7}\, y(8)

is equal to :

A

11+6 \log _{e} 3

B 19

C

12-2 \log _{\mathrm{e}} 3

D

19-6 \log _{\mathrm{e}} 3

Correct Answer

Option D

Solution

{{dy} \over {dx}} + {1 \over {{x^2} - 1}}y = \sqrt {{{x - 1} \over {x + 1}}} ,\,x > 1

Integrating factor I.F.

= {e^{\int {{1 \over {{x^2} - 1}}dx} }} = {e^{{1 \over 2}\ln \left| {{{x - 1} \over {x + 1}}} \right|}}

= \sqrt {{{x - 1} \over {x + 1}}}

Solution of differential equation

y\sqrt {{{x - 1} \over {x + 1}}} = \int {{{x - 1} \over {x + 1}}dx = \int {\left( {1 - {2 \over {x + 1}}} \right)dx} }

y\sqrt {{{x - 1} \over {x + 1}}} = x - 2\ln |x + 1| + C

Curve passes through

\left( {2,\sqrt {{1 \over 3}} } \right)

{1 \over {\sqrt 3 }} \times {1 \over {\sqrt 3 }} = 2 - 2\ln 3 + C

C = 2\ln 3 - {5 \over 3}

y(8) \times {{\sqrt 7 } \over 3} = 8 - 2\ln 9 + 2\ln 3 - {5 \over 3}

\sqrt 7 \,.\,y(8) = 19 - 6\ln 3

Q104

The differential equation of the family of circles passing through the points

(0,2)

and

(0,-2)

is :

A

2 x y \dfrac{d y}{d x}+\left(x^{2}-y^{2}+4\right)=0

B

2 x y \dfrac{d y}{d x}+\left(x^{2}+y^{2}-4\right)=0

C

2 x y \dfrac{d y}{d x}+\left(y^{2}-x^{2}+4\right)=0

D

2 x y \dfrac{d y}{d x}-\left(x^{2}-y^{2}+4\right)=0

Correct Answer

Option A

Solution

Family of circles passing through the points (0, 2) and (0, $-$ 2)

{x^2} + (y - 2)(y + 2) + \lambda x = 0,\,\lambda \in R

{x^2} + {y^2} + \lambda x - 4 = 0

...... (1) Differentiate w.r.t x

2x + 2y{{dy} \over {dx}} + \lambda = 0

....... (2) Using (1) and (2), eliminate $\lambda$

{x^2} + {y^2} - \left( {2x + 2y{{dy} \over {dx}}} \right)x - 4 = 0

2xy{{dy} \over {dx}} + {x^2} - {y^2} + 4 = 0

Q105

Let the solution curve

y=y(x)

of the differential equation

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

pass through the point

\left(0, \dfrac{\pi}{2}\right)

. Then,

\lim\limits_{x \rightarrow \infty} \mathrm{e}^{x} y(x)

is equal to :

A

\dfrac{\pi}{4}

B

\dfrac{3\pi}{4}

C

\dfrac{\pi}{2}

D

\dfrac{3\pi}{2}

Correct Answer

Option B

Solution

D.E.

(1 + {e^{2x}})\left( {{{dy} \over {dx}} + y} \right) = 1

\Rightarrow {{dy} \over {dx}} + y = {1 \over {1 + {e^{2x}}}}

I.F.

= {e^{\int {1\,.\,dx} }} = {e^x}

$\therefore$ Solution

{e^x}y(x) = \int {{{{e^x}} \over {1 + {e^{2x}}}}dx}

\Rightarrow {e^x}y(x) = {\tan ^{ - 1}}({e^x}) + C

$\because$ It passes through

\left( {0,{\pi \over 2}} \right),\,C = {\pi \over 2} - {\pi \over 4} = {\pi \over 4}

$\therefore$

\mathop {\lim }\limits_{x \to \infty } {e^x}y(x) = \mathop {\lim }\limits_{x \to \infty } {\tan ^{ - 1}}({e^x}) + {\pi \over 4}

= {{3\pi } \over 4}

Q106

If the solution curve of the differential equation

\dfrac{d y}{d x}=\dfrac{x+y-2}{x-y}

passes through the points

(2,1)

and

(\mathrm{k}+1,2), \mathrm{k}>0

, then

A

2 \tan ^{-1}\left(\dfrac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)

B

\tan ^{-1}\left(\dfrac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)

C

2 \tan ^{-1}\left(\dfrac{1}{k+1}\right)=\log _{e}\left(k^{2}+2 k+2\right)

D

2 \tan ^{-1}\left(\dfrac{1}{k}\right)=\log _{e}\left(\dfrac{k^{2}+1}{k^{2}}\right)

Correct Answer

Option A

Solution

$\dfrac{d y}{d x}=\dfrac{x+y-2}{x-y}=\dfrac{(x-1)+(y-1)}{(x-1)-(y-1)}$ Let $x-1=X, y-1=Y$

