Differential Equations — Page 17 | JEE Mathematics

Q161

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

(x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x

,

y(0) = 1

. Then

\int\limits_{-3}^{3} y(x) \, dx

is :

A 36

B 24

C 18

D 30

Correct Answer

Option B

Solution

\begin{aligned} &\begin{aligned} & \left(x^2+1\right) \frac{d y}{d x}-2 x y=\left(x^4+2 x^2+1\right) \cos x \\ & \frac{d y}{d x}-\left(\frac{2 x}{x^2+1}\right) y=\frac{\left(x^2+1\right)^2 \cos x}{c x^2+1}=\left(x^2+1\right) \cos x \end{aligned}\\ &\text{ (Linear D.E) }\\ &\mathrm{P}=\frac{-2 \mathrm{x}}{\mathrm{x}^2+1}, \mathrm{Q}=\left(\mathrm{x}^2+1\right) \cos \mathrm{x} \end{aligned}

\begin{aligned} & \text{ I.F }=\mathrm{e}^{\int P d x}=\mathrm{e}^{\int \frac{-2 x}{x^2+1} d x}=\frac{1}{x^2+1} \\ & y \cdot \frac{1}{x^2+1}=\int\left(x^2+1\right) \cos x \cdot \frac{1}{x^2+1} d x \\ & \frac{y}{x^2+1}=\sin x+c \Rightarrow y \cos =1 \Rightarrow c=1 \end{aligned}

\begin{aligned} & y=\left(x^2+1\right)(\sin x+1) \\ & \int_{-3}^3 y d x=\int_{-3}^3\left(x^2+1\right)(\sin x+1) \\ & d x=\int_{-3}^3 x^2 \sin x+x^2 \sin x+1 d x \\ & \Rightarrow \int_{-3}^3 x^2 \sin x d x+\int_{-3}^3 x^2 d x+\int_{-3}^3 \sin x d x+\int_{-3}^3 1 d x \\ & =0+18+0+6=24 \end{aligned}

Q162

Let

y=y(x)

be the solution curve of the differential equation

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

, passing through the point

(1,0)

. Then

y(2)

is equal to :

A

\dfrac{2}{2+e^2}

B

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

C

\dfrac{4}{4+e^2}

D

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

Correct Answer

Option C

Solution

\begin{aligned} & x\left(x^2+e^x\right) d y+\left(e^x(x-2) y-x^3\right) d x=0 \\ & \frac{d y}{d x}+\frac{e^x(x-2)}{x\left(x^2+e^x\right)} y=\frac{x^3}{x\left(x^2+e^x\right)} \\ & \text{ I.F. }=e^{\iint \frac{e^x(x-2)}{x\left(x^2+e^x\right)} d x} \\ & =e^{\int \frac{e^x+2 x}{e^x+x^2} d x-\int \frac{2}{x} d x} \\ & =e^{\ln e^{-x}+x^2-2 \ln x} \end{aligned}

\begin{aligned} & =\frac{e^x+x^2}{x^2} \\ & \therefore \quad y\left(\frac{e^2+x^2}{x^2}\right)=\int d x+c \\ & \Rightarrow \quad y\left(\frac{e^x+x^2}{x^2}\right)=x+c \end{aligned}

Also, $y(1)=0$

\begin{aligned} & \Rightarrow \quad c=-1 \\ & \therefore \quad y\left(\frac{e^2+x^2}{x^2}\right)=x-1 \end{aligned}

Hence, $y(2)=\dfrac{4}{e^2+4}$

Q163

Let

g

be a differentiable function such that

\int_0^x g(t) d t=x-\int_0^x \operatorname{tg}(t) d t, x \geq 0

and let

y=y(x)

satisfy the differential equation

\dfrac{d y}{d x}-y \tan x=2(x+1) \sec x g(x), x \in\left[0, \dfrac{\pi}{2}\right)

