Differential Equations — Page 6 | JEE Mathematics

Q51

If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is

{{{x^2} - 4x + y + 8} \over {x - 2}}

, then this curve also passes through the point :

A (4, 4)

B (5, 5)

C (5, 4)

D (4, 5)

Correct Answer

Option B

Solution

Given y (0) = 0 &

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

\Rightarrow {{dy} \over {dx}} - {y \over {x - 2}} = (x - 2) + {4 \over {x - 2}}

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

Solution of D.E.

\Rightarrow y.{1 \over {x - 2}} = \int {{1 \over {x - 2}}\left( {(x - 2) + {4 \over {x - 2}}} \right)} \,.\,dx

\Rightarrow {y \over {x - 2}} = x - {4 \over {x - 2}} + C

Now, at x = 0, y = 0 $\Rightarrow$ C = $-$ 2 $\therefore$ y = x (x $-$ 2) $-$ 4 $-$ 2 (x $-$ 2) $\Rightarrow$ y = x2 $-$ 4x This curve passes through (5, 5)

Q52

The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t = 0. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after

{k \over {{{\log }_e}\left( {{6 \over 5}} \right)}}

hours, then

{\left( {{k \over {{{\log }_e}2}}} \right)^2}

is equal to :

A 16

B 8

C 2

D 4

Correct Answer

Option D

Solution

{{dx} \over {dt}} \propto x

{{dx} \over {dt}} = \lambda x

\int\limits_{1000}^x {{{dx} \over x} = \int\limits_0^t {\lambda dt} }

\ln x - \ln 1000 = \lambda t

\ln \left( {{x \over {1000}}} \right) = \lambda t

Put t = 2, x = 1200

\ln \left( {{{12} \over {10}}} \right) = 2\lambda \Rightarrow \lambda = {1 \over 2}\ln {6 \over 5}

Now,

\ln \left( {{x \over {1000}}} \right) = {t \over 2}\ln \left( {{6 \over 5}} \right)

x = 1000{e^{{t \over 2}\ln \left( {{6 \over 5}} \right)}}

x = 2000

at

t = {k \over {\ln \left( {{6 \over 5}} \right)}}

\Rightarrow 2000 = 1000{e^{{k \over {2\ln (6/5)}} \times \ln (6/5)}}

\Rightarrow 2 = {e^{k/2}}

\Rightarrow \ln 2 = {k \over 2}

\Rightarrow {k \over {\ln 2}} = 2

\Rightarrow {\left( {{k \over {\ln 2}}} \right)^2} = 4

Q53

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

{{dy} \over {dx}} + 2y\tan x = \sin x,y\left( {{\pi \over 3}} \right) = 0

, then the maximum value of the function y(x) over R is equal to:

A

{1 \over 8}

B 8

C

-

{15 \over 4}

D

{1 \over 2}

Correct Answer

Option A

Solution

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

I.F. = {e^{2\ln (\sec x)}} = {\sec ^2}x

y.{\sec ^2}x = \int {\sin x{{\sec }^2}xdx = \int {\tan x\sec x\,dx + c} }

y{\sec ^2}x = \sec x + c

y = \cos x + c{\cos ^2}x

x = {\pi \over 3},y = 0

\Rightarrow {1 \over 2} + {c \over 4} \Rightarrow c = - 2

$\therefore$

y = \cos x - 2{\cos ^2}x

y = - 2\left( {{{\cos }^2}x - {1 \over 2}\cos x} \right) = - 2\left( {{{\left( {\cos x - {1 \over 4}} \right)}^2} - {1 \over {16}}} \right)

y = {1 \over 8} - 2{\left( {\cos x - {1 \over 4}} \right)^2}

$\therefore$

{y_{\max }} = {1 \over 8}

Q54

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

{{dy} \over {dx}}

+ (tan x) y = sin x,

0 \le x \le {\pi \over 3}

, with y(0) = 0, then

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

equal to :

