Differential Equations — Page 4 | JEE Mathematics

Q31

If

\cos x{{dy} \over {dx}} - y\sin x = 6x

, (0 < x <

{\pi \over 2}

) and

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

= 0 then

y\left( {{\pi \over 6}} \right)

is equal to :-

A

- {{{\pi ^2}} \over {2 }}

B

- {{{\pi ^2}} \over {4\sqrt 3 }}

C

{{{\pi ^2}} \over {2\sqrt 3 }}

D

- {{{\pi ^2}} \over {2\sqrt 3 }}

Correct Answer

Option D

Solution

\cos x{{dy} \over {dx}} - y\sin x = 6x

$\Rightarrow$

{{dy} \over {dx}} - y\tan x = 6x.\sec x

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

{e^{\int { - \tan xdx} }}

=

\left| {\cos x} \right|

As 0 < x <

{\pi \over 2}

, so I.F =

\cos x

So solution is y(I.F) =

\int {6x.\sec x\left( {I.F} \right)} dx

$\Rightarrow$ y cosx =

\int {6x.\sec x\cos x} dx

$\Rightarrow$ y cosx =

6\int x dx

$\Rightarrow$ y cosx = 3x2 + C As

y\left( {{\pi \over 3}} \right) = 0

$\therefore$

\left( 0 \right) \times \left( {{1 \over 2}} \right) = {{3{\pi ^2}} \over 9} + C

\Rightarrow C = {{ - {\pi ^2}} \over 3}

\Rightarrow y\cos x = 3{x^2} - {{{\pi ^2}} \over 3}

For

y\left( {{\pi \over 6}} \right)

y{{\sqrt 3 } \over 2} = {{3{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 3}

$\Rightarrow$

{{\sqrt 3 y} \over 2} = {{ - 3{\pi ^2}} \over {12}}

\Rightarrow y = {{ - {\pi ^2}} \over {2\sqrt 3 }}

Q32

The solution of the differential equation

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

= x2 (x

\ne

0) with y(1) = 1, is :

A

y = {4 \over 5}{x^3} + {1 \over {5{x^2}}}

B

y = {3 \over 4}{x^2} + {1 \over {4{x^2}}}

C

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

D

y = {{{x^3}} \over 5} + {1 \over {5{x^2}}}

Correct Answer

Option C

Solution

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

= x2 $\Rightarrow$

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

$\therefore$ I.F =

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

= x2 $\therefore$ The solution is yx2 =

\int {{x^3}dx}

$\Rightarrow$ yx2 =

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

.....(1) As y(1) = 1 $\therefore$ when x = 1 then y = 1. Putting the value of x and y in equation (1), we get 1 =

{1 \over 4} + C

$\Rightarrow$ C =

{3 \over 4}

$\therefore$ Required solution yx2 =

{{{x^4}} \over 4} + {3 \over 4}

$\Rightarrow$

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

Q33

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

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

such that y(0) = 0. If

\sqrt ay(1)

=

\pi \over 32

, then the value of 'a' is :

A

{1 \over 2}

B

{1 \over 16}

C 1

D

{1 \over 4}

Correct Answer

Option B

Solution

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

$\Rightarrow$

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

I.F =

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

$\therefore$

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

Integrating both sides we get,

{\left( {1 + {x^2}} \right)y}

=

{\tan ^{ - 1}}x

+ C When x = 0 then y = 0 So, 0 = 0 + C $\Rightarrow$ C = 0 $\Rightarrow$

{\left( {1 + {x^2}} \right)y}

=

{\tan ^{ - 1}}x

Put x = 1, y.(2) =

{\pi \over 4}

$\Rightarrow$ y =

{\pi \over 8}

= y(1) Given,

\sqrt ay(1)

=

\pi \over 32

$\Rightarrow$

\sqrt a. {\pi \over 8}

=

\pi \over 32

$\Rightarrow$

\sqrt a = {1 \over 4}

$\Rightarrow$

a = {1 \over {16}}

Q34

If

y = \left( {{2 \over \pi }x - 1} \right) cosec\,x

is the solution of the differential equation,

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

,

0 < x < {\pi \over 2}

, then the function p(x) is equal to :

