Differential Equations — Page 8 | JEE Mathematics

Q71

If the solution curve of the differential equation (2x

-

10y3)dy + ydx = 0, passes through the points (0, 1) and (2,

\beta

), then

\beta

is a root of the equation :

A y5

-

2y

-

2 = 0

B 2y5

-

2y

-

1 = 0

C 2y5

-

y2

-

2 = 0

D y5

-

y2

-

1 = 0

Correct Answer

Option D

Solution

(2x - 10{y^3})dy + ydx = 0

\Rightarrow {{dx} \over {dy}} + \left( {{2 \over y}} \right)x = 10{y^2}

I.F. = {e^{\int {{2 \over y}dy} }} = {e^{2\ln (y)}} = {y^2}

Solution of D.E. is $\therefore$

x\,.\,y = \int {(10{y^2}){y^2}.\,dy}

x{y^2} = {{10{y^5}} \over 5} + C \Rightarrow x{y^2} = 2{y^5} + C

It passes through (0, 1) $\to$ 0 = 2 + C $\Rightarrow$ C = $-$ 2 $\therefore$ Curve is

x{y^2} = 2{y^5} - 2

Now, it passes through (2, $\beta$ )

2{\beta ^2} = 2{\beta ^5} - 2 \Rightarrow {\beta ^5} - {\beta ^2} - 1 = 0

$\therefore$ $\beta$ is root of an equation

{y^5} - {y^2} - 1 = 0

Q72

If

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

, y(0) = 1, then y(1) is equal to :

A log2(2 + e)

B log2(1 + e)

C log2(2e)

D log2(1 + e2)

Correct Answer

Option B

Solution

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

{2^y}{{dy} \over {dx}} = {2^x}({2^y} - 1)

\int {{{{2^y}} \over {{2^y} - 1}}dy = \int {{2^x}\,dx} }

{{\ln ({2^y} - 1)} \over {\ln 2}} = {{{2^x}} \over {\ln 2}} + C

\Rightarrow {\log _2}({2^y} - 1) = {2^x}{\log _2}e + C

$\because$

y(0) = 1 \Rightarrow 0 = {\log _2}e + C

C = - {\log _2}e

\Rightarrow {\log _2}({2^y} - 1) = ({2^x} - 1){\log _2}e

put x = 1,

{\log _2}({2^y} - 1) = {\log _2}e

2y = e + 1 y = log2(e + 1) Ans.

Q73

If

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

, y(0) = 0, then for y = 1, the value of x lies in the interval :

A (1, 2)

B

\left( {{1 \over 2},1} \right]

C (2, 3)

D

\left( {0,{1 \over 2}} \right]

Correct Answer

Option A

Solution

{{dy} \over {dx}} = {{{2^x}(y + {2^y})} \over {{2^x}(1 + {2^y}\ln 2)}}

\Rightarrow \int {{{(1 + {2^y})\ln 2} \over {(y + {2^y})}}dy = \int {dx} }

\Rightarrow \ln \left| {y + {2^y}} \right| = x + c

x = 0; y = 0 $\Rightarrow$ c = 0

\Rightarrow x = \ln \left| {y + {2^y}} \right|

$\Rightarrow$ at y = 1, x = ln3 $\because$

3 \in (e,{e^2}) \Rightarrow x \in (1,2)

Q74

If the solution of the differential equation

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

satisfies

y(0) = 0

, then the value of y(2) is _______________.

A

-

1

B 1

C 0

D e

Correct Answer

Option C

Solution

$\because$

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

Here,

I.F. = {e^{\int {{e^x}({x^2} - 2)dx} }}

= {e^{({x^2} - 2x){e^x}}}

$\therefore$ Solution of the differential equation is

y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {({x^2} - 2x)({x^2} - 2){e^{2x}}\,.\,{e^{({x^2} - 2x){e^x}}}dx}

= \int {({x^2} - 2x){e^x}\,.\,({x^2} - 2){e^x}\,.\,{e^{({x^2} - 2x){e^x}}}dx}

Let

({x^2} - 2x){e^x} = t

$\therefore$

({x^2} - 2){e^x}dx = dt

y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {t\,.\,{e^t}dt}

y\,.\,{e^{({x^2} - 2x){e^x}}} = ({x^2} - 2x - 1){e^{({x^2} - 2x){e^x}}} + c

$\therefore$

y(0) = 0

$\therefore$

c = 1

$\therefore$

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

$\therefore$

y(2) = - 1 + 1 = 0

Q75

If

y{{dy} \over {dx}} = x\left[ {{{{y^2}} \over {{x^2}}} + {{\phi \left( {{{{y^2}} \over {{x^2}}}} \right)} \over {\phi '\left( {{{{y^2}} \over {{x^2}}}} \right)}}} \right]

