Differential Equations — Page 7 | JEE Mathematics

Q61

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

{\log _{}}\left( {{{dy} \over {dx}}} \right) = 3x + 4y

, with y(0) = 0. If

y\left( { - {2 \over 3}{{\log }_e}2} \right) = \alpha {\log _e}2

, then the value of

\alpha

is equal to :

A

- {1 \over 4}

B

{1 \over 4}

C

2

D

- {1 \over 2}

Correct Answer

Option A

Solution

{{dy} \over {dx}} = {e^{3x}}.{e^{4y}} \Rightarrow \int {{e^{ - 4y}}dy = \int {{e^{3x}}dx} }

{{{e^{ - 4y}}} \over { - 4}} = {{{e^{3x}}} \over 3} + C \Rightarrow - {1 \over 4} - {1 \over 3} = C \Rightarrow C = - {7 \over {12}}

{{{e^{ - 4y}}} \over { - 4}} = {{{e^{3x}}} \over 3} - {7 \over {12}} \Rightarrow {e^{ - 4y}} = {{4{e^{3x}} - 7} \over { - 3}}

{e^{4y}} = {3 \over {7 - 4{e^{3x}}}} \Rightarrow 4y = \ln \left( {{3 \over {7 - 4{e^{3x}}}}} \right)

4y = \ln \left( {{3 \over 6}} \right)

when

x = - {2 \over 3}\ln 2

y = {1 \over 4}\ln \left( {{1 \over 2}} \right) = - {1 \over 4}\ln 2

Q62

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

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

, 0 < x < 2.1, with y(2) = 0. Then the value of

{{dy} \over {dx}}

at x = 1 is equal to :

A

{{{e^{5/2}}} \over {{{(1 + {e^2})}^2}}}

B

{{5{e^{1/2}}} \over {{{({e^2} + 1)}^2}}}

C

- {{2{e^2}} \over {{{(1 + {e^2})}^2}}}

D

{{ - {e^{3/2}}} \over {{{({e^2} + 1)}^2}}}

Correct Answer

Option D

Solution

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

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

Put,

{1 \over {y + 1}} = z

- {1 \over {{{(y + 1)}^2}}}.{{dy} \over {dx}} = {{dz} \over {dx}}

$\therefore$

{{dz} \over {dx}} + z( - x) = - {e^{{{{x^2}} \over 2}}}

I.F = {e^{\int { - xdx} }} = {e^{{{ - {x^2}} \over 2}}}

z.\left( {{e^{ - {{{x^2}} \over 2}}}} \right) = - \int {{e^{ - {{{x^2}} \over 2}}}.{e^{{{{x^2}} \over 2}}}dx} = - \int {1.dx = - x + C}

\Rightarrow {{{e^{ - {{{x^2}} \over 2}}}} \over {y + 1}} = - x + C

.... (1) Given y = 0 at x = 2 Put in (1)

{{{e^{ - 2}}} \over {0 + 1}} = - 2 + C

C = {e^{ - 2}} + 2

.... (2) From (1) and (2)

y + 1 = {{{e^{ - {{{x^2}} \over 2}}}} \over {{e^{ - 2}} + 2 - x}}

Again, at x = 1

\Rightarrow y + 1 = {{{e^{3/2}}} \over {{e^2} + 1}}

\Rightarrow y + 1 = {{{e^{3/2}}} \over {{e^2} + 1}}

$\therefore$

{\left. {{{dy} \over {dx}}} \right|_{x = 1}} = {{{e^{3/2}}} \over {{e^2} + 1}}\left( {{{{e^{3/2}}} \over {{e^2} + 1}} \times {e^{1/2}} - 1} \right)

= - {{{e^{3/2}}} \over {{{({e^2} + 1)}^2}}}

Q63

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

x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx

,

- 1 \le x \le 1

,

y\left( {{1 \over 2}} \right) = {\pi \over 6}

. Then the area of the region bounded by the curves x = 0,

x = {1 \over {\sqrt 2 }}

and y = y(x) in the upper half plane is :

A

{1 \over 8}(\pi - 1)

B

{1 \over {12}}(\pi - 3)

C

{1 \over 4}(\pi - 2)

D

{1 \over 6}(\pi - 1)

Correct Answer

Option A

Solution

We have,

{{dy} \over {dx}} = {{x\left( {{y \over x}.\tan {y \over x} - 1} \right)} \over {x\tan {y \over x}}}

$\therefore$

{{dy} \over {dx}} = {y \over x} - \cot \left( {{y \over x}} \right)