A cot x

B sec x

C tan x

D cosec x

Correct Answer

Option A

Solution

y = \left( {{2 \over \pi }x - 1} \right) cosec\,x

Differentiate w.r.t x

{{dy} \over {dx}} = {2 \over \pi }

cosec x -

\left( {{2 \over \pi }x - 1} \right)

cosec x.cot x $\Rightarrow$

{{dy} \over {dx}}

+

\left( {{2 \over \pi }x - 1} \right)

cosec x.cot x =

{2 \over \pi }

cosec x $\Rightarrow$

{{dy} \over {dx}}

+ ycot x =

{2 \over \pi }

cosec x Compare this differential equation with given differential equation, we get p(x) = cot x

Q35

The general solution of the differential equation

\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}}

+ xy

{{dy} \over {dx}}

= 0 is : (where C is a constant of integration)

A

\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C

B

\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C

C

\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C

D

\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C

Correct Answer

Option C

Solution

\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}}

+ xy

{{dy} \over {dx}}

= 0 $\Rightarrow$

\sqrt {\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)}

+ xy

{{dy} \over {dx}}

= 0 $\Rightarrow$

\sqrt {\left( {1 + {x^2}} \right)} \sqrt {\left( {1 + {y^2}} \right)}

= -xy

{{dy} \over {dx}}

$\Rightarrow$

\int {{{ydy} \over {\sqrt {1 + {y^2}} }}} = - \int {{{\sqrt {1 + {x^2}} } \over x}} dx

.....(

1) Now put 1 + x2 = u2 and 1 + y2 = v2 2xdx = 2udu and 2ydy = 2vdv $\Rightarrow$ xdx = udu and ydy = vdv Substitute these values in equation (1)

\int {{{vdv} \over v}} = - \int {{{{u^2}du} \over {{u^2} - 1}}}

$\Rightarrow$

\int {dv} = - \int {{{{u^2} - 1 + 1} \over {{u^2} - 1}}} du

$\Rightarrow$ v =

- \int {\left( {1 + {1 \over {{u^2} - 1}}} \right)} du

$\Rightarrow$ v = -u -

{1 \over 2}{\log _e}\left| {{{u - 1} \over {u + 1}}} \right|

+ C $\Rightarrow$

\sqrt {1 + {y^2}}

=

- \sqrt {1 + {x^2}}

+

{1 \over 2}{\log _e}\left| {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right|

+ C $\Rightarrow$

\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C

Q36

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

{{dy} \over {dx}}

+ 2ysinx = sin2x, x

\in

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

. If y

\left( {{\pi \over 3}} \right)

= 0, then y

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

is equal to :

A

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

B

{\sqrt 2 - 2}

C

{2 - \sqrt 2 }

D

{2 + \sqrt 2 }

Correct Answer

Option B

Solution

cosx

{{dy} \over {dx}}

+ 2ysinx = sin2x $\Rightarrow$

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

I.F =

{e^{\int {2\tan xdx} }}

= sec2 x y.sec2 x =

{\int {2\sin x{{\sec }^2}xdx} }

$\Rightarrow$ ysec2x =

{\int {2\tan x\sec xdx} }

$\Rightarrow$ ysec2x = 2secx + c Given at x =

{\pi \over 3}

, y = 0 $\Rightarrow$ 0 =

2\sec {\pi \over 3} + c

$\Rightarrow$ c = -4 ysec2x = 2secx - 4 Here put x =

{\pi \over 4}

y.2 =

2\sqrt 2

- 4 $\Rightarrow$ y =

\sqrt 2 - 2

Q37

The solution of the differential equation

{{dy} \over {dx}} - {{y + 3x} \over {{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0

is: (where c is a constant of integration)

A

x - {1 \over 2}{\left( {{{\log }_e}\left( {y + 3x} \right)} \right)^2} = C

B

y + 3x - {1 \over 2}{\left( {{{\log }_e}x} \right)^2} = C

C x – loge(y+3x) = C

D x – 2loge(y+3x) = C

Correct Answer

Option A

Solution

{{dy} \over {dx}} - {{y + 3x} \over {\ln (y + 3x)}} + 3 = 0

Let

\ln (y + 3x) = t

{1 \over {y + 3x}}.\left( {{{dy} \over {dx}} + 3} \right) = {{dt} \over {dx}}

\Rightarrow \left( {{{dy} \over {dx}} + 3} \right) = {{y + 3x} \over {\ln (y + 3x)}}