. If

y(0)=0

, then

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

is equal to

A

\dfrac{4 \pi}{3}

B

\dfrac{2 \pi}{3}

C

\dfrac{2 \pi}{3 \sqrt{3}}

D

\dfrac{4 \pi}{3 \sqrt{3}}

Correct Answer

Option A

Solution

To solve the given problem, let's start by considering the equation: $\int_0^x g(t) \, dt = x - \int_0^x \tan g(t) \, dt$ Differentiate both sides with respect to $x$ : $g(x) = 1 - xg(x)$ Rearranging gives: $g(x)(1 + x) = 1 \implies g(x) = \dfrac{1}{1 + x}$ Now, consider the differential equation: $\dfrac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x)$ Substitute $g(x) = \dfrac{1}{1 + x}$ : $\dfrac{dy}{dx} - y \tan x = 2(x+1) \sec x \cdot \dfrac{1}{1 + x}$ This simplifies to: $\dfrac{dy}{dx} - y \tan x = 2 \sec x$ To solve this, we use an Integrating Factor (I.F): $\text{I.F} = e^{\int \tan x \, dx} = e^{-\ln \cos x} = \cos x$ Multiply the entire differential equation by the Integrating Factor: $\cos x \cdot \dfrac{dy}{dx} - y \cdot \sin x = 2$ This implies: $\dfrac{d}{dx}(y \cos x) = 2$ Integrate both sides with respect to $x$ : $y \cos x = \int 2 \, dx = 2x + c$ Given the initial condition $y(0) = 0$ : $0 \cdot 1 = 2 \cdot 0 + c \implies c = 0$ Thus: $y \cos x = 2x$ Evaluate at $x = \dfrac{\pi}{3}$ : $y \cdot \dfrac{1}{2} = 2 \cdot \dfrac{\pi}{3}$ Solving for $y$ : $y = \dfrac{4 \pi}{3}$ Therefore, $y\left(\dfrac{\pi}{3}\right) = \dfrac{4 \pi}{3}$ .

Q164

If a curve

y=y(x)

passes through the point

\left(1, \dfrac{\pi}{2}\right)

and satisfies the differential equation

\left(7 x^4 \cot y-\mathrm{e}^x \operatorname{cosec} y\right) \dfrac{\mathrm{d} x}{\mathrm{~d} y}=x^5, x \geq 1

, then at

x=2

, the value of

\cos y

is :

A

\dfrac{2 \mathrm{e}^2+\mathrm{e}}{64}

B

\dfrac{2 \mathrm{e}^2-\mathrm{e}}{64}

C

\dfrac{2 \mathrm{e}^2-\mathrm{e}}{128}

D

\dfrac{2 \mathrm{e}^2+\mathrm{e}}{128}

Correct Answer

Option C

Solution

\begin{aligned} & \left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5 \\ & x^5 \frac{d y}{d x}-7 x^4 \cot y=-e^x \operatorname{cosec} y \\ & \frac{d y}{d x}-\frac{7}{x} \cot y=-\frac{e^x}{x^5} \operatorname{cosec} y \\ & \sin y \frac{d y}{d x}-\frac{7}{x} \cos y=-\frac{e^x}{x^5} \\ & \text{ Let }-\cos y=t \\ & \sin y \frac{d y}{d x}=\frac{d t}{d x} \\ & \therefore \frac{d t}{d x}+\frac{7}{x} t=-\frac{e^x}{x^5} \end{aligned}

\begin{aligned} & \therefore \text{ I.F. }=e^{\int \frac{7}{x} d x}=x^7 \\ & t \cdot x^7=\int \frac{-e^x}{x^5} \cdot x^7 d x \\ & -\cos y \cdot x^7=-\int e^x x^2 d x \\ & \cos y x^7=e^x\left(x^2-2 x+2\right)+c \\ & \because \quad x=1 \text{ then } y=\frac{\pi}{2} \Rightarrow c=-e \\ & \therefore \quad \cos y \cdot x^7=e^x\left(x^2-2 x+2\right)-e \\ & \text{ When } x=2 \text{ then } \cos y=\frac{2 e^2-e}{128} \end{aligned}