A

{1 \over 2}

loge 2

B

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

loge 2

C loge 2

D

{1 \over 4}

loge 2

Correct Answer

Option B

Solution

Integrating Factor

= {e^{\int {\tan x\,dx} }} = {e^{\ln (\sec x)}} = \sec x

y\sec x = \int {(\sin x)\sec x\,dx = \ln (\sec x) + C}

y(0) = 0 \Rightarrow C = 0

$\therefore$

y = \cos x\ln |\sec x|

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

Q55

Let y = y(x) be the solution of the differential equation xdy = (y + x3 cosx)dx with y(

\pi

) = 0, then

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

is equal to :

A

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

B

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

C

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

D

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

Correct Answer

Option A

Solution

xdy = (y + {x^3}\cos x)dx

$\Rightarrow$

xdy = ydx + {x^3}\cos xdx

$\Rightarrow$

{{xdy - ydx} \over {{x^2}}} = {{{x^3}coxdx} \over {{x^2}}}

$\Rightarrow$

{d \over {dx}}\left( {{y \over x}} \right) = \int {x\cos xdx}

\Rightarrow {y \over x} = x\sin x - \int {1.\sin xdx}

$\Rightarrow$

{y \over x} = x\sin x + \cos x + C

\Rightarrow 0 = - 1 + C \Rightarrow C = 1,x = \pi ,y = 0

so,

{y \over x} = x\sin x + \cos x + 1

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

x = {\pi \over 2}

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

Q56

Let C1 be the curve obtained by the solution of differential equation

2xy{{dy} \over {dx}} = {y^2} - {x^2},x > 0

. Let the curve C2 be the solution of

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

. If both the curves pass through (1, 1), then the area enclosed by the curves C1 and C2 is equal to :

A

{\pi \over 4}

+ 1

B

\pi

+ 1

C

\pi

-

1

D

{\pi \over 2}

-

1

Correct Answer

Option D

Solution

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

Put

y = vx

v + x{{dv} \over {dx}} = {{{v^2}{x^2} - {x^2}} \over {2v{x^2}}} = {{{v^2} - 1} \over {2v}}

x{{dv} \over {dx}} = {{{v^2} - 1 - 2{v^2}} \over {2v}} = - {{({v^2} + 1)} \over {2v}}

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

\ln ({v^2} + 1) = - \ln x + \ln c \Rightarrow {v^2} + 1 = {c \over x}

\Rightarrow {{{y^2}} \over {{x^2}}} + 1 = {c \over x} \Rightarrow {x^2} + {y^2} = cx

It pass through (1, 1) $\therefore$

{x^2} + {y^2} - 2x = 0

Similarly for second differential equation

{{dy} \over {dx}} =

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

Equation of curve is x2 + y2 $-$ 2y = 0 Now required area is

= 2\int\limits_0^1 {\left( {\sqrt {2x - {x^2}} - x} \right)} dx

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

sq. units

Q57

Which of the following is true for y(x) that satisfies the differential equation

{{dy} \over {dx}}

= xy

-

1 + x

-

y; y(0) = 0 :

A y(1) = 1

B y(1) = e

-

{1 \over 2}

-

1

C y(1) = e

{1 \over 2}

-

e

-

{1 \over 2}

D y(1) = e

{1 \over 2}

-

1

Correct Answer

Option B

Solution

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

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

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

Integrating both sides, we get

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

x = 0,y = 0

\Rightarrow c = 0

$\therefore$

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

putting

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

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

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

$\therefore$

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

Q58

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

2({x^2} + {x^{5/4}})dy - y(x + {x^{1/4}})dx = {2x^{9/4}}dx

, x > 0 which passes through the point

\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)

, then the value of y(16) is equal to :

A

4\left( {{{31} \over 3} - {8 \over 3}{{\log }_e}3} \right)

B

\left( {{{31} \over 3} - {8 \over 3}{{\log }_e}3} \right)

C

\left( {{{31} \over 3} + {8 \over 3}{{\log }_e}3} \right)

D

4\left( {{{31} \over 3} + {8 \over 3}{{\log }_e}3} \right)

Correct Answer

Option A

Solution

{{dy} \over {dx}} - {y \over {2x}} = {{{x^{9/4}}} \over {{x^{5/4}}({x^{3/4}} + 1)}}