\frac{d Y}{d X}=\frac{X+Y}{X-Y}

Let $Y=t X \Rightarrow \dfrac{d Y}{d X}=t+X \dfrac{d t}{d X}$ $t+X \dfrac{d t}{d X}=\dfrac{1+t}{1-t}$ $X \dfrac{d t}{d X}=\dfrac{1+t}{1-t}-t=\dfrac{1+t^{2}}{1-t}$ $\int \dfrac{1-t}{1+t^{2}} d t=\int \dfrac{d X}{X}$

\begin{aligned} &\tan ^{-1} t-\frac{1}{2} \ln \left(1+t^{2}\right)=\ln |X|+c \\\\ &\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^{2}\right)=\ln |x-1|+c \end{aligned}

Curve passes through $(2,1)$ $0-0=0+c \Rightarrow c=0$ If $(k+1,2)$ also satisfies the curve

\begin{aligned} &\tan ^{-1}\left(\frac{1}{k}\right)-\frac{1}{2} \ln \left(\frac{1+k^{2}}{k^{2}}\right)=\ln k \\\\ &2 \tan ^{-1}\left(\frac{1}{k}\right)=\ln \left(1+k^{2}\right) \end{aligned}

Q107

Let

y=y(x)

be the solution curve of the differential equation

\dfrac{d y}{d x}+\left(\dfrac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\dfrac{(x+3)}{x+1}, x>-1

, which passes through the point

(0,1)

. Then

y(1)

is equal to :

A

\dfrac{1}{2}

B

\dfrac{3}{2}

C

\dfrac{5}{2}

D

\dfrac{7}{2}

Correct Answer

Option B

Solution

$\dfrac{d y}{d x}+\left(\dfrac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\dfrac{(x+3)}{x+1}, x>-1$ , Integrating factor I.F.

$=e^{\int \dfrac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6} d x}$

\begin{aligned} & \text{ Let } \frac{2 x^{2}+11 x+13}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3} \\\\ & A=2, B=1, C=-1 \\\\ & \text{ I.F. }=e^{(2 \ln |x+1|+\ln |x+2|-\ln |x+3|)} \\\\ & =\frac{(x+1)^{2}(x+2)}{x+3} \end{aligned}

Solution of differential equation

\begin{aligned} &y \cdot \frac{(x+1)^{2}(x+2)}{x+3}=\int(x+1)(x+2) d x \\\\ &y \frac{(x+1)^{2}(x+2)}{x+3}=\frac{x^{3}}{3}+\frac{3 x^{2}}{2}+2 x+c \end{aligned}

Curve passes through $(0,1)$

\begin{gathered} 1 \times \frac{1 \times 2}{3}=0+c \Rightarrow c=\frac{2}{3} \\\\ \text{ So, } y(1)=\frac{\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}}{\frac{\left(2^{2} \times 3\right)}{4}}=\frac{3}{2} \end{gathered}

Q108

Let

\alpha x=\exp \left(x^{\beta} y^{\gamma}\right)

be the solution of the differential equation

2 x^{2} y \mathrm{~d} y-\left(1-x y^{2}\right) \mathrm{d} x=0, x > 0,y(2)=\sqrt{\log _{e} 2}

. Then

\alpha+\beta-\gamma

equals :

A 1

B 0

C 3

D

-1

Correct Answer

Option A

Solution

$\begin{aligned} & 2 x^2 y d y-\left(1-x y^2\right) d x=0 \\\\ & \Rightarrow 2 x^2 y \dfrac{d y}{d x}-1+x y^2=0 \\\\ & \Rightarrow 2 y \dfrac{d y}{d x}-\dfrac{1}{x^2}+\dfrac{y^2}{x}=0 \\\\ & \Rightarrow 2 y \dfrac{d y}{d x}+\dfrac{y^2}{x}=\dfrac{1}{x^2} \text{ (L.D.E) } \\\\ & Let, y^2=t \Rightarrow 2 y \dfrac{d y}{d x}=\dfrac{d t}{d x} \\\\ & \dfrac{d t}{d x}+\dfrac{t}{x}=\dfrac{1}{x^2} \\\\ & \text{ I.F }=e^{\int \dfrac{1}{x} d x}=x\end{aligned}$ So, the general solution is

\begin{aligned} & \Rightarrow t \times x=\int x \cdot \frac{1}{x^2} d x \\\\ & \Rightarrow y^2 \cdot x=\ln x+c \\\\ & \text{ Also,y(2) }=\sqrt{\log _e 2} \\\\ & \log _e^2 2=\log _e 2+c \Rightarrow c=\log _e 2 \\\\ & \Rightarrow y^2 x=\ln x+\ln 2 \\\\ & \Rightarrow y^2 x=\ln 2 x \\\\ & 2 x=\exp \left(x^1 y^2\right) \end{aligned}

By compare it with given solution we get,

\begin{aligned} & \alpha=2,\beta=1, \gamma=2 \\\\ & \Rightarrow \alpha+\beta-\gamma=2+1-2=1 \end{aligned}