$\therefore$

\left( {y + 3x} \right){{dt} \over {dx}} = {{y + 3x} \over t}

\Rightarrow tdt = dx

{{{t^2}} \over 2} = x + c

{1 \over 2}{\left( {\ln (y + 3x)} \right)^2} = x + c

Q38

Let y = y(x) be the solution of the differential equation, xy'- y = x2(xcosx + sinx), x > 0. if y (

\pi

) =

\pi

then

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

is equal to :

A

1 + {\pi \over 2}

B

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

C

2 + {\pi \over 2}

D

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

Correct Answer

Option C

Solution

xy' - y = {x^2}(x{\mathop{\rm cosx}\nolimits} + sinx),\,x > 0,\,y(\pi ) = \pi

y' - {1 \over x}y = x\{ x\cos x\, + \,\sin x\}

I.F. = {e^{ - \int {{1 \over 2}dx} }} = {e^{ - \ln x}} = {1 \over x}

$\therefore$

y.{1 \over x} = \int {{1 \over x}.x(x\,cos\,x + sin\,x) dx}

{{y \over x}}

=

\int ( x\,\cos \,x + \sin \,x)dx

= x\,\sin \,x + C

{{y \over x}}

=

x\,\sin \,x + C

\Rightarrow y = {x^2} = \sin \,x + cx

x = $\pi$ , y = $\pi$ $\pi$ = $\pi$ c $\Rightarrow$ C = 1

y = {x^2}\sin x + x \Rightarrow y\left( {{\pi \over 2}} \right) = {{{\pi ^2}} \over 4} + {\pi \over 2}

y' = 2x\sin x + {x^2}\cos x + 1

y" =

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

y"

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

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

+ y"

\left( {{\pi \over 2}} \right) = 2 + {\pi \over 2}

Q39

If x3dy + xy dx = x2dy + 2y dx; y(2) = e and x > 1, then y(4) is equal to :

A

{{\sqrt e } \over 2}

B

{1 \over 2} + \sqrt e

C

{3 \over 2} + \sqrt e

D

{3 \over 2}\sqrt e

Correct Answer

Option D

Solution

{x^3}dy + xy\,dx = {x^2}dy + 2y\,dx

\Rightarrow dy({x^3} - {x^2}) = dx(2y - xy)

$\Rightarrow$

- \int {{1 \over y}dx}

=

\int {{{x - 2} \over {{x^2}(x - 1)}}dx}

$\Rightarrow$

- \ln y = \int {\left( {{A \over x} + {B \over {{x^2}}} + {C \over {(x - 1)}}} \right)dx}

Where A = 1, B = +2, C = $-$ 1

\Rightarrow - \ln y = \ln x - {2 \over x} - \ln (x - 1) + \lambda

As

y(2) = e

\Rightarrow - 1 = \ln 2 - 1 - 0 + \lambda

$\therefore$

\lambda = - \ln 2

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

Now put x = 4 in equation

\Rightarrow \ln y = - \ln 4 + {1 \over 2} + \ln 3 + \ln 2

\Rightarrow {\mathop{\rm lny}\nolimits} = ln\left( {{3 \over 2}} \right) + {1 \over 2}\ln e

\Rightarrow y = {3 \over 2}\sqrt e

Q40

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

-

x3)dy = (y + yx2

-

3x4)dx, x > 2. If y(3) = 3, then y(4) is equal to :

A 4

B 12

C 8

D 16

Correct Answer

Option B

Solution

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

\Rightarrow xdy - ydx = (y{x^2} - 3{x^4})dx + {x^3}dy

\Rightarrow {{xdy - ydx} \over {{x^2}}} = (ydx + xdy) - 3{x^2}dx

\Rightarrow d\left( {{y \over x}} \right) = d(xy) - d({x^3})

Integrate

\Rightarrow {y \over x} = xy - {x^3} + c

given f(3) = 3

\Rightarrow {3 \over 3} = 3 \times 3 - {3^3} + c

\Rightarrow c = 19

$\therefore$

{y \over x} = xy - {x^3} + 19

at

x = 4,{y \over 4} = 4y - 64 + 19

15y = 4 \times 45

\Rightarrow y = 12