, x > 0,

\phi

> 0, and y(1) =

-

1, then

\phi \left( {{{{y^2}} \over 4}} \right)

is equal to :

A 4

\phi

(2)

B 4

\phi

(1)

C 2

\phi

(1)

D

\phi

(1)

Correct Answer

Option B

Solution

Let,

y = tx

{{dy} \over {dx}} = t + x{{dt} \over {dx}}

$\therefore$

tx\left( {t + x{{dt} \over {dx}}} \right) = x\left( {{t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}} \right)

{t^2} + xt{{dt} \over {dx}} = {t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}

\int {{{t\varphi '({t^2})} \over {\varphi ({t^2})}}} dt = \int {{{dx} \over x}}

Let

\varphi ({t^2}) = p

$\therefore$

\varphi '({t^2})2tdt = dp

\Rightarrow \int {{{dy} \over {2p}}} = \int {{{dx} \over x}}

{1 \over 2}\ln \varphi ({t^2}) = \ln x + \ln c

\varphi ({t^2}) = {x^2}k

\varphi \left( {{{{y^2}} \over {{x^2}}}} \right) = k{x^2},\varphi (1) = k

\varphi \left( {{{{y^2}} \over 4}} \right) = 4\varphi (1)

Q76

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

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

; x > 0 and y(1) = 1, then

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

is equal to :

A

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

B

3 + {1 \over {\sqrt e }}

C 3 + e

D 3

-

e

Correct Answer

Option D

Solution

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

; x > 0, y(1) = 1

{x^2}dy + {{(xy - 1)} \over x}dx = 0

{x^2}dy = {{(xy - 1)} \over x}dx

{{dy} \over {dx}} = {{1 - xy} \over {{x^3}}}

{{dy} \over {dx}} = {1 \over {{x^3}}} - {y \over {{x^2}}}

{{dy} \over {dx}} = {1 \over {{x^2}}}.y = {1 \over {{x^3}}}

If

{e^{\int {{1 \over {{x^2}}}dx} }} = {e^{ - {1 \over x}}}

y{e^{ - {1 \over x}}} = \int {{1 \over {{x^3}}}.{e^{ - {1 \over x}}}}

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

1.\,{e^{ - 1}} = {e^{ - 1}}(2) + C

C = - {e^{ - 1}} = - {1 \over e}

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

y\left( {{1 \over 2}} \right) = 3 - {1 \over e} \times {e^2}

y\left( {{1 \over 2}} \right) = 3 - e

Q77

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

x(1 - {x^2}){{dy} \over {dx}} + (3{x^2}y - y - 4{x^3}) = 0

,

x > 1

, with

y(2) = - 2

. Then y(3) is equal to :

A

-

18

B

-

12

C

-

6

D

-

3

Correct Answer

Option A

Solution

{{dy} \over {dx}} + {{y(3{x^2} - 1)} \over {x(1 - {x^2})}} = {{4{x^3}} \over {x(1 - {x^2})}}

IF = {e^{\int {{{3{x^2} - 1} \over {x - {x^3}}}dx} }} = {e^{ - \ln |{x^3} - x|}} = {e^{ - \ln ({x^3} - x)}} = {1 \over {{x^3} - x}}

Solution of D.E. can be given by

y.\,{1 \over {{x^3} - x}} = \int {{{4{x^3}} \over {x(1 - {x^2})}}.\,{1 \over {x({x^2} - 1)}}dx}

\Rightarrow {y \over {{x^3} - x}} = \int {{{ - 4x} \over {{{({x^2} - 1)}^2}}}dx}

\Rightarrow {y \over {{x^3} - x}} = {2 \over {({x^2} - 1)}} + c

at x = 2, y = $-$ 2

{{ - 2} \over 6} = {2 \over 3} + c \Rightarrow c = - 1

at

x = 3 \Rightarrow {y \over {24}} = {2 \over 8} - 1 \Rightarrow y = - 18

Q78

Let the solution curve of the differential equation

x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}}

,

y(1) = 3

be

y = y(x)

. Then y(2) is equal to:

A 15

B 11

C 13

D 17

Correct Answer

Option A

Solution

Given,

x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16x}

\Rightarrow x{{dy} \over {dx}} = y + \sqrt {{y^2} + 16x}

\Rightarrow {{dy} \over {dx}} = {y \over x} + \sqrt {{{\left( {{y \over x}} \right)}^2} + 16}