Q165

Let

y=y(x)

be the solution of the differential equation

\dfrac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x, y(0)=\dfrac{1}{3}+e^3

. Then

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

is equal to :

A

\dfrac{4}{3}

B

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

C

\dfrac{4}{3}+e^3

D

\dfrac{2}{3}

Correct Answer

Option A

Solution

\begin{aligned} & \frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x \\ & \Rightarrow \frac{d y}{d x}+3 \sec ^2 x y=\sec ^2 x \\ & \text{ I.F }=e^{\int 3 \sec ^2 x d x} \\ & \quad=e^{3 \tan x} \end{aligned}

\begin{aligned} & y \cdot e^{\tan x}=\int e^{3 \tan x} \cdot \sec ^2 x d x+c \\ & y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+c \\ & \text{ Also } f(0)=\frac{1}{3}+e^3 \\ & \Rightarrow\left(\frac{1}{3}+e^3\right)=\frac{1}{3}+c \\ & \Rightarrow c=e^3 \\ & \therefore y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+e^3 \\ & \text{ Put } x=\frac{\pi}{4} \end{aligned}

$y e^3=\dfrac{e^3}{3}+e^3 \Rightarrow y=\dfrac{4}{3}$

Q166

If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as

{{{x^2} - 2y} \over x}

, then the curve also passes through the point :

A (–1, 2)

B

\left( { - \sqrt 2 ,1} \right)

C

\left( { \sqrt 3 ,0} \right)

D (3, 0)

Correct Answer

Option C

Solution

{{dy} \over {dx}} = {{{x^2} - 2y} \over x}

(Given)

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

I.F =

{e^{\int {{2 \over x}dx} }} = {x^2}

$\therefore$ y.x2 =

\int {x.{x^2}} dx + C

=

{{{x^4}} \over y} + C

This curve passes through (1, $-$ 2) $\Rightarrow$ C= $-$

{9 \over 4}

$\therefore$ yx2 =

{{{x^4}} \over 4} - {9 \over 4}

Now check option(s), Which is satisly by option (ii)

Q167

The curve satifying the differeial equation, (x2

-

y2) dx + 2xydy = 0 and passing through the point (1, 1) is :

A a circle of radius one.

B a hyperbola.

C an ellipse.

D a circle of radius two.

Correct Answer

Option A

Solution

(x2 $-$ y2) dx + 2xydy = 0 $\Rightarrow$

{{dy} \over {dx}}

=

{{{y^2} - {x^2}} \over {2xy}}

Let y = vx

{{dy} \over {dx}}

= v + x

{{dv} \over {dx}}

$\Rightarrow$ v + x

{{dv} \over {dx}}

=

{{{v^2}{x^2} - {x^2}} \over {2v{x^2}}}

$\Rightarrow$ v + x

{{dv} \over {dx}}

=

{{{v^2} - 1} \over {2v}}

$\Rightarrow$ x

{{dv} \over {dx}}

=

{{ - {v^2} - 1} \over {2v}}

$\Rightarrow$

{{2vdv} \over {{v^2} + 1}}

= $-$

{{dx} \over x}

After intergrating, we get

\ln \left| {{v^2} + 1} \right|

= $-$ ln

\left| x \right|

+ lnc

{{y{}^2} \over {{x^2}}}

+ 1 =

{c \over x}

As curve passes through the point (1, 1), so 1 + 1 = c $\Rightarrow$ c = 2 x2 + y2 $-$ 2x = 0, which is a circle of radius one.

Q168

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

{{dy} \over {dx}} + 2y = f\left( x \right),

where

f\left( x \right) = \left\{ \begin{array}{ll}{1,} & {x \in \left[ {0,1} \right]} \\ {0,} & {otherwise} \end{array} \right.