Q109

Let

y=y(x)

be the solution of the differential equation

\left(3 y^{2}-5 x^{2}\right) y \mathrm{~d} x+2 x\left(x^{2}-y^{2}\right) \mathrm{d} y=0

such that

y(1)=1

. Then

\left|(y(2))^{3}-12 y(2)\right|

is equal to :

A 64

B

16 \sqrt{2}

C 32

D

32 \sqrt{2}

Correct Answer

Option D

Solution

$\left(3 y^{2}-5 x^{2}\right) y \cdot d x+2 x\left(x^{2}-y^{2}\right) d y=0$

\Rightarrow \frac{d y}{d x}=\frac{y\left(5 x^{2}-3 y^{2}\right)}{2 x\left(x^{2}-y^{2}\right)}

Put $\mathrm{y}=\mathrm{mx}$

\Rightarrow m+x \cdot \frac{d m}{d x}=\frac{m\left(5-3 m^{2}\right)}{2\left(1-m^{2}\right)}

\begin{aligned} & x \cdot \frac{d m}{d x}=\frac{\left(5-3 m^{2}\right) m-2 m\left(1-m^{2}\right)}{2\left(1-m^{2}\right)} \\\\ & \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\frac{2\left(\mathrm{~m}^{2}-1\right)}{\mathrm{m}\left(\mathrm{m}^{2}-3\right)} \mathrm{dm} \\\\ & \Rightarrow \frac{d x}{x}=\left(\frac{2}{m}-\frac{\frac{4}{3}}{m}+\frac{\frac{4 m}{3}}{\mathrm{~m}^{2}-3}\right) d m \end{aligned}

$\Rightarrow \int \dfrac{d x}{x}=\int \dfrac{\left(\dfrac{2}{3}\right)}{m}+\int \dfrac{2}{3}\left(\dfrac{2 m}{m^{2}-3}\right) d m$ $\Rightarrow \ln |\mathrm{x}|=\dfrac{2}{3} \ln |\mathrm{m}|+\dfrac{2}{3} \ln \left|\mathrm{m}^{2}-3\right|+\mathrm{C}$ Or, $\ln |\mathrm{x}|=\dfrac{2}{3} \ln \left|\dfrac{\mathrm{y}}{\mathrm{x}}\right|+\dfrac{2}{3} \ln \left|\left(\dfrac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right|+\mathrm{C}$ Put $(\mathrm{x}=1, \mathrm{y}=1)$ : we get $\mathrm{c}=-\dfrac{2}{3} \ln (2)$ $\Rightarrow \ln |\mathrm{x}|=\dfrac{2}{3} \ln \left|\dfrac{\mathrm{y}}{\mathrm{x}}\right|+\dfrac{2}{3} \ln \left|\left(\dfrac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right|-\dfrac{2}{3} \ln (2)$ $\Rightarrow\left(\dfrac{\mathrm{y}}{\mathrm{x}}\right)\left[\left(\dfrac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right]=2 .\left(\mathrm{x}^{3 / 2}\right)$ ........(

1) Put $\mathrm{x}=2$ in equation (1), we get $\Rightarrow \mathrm{y}\left(\mathrm{y}^{2}-12\right)=4 \times 2 \times 2 \times 2 \sqrt{2}$ $\Rightarrow \mathrm{y}^{3}-12 \mathrm{y}=32 \sqrt{2}$ $\Rightarrow\left|\mathrm{y}^{3}(2)-12 \mathrm{y}(2)\right|=32 \sqrt{2}$

Q110

If the solution

y=y(x)

of the differential equation

(x^4+2 x^3+3 x^2+2 x+2) \mathrm{d} y-(2 x^2+2 x+3) \mathrm{d} x=0

satisfies

y(-1)=-\dfrac{\pi}{4}

, then

y(0)

is equal to :

A

-\dfrac{\pi}{12}

B

\dfrac{\pi}{2}

C 0

D

\dfrac{\pi}{4}

Correct Answer

Option D

Solution

\begin{aligned} & \left(x^4+2 x^3+3 x^2+2 x+2\right) d y-\left(2 x^2+2 x+3\right) d x=0 \\ & \int d y=\int\left(\frac{2 x^2+2 x+3}{x^4+2 x^3+3 x^2+2 x+2} d x\right. \\ & \int d y=\int \frac{1}{x^2+1} d x+\int \frac{}{x^2+2 x+2} d x \\ & y=\tan ^{-1}(x)+\tan ^{-1}(1+x)+C \\ & y(-1)=\tan ^{-1}(-1)+\tan ^{-1}(1-1)+C \\ & y(-1)=-\frac{\pi}{4}+C=\left(\frac{-\pi}{4}\right)-\{\text{ given }\} \\ & \Rightarrow C=0 \\ & \text{ So, } y(x)=\tan ^{-1}(x)+\tan ^{-1}(1+x) \\ & y(0)=\tan ^{-1}(0)+\tan ^{-1}(1+0) \\ & y(0)=\frac{\pi}{4} \end{aligned}