This is a homogenous different equation. Let

{y \over x} = v

\Rightarrow y = vx

\Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}

$\therefore$

v + x{{dv} \over {dx}} =v+ \sqrt {{v^2} + 16}

$\Rightarrow$

x{{dv} \over {dx}} = \sqrt {{v^2} + 16}

\Rightarrow {{dv} \over {\sqrt {{v^2} + 16} }} = {{dx} \over x}

Integrating both sides, we get

\int {{{dv} \over {\sqrt {{v^2} + 16} }} = \int {{{dx} \over x}} }

\Rightarrow \ln \left| {v + \sqrt {{v^2} + 16} } \right| = \ln x + \ln c

\Rightarrow v + \sqrt {{v^2} + 16} = cx

Now putting,

v = {y \over x}

, we get

{y \over x} + \sqrt {{{{y^2}} \over {{x^2}}} + 16} = cx

\Rightarrow {y \over x} + \sqrt {{{{y^2} + 16{x^2}} \over {{x^2}}}} = cx

\Rightarrow y + \sqrt {{y^2} + 16{x^2}} = c{x^2}

...... (1) Given,

y(1) = 3

$\therefore$ When x = 1 then y = 3. Putting in equation (1) we get,

3 + \sqrt {9 + 16} = c.\,1

\Rightarrow c = 8

$\therefore$ Solution of equation,

y + \sqrt {{y^2} + 16{x^2}} = 8{x^2}

Now, y(2) means when x = 2 then y = ? $\therefore$

y + \sqrt {{y^2} + 16 \times 4} = 8 \times 4

\Rightarrow y = 15

Q79

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

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

and y (0) = 0, then

6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)

is equal to

A 2

B

-

2

C

-

4

D

-

1

Correct Answer

Option C

Solution

Given,

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

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

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

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

Now, Let

{e^x} = t

\Rightarrow {e^x}dx = dt

\Rightarrow {\tan ^{ - 1}}(y) = \int {{{ - 2dt} \over {1 + {t^2}}}}

\Rightarrow {\tan ^{ - 1}}(y) = - 2{\tan ^{ - 1}}(t) + C

\Rightarrow {\tan ^{ - 1}}(y) = - 2{\tan ^{ - 1}}({e^x}) + C

[Putting value of t] Given, y(0) = 0 means when x = 0 the y = 0 $\therefore$

{\tan ^{ - 1}}(0) = - 2{\tan ^{ - 1}}({e^o}) + C

\Rightarrow 0 = - 2 \times {\pi \over 4} + C

\Rightarrow C = {\pi \over 2}

$\therefore$

{\tan ^{ - 1}}(y) = 2{\tan ^{ - 1}}({e^x}) + {\pi \over 2}

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

Differentiating both sides, we get

y' = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right) - 2\,.\,{{{e^x}} \over {1 + {e^{2x}}}}

= {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right) \times {{ - 2{e^x}} \over {1 + {e^{2x}}}}

$\therefore$

y'(0) = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^o}) + {\pi \over 2}} \right) \times {{ - 2{e^o}} \over {1 + {e^o}}}

= {\sec ^2}\left( { - 2 \times {\pi \over 4} + {\pi \over 2}} \right) \times {{ - 2} \over 2}

= {\sec ^2}(0)x - 1

= 1 \times 1

= - 1

And

y\left( {\log _e^{\sqrt 3 }} \right) = \tan \left( { - 2{{\tan }^{ - 1}}\left( {{e^{\log _e^{\sqrt 3 }}}} \right) + {\pi \over 2}} \right)

= \tan \left( { - 2{{\tan }^{ - 1}}(\sqrt 3 ) + {\pi \over 2}} \right)

= \tan \left( { - 2 \times {\pi \over 3} + {\pi \over 2}} \right)

= \tan \left( { - {\pi \over 6}} \right)

= - \tan \left( {{\pi \over 6}} \right)

= - {1 \over {\sqrt 3 }}

$\therefore$

6\left( {y'(0) + {{\left( {y(\log _e^{\sqrt 3 })} \right)}^2}} \right)

= 6\left( { - 1 + {{\left( {{{ - 1} \over {\sqrt 3 }}} \right)}^2}} \right)

= 6\left( { - 1 + {1 \over 3}} \right) = 6 \times {{ - 2} \over 3} = - 4

Q80

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

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

such that x(1) = 0. Then, x(e) is equal to :

A

e{\log _e}(2)

B

- e{\log _e}(2)

C

{e^2}{\log _e}(2)

D

- {e^2}{\log _e}(2)

Correct Answer

Option D

Solution

Given differential equation

2y{e^{{x \over {{y^2}}}}}dx + \left( {{y^2} - 4x{e^{{x \over {{y^2}}}}}} \right)dy = 0,\,x(1) = 0

\Rightarrow {e^{{x \over {{y^2}}}}}[2ydx - 4xdy] = - {y^2}dy

\Rightarrow {e^{{x \over {{y^2}}}}}\left[ {{{2{y^2}dx - 4xydy} \over {{y^4}}}} \right] = {{ - 1} \over y}dy

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

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

...... (i) Now, using x(1) = 0, c = 2 So, for x(e), Put y = e in (i)

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

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