If y(0) = 0, then

y\left( {{3 \over 2}} \right)

is :

A

{{{e^2} + 1} \over {2{e^4}}}

B

{1 \over {2e}}

C

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

D

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

Correct Answer

Option D

Solution

When x

\in

[0, 1], then

{{dy} \over {dx}}

+ 2y = 1 $\Rightarrow$ y =

{1 \over 2}

+ C1e $-$ 2x $\because$ y(0) = 0 $\Rightarrow$ y(x) =

{1 \over 2}

$-$

{1 \over 2}

e $-$ 2x Here, y(1) =

{1 \over 2}

$-$

{1 \over 2}

e $-$ 2 =

{{{e^2} - 1} \over {2{e^2}}}

When

x \notin \left[ {0,1} \right]

, then

{{dy} \over {dx}}

+ 2y = 0 $\Rightarrow$ y = c2 e $-$ 2x $\because$ y(1) =

{{{e^2} - 1} \over 2}

$\Rightarrow$

{{{e^2} - 1} \over 2}

= c2e $-$ 2 $\Rightarrow$ C2 =

{{{e^2} - 1} \over 2}

$\therefore$ y(x)

\left( {{{{e^2} - 1} \over 2}} \right){e^{ - 2x}} \Rightarrow y\left( {{3 \over 2}} \right)

=

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

Q169

The differential equation representing the family of ellipse having foci eith on the x-axis or on the

y

-axis, center at the origin and passing through the point (0, 3) is :

A xy y'' + x (y')2

-

y y' = 0

B x + y y'' = 0

C xy y'+ y2

-

9 = 0

D xy y'

-

y2 + 9 = 0

Correct Answer

Option D

Solution

Equation of ellipse,

{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1

As ellipse passes through (0, 3)

\therefore\,\,\,

{{{0^2}} \over {{a^2}}} + {{{3^2}} \over {{b^2}}} = 1

$\Rightarrow$ b2 = 9

\therefore\,\,\,

Equation of ellipse becomes,

{{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9} = 1

Differentiating w.r.t x, we get,

{{2x} \over {a{}^2}}

+

{{2y} \over 9}

.

{{dy} \over {dx}} = 0

$\Rightarrow$

{x \over {{a^2}}}

= $-$

{y \over 9}.{{dy} \over {da}}

$\Rightarrow$

{x \over {{a^2}}} = - {y \over 9}.y'......

(1) We got earlier,

{{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9}

= 1 $\Rightarrow$

{x \over {{a^2}}}.x + {{{y^2}} \over 9} = 1

putting value of equation (1) here,

{-{y\,y'} \over 9}.x + {{{y^2}} \over 9} = 1

$\Rightarrow$ $-$ xyy' + y2 = 9 $\Rightarrow$ xyy' $-$ y2 + 9 = 0

Q170

If y = y(x) is the solution of the differential equation, x

dy \over dx

+ 2y = x2, satisfying y(1) = 1, then y(

1\over2

) is equal to :

A

{{7} \over {64}}

B

{{49} \over {16}}

C

{{1} \over {4}}

D

{{13} \over {16}}

Correct Answer

Option B

Solution

Given,

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

$\Rightarrow$

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

This is a linear differential equation. $\therefore$ I.F

= {e^{\int {{2 \over x}dx} }}

= {e^{2\ln x}}

= {x^2}

$\therefore$ Solution is,

y \cdot {x^2} = \int {x \cdot {x^2}dx}

$\Rightarrow$

y{x^2} = {{{x^4}} \over 4} + C

given

y\left( 1 \right) = 1

$\therefore$

1.1 = {4 \over 4} + C

$\Rightarrow$

C = {3 \over 4}

$\therefore$ Equation is

y{x^2} = {{{x^4}} \over 4} + {3 \over 4}

$\therefore$

y\left( {{1 \over 2}} \right)

means

x = {1 \over 2}

$\therefore$

y \cdot {\left( {{1 \over 2}} \right)^2} = {1 \over 4} \times {\left( {{1 \over 2}} \right)^4} + {3 \over 4}

$\Rightarrow$

{y \over 4} = {1 \over {64}} + {3 \over 4}

$\Rightarrow$

{y \over 4} = {{1 + 48} \over {64}}

$\Rightarrow$ y =

{{49} \over {16}}