Put

{y \over x} = v

\Rightarrow y = vn

$\therefore$

{{dy} \over {dx}} = v + {{ndv} \over {dx}}

Now, we get

v + n{{dv} \over {dx}} = v - \cot (v)

\Rightarrow \int {(\tan )dv} = - \int {{{dx} \over x}}

$\therefore$

\ln \left| {\sec \left( {{y \over x}} \right)} \right| = - \ln \left| x \right| + c

As,

\left( {{1 \over 2}} \right) = \left( {{y \over x}} \right) \Rightarrow C = 0

$\therefore$

\sec \left( {{y \over x}} \right) = {1 \over x}

\Rightarrow \cos \left( {{y \over x}} \right) = x

$\therefore$

y = x{\cos ^{ - 1}}(x)

So, required bounded area

= \int\limits_0^{{1 \over {\sqrt 2 }}} {\mathop x\limits_{(II)} (\mathop {{{\cos }^{ - 1}}x}\limits_{(I)} )dx = \left( {{{\pi - 1} \over 8}} \right)}

(I. B. P.) $\therefore$ Option (1) is correct.

Q64

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

{e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0

, y(1) =

-

1. Then the value of (y(3))2 is equal to :

A 1

-

4e3

B 1

-

4e6

C 1 + 4e3

D 1 + 4e6

Correct Answer

Option B

Solution

{e^x}\sqrt {1 - {y^2}} dx + {y \over x}dy = 0

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

\Rightarrow \int {{{ - y} \over {\sqrt {1 - {y^2}} }}} dy = \int_{II}^{{e^x}} {_1^xdx}

\Rightarrow \sqrt {1 - {y^2}} = {e^x}(x - 1) + c

Given : At x = 1, y = $-$ 1 $\Rightarrow$ 0 = 0 + c $\Rightarrow$ c = 0 $\therefore$

\sqrt {1 - {y^2}} = {e^x}(x - 1)

At x = 3

1 - {y^2} = {({e^3}2)^2} \Rightarrow {y^2} = 1 - 4{e^6}

Q65

Let y = y(x) satisfies the equation

{{dy} \over {dx}} - |A| = 0

, for all x > 0, where

A = \left[ \begin{array}{lll}y & {\sin x} & 1 \\ 0 & { - 1} & 1 \\ 2 & 0 & {{1 \over x}} \end{array} \right]

. If

y(\pi ) = \pi + 2

, then the value of

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

is :

A

{\pi \over 2} + {4 \over \pi }

B

{\pi \over 2} - {1 \over \pi }

C

{{3\pi } \over 2} - {1 \over \pi }

D

{\pi \over 2} - {4 \over \pi }

Correct Answer

Option A

Solution

|A| = - {y \over x} + 2\sin x + 2

{{dy} \over {dx}} = |A|

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

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

I.F. = {e^{\int {{1 \over x}dx} }} = x

\Rightarrow yx = \int {x(2\sin x + 2)dx}

xy = {x^2} - 2x\cos x + 2\sin x + c

..... (i) Now, x = $\pi$ , y = $\pi$ + 2 Use in (i) c = 0 Now, (i) becomes

xy = {x^2} - 2x\cos x + 2\sin x

put

x = \pi /2

{\pi \over 2}y = {\left( {{\pi \over 2}} \right)^2} - 2.{\pi \over 2}\cos {\pi \over 2} + 2\sin {\pi \over 2}

y({\pi \over 2}) = {{{\pi ^2}} \over 4} + 2

Q66

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

\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdx

, with

y\left( {{\pi \over 4}} \right) = 0

. Then, the value of

{(y(0) + 1)^2}

is equal to :

A e1/2

B e

-

1/2

C e

-

1

D e

Correct Answer

Option C

Solution

{{dy} \over {dx}} + 2{\sin ^2}x = 1 + y\cos 2x

\Rightarrow {{dy} \over {dx}} + ( - \cos 2x)y = \cos 2x

I.F. = {e^{\int { - \cos 2xdx} }} = {e^{ - {{\sin 2x} \over 2}}}

Solution of D.E.

y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = \int {(\cos 2x)\left( {{e^{ - {{\sin 2x} \over 2}}}} \right)} dx + c

\Rightarrow y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = - {e^{ - {{\sin 2x} \over 2}}} + c

Given

y\left( {{\pi \over 4}} \right) = 0

\Rightarrow 0 = - {e^{{{ - 1} \over 2}}} + c \Rightarrow c = {e^{{{ - 1} \over 2}}}