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

y.{x^{ - 1/2}} = \int {{{{x^{9/4}}.{x^{ - 1/2}}} \over {{x^{5/4}}({x^{3/4}} + 1)}}} dx

=

\int {{{{x^{1/2}}} \over {({x^{3/4}} + 1)}}} dx

Let,

x = {t^4} \Rightarrow dx = 4{t^3}dt

=

\int {{{{t^2}.4{t^3}dt} \over {({t^3} + 1)}}}

=

4\int {{{{t^2}({t^3} + 1 - 1)} \over {({t^3} + 1)}}} d

=

4\int {{t^2}dt - 4\int {{{{t^2}} \over {{t^3} + 1}}dt} }

=

{{4{t^3}} \over 3} - {4 \over 3}\ln ({t^3} + 1) + C

y{x^{ - 1/2}} = {{4{x^{3/4}}} \over 3} - {4 \over 3}\ln ({x^{3/4}} + 1) + C

It passes through the point

\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)

$\therefore$

1 - {4 \over 3}{\log _e}2 = {4 \over 3} - {4 \over 3}{\log _e}2 + C

\Rightarrow C = - {1 \over 3}

y = {4 \over 3}{x^{5/4}} - {4 \over 3}\sqrt x \ln ({x^{3/4}} + 1) - {{\sqrt x } \over 3}

y(16) = {4 \over 3} \times 32 - {4 \over 3} \times 4\ln 9 - {4 \over 3}

= {{124} \over 3} - {{32} \over 3}\ln 3 = 4\left( {{{31} \over 3} - {8 \over 3}\ln 3} \right)

Q59

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

\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx,0 \le x \le {\pi \over 2},y(0) = 0

. Then,

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

is equal to :

A

2{\log _e}\left( {{{\sqrt 3 + 7} \over 2}} \right)

B

2{\log _e}\left( {{{3\sqrt 3 - 8} \over 4}} \right)

C

2{\log _e}\left( {{{2\sqrt 3 + 10} \over {11}}} \right)

D

2{\log _e}\left( {{{2\sqrt 3 + 9} \over 6}} \right)

Correct Answer

Option C

Solution

\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx

..... (1)

(3\sin x + \cos x + 3)(\cos x\,dy - y\sin x\,dx) = dx

\int {d(y.\cos x) = \int {{{dx} \over {3\sin x + \cos x + 3}}} }

y\cos x = \int {{1 \over {3\left( {{{2 + \tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \right) + \left( {{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \right) + 3}}}

y\cos x = \int {{{{{\sec }^2}{x \over 2}} \over {6\tan {x \over 2} + 1 - {{\tan }^2}{x \over 2} + 3 + 3{{\tan }^2}{x \over 2}}}}

y\cos x = \int {{{{{\sec }^2}{x \over 2}} \over {2{{\tan }^2}{x \over 2} + 6\tan {x \over 2} + 4}}} = \int {{{{1 \over 2}{{\sec }^2}{x \over 2}dx} \over {{{\tan }^2}{x \over 2} + 3\tan {x \over 2} + 2}}}

y\cos x = \ln \left| {{{\tan {x \over 2} + 1} \over {\tan {x \over 2} + 2}}} \right| + c

Put x = 0 & y = 0

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

y\left( {{\pi \over 3}} \right) = 2\ln \left| {{{1 + \sqrt 3 } \over {1 + 2\sqrt 3 }}} \right| + \ln 2

= 2\ln \left| {{{5 + \sqrt 3 } \over {11}}} \right| + \ln 2

= 2\ln \left| {{{2\sqrt 3 + 10} \over {11}}} \right|

Q60

The differential equation satisfied by the system of parabolas y2 = 4a(x + a) is :

A

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

B

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

C

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

D

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

Correct Answer

Option C

Solution

{y^2} = 4ax + 4{a^2}

differentiate with respect to x

\Rightarrow 2y{{dy} \over {dx}} = 4a

\Rightarrow a = \left( {{y \over 2}{{dy} \over {dx}}} \right)

So, required differential equation is

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

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

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