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

at x = 0

y = - 1 + {e^{{{ - 1} \over 2}}}

\Rightarrow y(0) = - 1 + {e^{{{ - 1} \over 2}}} \Rightarrow {(y(0) + 1)^2} = {e^{ - 1}}

Q67

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

{{dy} \over {dx}} = 1 + x{e^{y - x}}, - \sqrt 2 < x < \sqrt 2 ,y(0) = 0

then, the minimum value of

y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)

is equal to :

A

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

B

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

C

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

D

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

Correct Answer

Option D

Solution

{{dy - dx} \over {{e^{y - x}}}} = xdx

\Rightarrow {{dy - dx} \over {{e^{y - x}}}} = xdx

\Rightarrow - {e^{x - y}} = {{{x^2}} \over 2} + c

At x = 0, y = 0 $\Rightarrow$ c = $-$ 1

\Rightarrow {e^{x - y}} = {{2 - {x^2}} \over 2}

\Rightarrow y = x - \ln \left( {{{2 - {x^2}} \over 2}} \right)

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

So minimum value occurs at

x = 1 - \sqrt 3

y\left( {1 - \sqrt 3 } \right) = \left( {1 - \sqrt 3 } \right) - \ln \left( {{{2 - \left( {4 - 2\sqrt 3 } \right)} \over 2}} \right)

= \left( {1 - \sqrt 3 } \right) - \ln \left( {\sqrt 3 - 1} \right)

Q68

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

(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0

,

x \in \left( {0,{\pi \over 2}} \right)

. If

\mathop {\lim }\limits_{x \to 0 + } xy(x) = 1

, then the value of

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

is :

A

- {\pi \over 4}

B

{\pi \over 4} - 1

C

{\pi \over 4} + 1

D

{\pi \over 4}

Correct Answer

Option D

Solution

(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0

or

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

IF = {e^{\int {{{{{\sec }^2}x} \over {\tan x}}dx} }} = {e^{\ln \tan x}} = \tan x

$\therefore$

y\tan x = - \int {{{\tan }^2}x\,dx}

or

y\tan x = - \tan x + x + C

or

y = - 1 + {x \over {\tan x}} + {C \over {\tan x}}

or

\mathop {\lim }\limits_{x \to 0} xy = - x + {{{x^2}} \over {\tan x}} + {{Cx} \over {\tan x}} = 1

or C = 1

y(x) = \cot x + x\cot x - 1

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

Q69

Let y(x) be the solution of the differential equation 2x2 dy + (ey

-

2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to :

A 0

B 2

C loge 2

D loge (2e)

Correct Answer

Option C

Solution

2{x^2}dy + ({e^y} - 2x)dx = 0

{{dy} \over {dx}} + {{{e^y} - 2x} \over {2{x^2}}} = 0 \Rightarrow {{dy} \over {dx}} + {{{e^y}} \over {2{x^2}}} - {1 \over x} = 0

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

Put

{e^{ - y}} = z

{{ - dz} \over {dx}} - {z \over x} = - {1 \over {2{x^2}}} \Rightarrow xdz + zdx = {{dx} \over {2x}}

d(xz) = {{dx} \over {2x}} \Rightarrow xz = {1 \over 2}{\log _e}x + c

x{e^{ - y}} = {1 \over 2}{\log _e}x + c

, passes through (e, 1)

\Rightarrow C = {1 \over 2}

x{e^{ - y}} = {{{{\log }_e}ex} \over 2}

{e^{ - y}} = {1 \over 2} \Rightarrow y = {\log _e}2

Q70

A differential equation representing the family of parabolas with axis parallel to y-axis and whose length of latus rectum is the distance of the point (2,

-

3) from the line 3x + 4y = 5, is given by :

A

10{{{d^2}y} \over {d{x^2}}} = 11

B

11{{{d^2}x} \over {d{y^2}}} = 10

C

10{{{d^2}x} \over {d{y^2}}} = 11

D

11{{{d^2}y} \over {d{x^2}}} = 10

Correct Answer

Option D

Solution

Length of latus rectum

= {{|3(2) + 4( - 3) - 5|} \over 5} = {{11} \over 5}

{(x - h)^2} = {{11} \over 5}(y - k)

differentiate w.r.t. 'x' :-

2(x - h) = {{11} \over 5}{{dy} \over {dx}}

again differentiate

2 = {{11} \over 5}{{{d^2}y} \over {d{x^2}}}

{{11{d^2}y} \over {d{x^2}}} = 10