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Differential Equations

JEE Mathematics · 179 questions · Page 3 of 18 · Click an option or "Show Solution" to reveal answer

Q21
Let the population of rabbits surviving at time tt be governed by the differential equation dp(t)dt=12p(t)−200.{{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 200. If p(0)=100,p(0)=100, then p(t)p(t) equals:
A 600−500 et/2600 - 500\,{e^{t/2}}
B 400−300 e−t/2400 - 300\,{e^{-t/2}}
C 400−300 et/2400 - 300\,{e^{t/2}}
D 300−200 e−t/2300 - 200\,{e^{-t/2}}
Correct Answer
Option C
Solution

Given differential equation is

dp(t)dt=12p(t)−200{{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 200

By separating the variable, we get

dp(t)=[12p(t)−200]dtdp\left( t \right) = \left[ {{1 \over 2}p\left( t \right) - 200} \right]dt
⇒dp(t)12p(t)−200=dt\Rightarrow {{dp\left( t \right)} \over {{1 \over 2}p\left( t \right) - 200}} = dt
              \,\,\,\,\,\,\,\,\,\,\,\,\,\,

Integrating on both the sides,

               ∫d(p(t))12p(t)−200=∫dt\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int {{{d\left( {p\left( t \right)} \right)} \over {{1 \over 2}p\left( t \right) - 200}}} = \int {dt}

Let

12p(t)−200=s⇒dp(t)2=ds{1 \over 2}p\left( t \right) - 200 = s \Rightarrow {{dp\left( t \right)} \over 2} = ds

So,

∫d p(t)(12p(t)−200)=∫dt\int {{{d\,p\left( t \right)} \over {\left( {{1 \over 2}p\left( t \right) - 200} \right)}}} = \int {dt}
⇒∫2dss=∫dt⇒2log⁡s=t+c\Rightarrow \int {{{2ds} \over s} = \int {dt} } \Rightarrow 2\log s = t + c
⇒2log⁡(p(t)2−200)=t+c\Rightarrow 2\log \left( {{{p\left( t \right)} \over 2} - 200} \right) = t + c
⇒p(t)2−200=e12k\Rightarrow {{p\left( t \right)} \over 2} - 200 = {e^{{1 \over 2}}}k

Using given condition

p(t)=400−300et/2p\left( t \right) = 400 - 300{e^{t/2}}
Q22
Let y(x)y(x) be the solution of the differential equation (x log⁡x)dydx+y=2x log⁡x,(x≥1).\left( {x\,\log x} \right){{dy} \over {dx}} + y = 2x\,\log x,\left( {x \ge 1} \right). Then y(e)y(e) is equal to :
A 22
B 2e2e
C ee
D 00
Correct Answer
Option A
Solution

Given,

dydx+(1x log⁡ x)y=2{{dy} \over {dx}} + \left( {{1 \over {x\,\log \,x}}} \right)y = 2
I.F.=e∫1xlog⁡xdx=elog⁡(log⁡x)=log⁡xI.F. = {e^{\int {{1 \over {x\log x}}dx} }} = {e^{\log \left( {\log x} \right)}} = \log x
y.log⁡x=∫2 log⁡xdx+cy.\log x = \int {2\,\log xdx + c}
ylog⁡x=2[xlog⁡x−x]+cy\log x = 2\left[ {x\log x - x} \right] + c

Put

x=1,y.0=−2+cx=1,y.0=-2+c
⇒c=2\Rightarrow c = 2

Put

x=ex=e
ylog⁡e=2e(log⁡e−1)+c⇒y(e)=c=2y\log e = 2e\left( {\log e - 1} \right) + c \Rightarrow y\left( e \right) = c = 2
Q23
If 2x = y15{^{{1 \over 5}}} + y−15{^{ - {1 \over 5}}} and (x2 −- 1) d2ydx2{{{d^2}y} \over {d{x^2}}} + λ\lambda x dydx{{dy} \over {dx}} + ky = 0, then λ\lambda + k is equal to :
A −- 23
B −- 24
C 26
D −- 26
Correct Answer
Option B
Solution

It is given that

2x=y1/5+y−1/52x = {y^{1/5}} + {y^{ - 1/5}}
⇒2x=y1/5+1/y1/5\Rightarrow 2x = {y^{1/5}} + 1/{y^{1/5}}

Therefore,

2x=a+1a⇒a2−2ax+1=02x = a + {1 \over a} \Rightarrow {a^2} - 2ax + 1 = 0
a=2x±4x2−42a = {{2x \pm \sqrt {4{x^2} - 4} } \over 2}
⇒a=2x±2x2−12\Rightarrow a = {{2x \pm 2\sqrt {{x^2} - 1} } \over 2}
⇒a=x±x2−1\Rightarrow a = x \pm \sqrt {{x^2} - 1}
⇒y1/5=x±x2−1\Rightarrow {y^{1/5}} = x \pm \sqrt {{x^2} - 1}
⇒y=(x±x2−1)5\Rightarrow y = {(x \pm \sqrt {{x^2} - 1} )^5}

Therefore,

dydx=5(x±x2−1)4(1±2x2x2−1){{dy} \over {dx}} = 5{(x \pm \sqrt {{x^2} - 1} )^4}\left( {1 \pm {{2x} \over {2\sqrt {{x^2} - 1} }}} \right)
=5(x+x2−1)4(x2−1±xx2−1)= 5{(x + \sqrt {{x^2} - 1} )^4}\left( {{{\sqrt {{x^2} - 1} \pm x} \over {\sqrt {{x^2} - 1} }}} \right)
⇒dydx=−5yx2−1\Rightarrow {{dy} \over {dx}} = {{ - 5y} \over {\sqrt {{x^2} - 1} }}

...... (1)

⇒d2ydx2=[x2−1(−5dydx)−5(−5y)122xx2−1](x2−1)\Rightarrow {{{d^2}y} \over {d{x^2}}} = {{\left[ {\sqrt {{x^2} - 1} \left( { - 5{{dy} \over {dx}}} \right) - 5( - 5y){1 \over 2}{{2x} \over {\sqrt {{x^2} - 1} }}} \right]} \over {({x^2} - 1)}}

Therefore,

(x2−1)d2ydx2=−5x2−1dydx+5yxx2−1({x^2} - 1){{{d^2}y} \over {d{x^2}}} = - 5\sqrt {{x^2} - 1} {{dy} \over {dx}} + 5y{x \over {\sqrt {{x^2} - 1} }}
⇒(x2−1)d2ydx2=25y−xdydx\Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = 25y - x{{dy} \over {dx}}
⇒(x2−1)d2ydx2+1xdydx−25y=0\Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} + 1x{{dy} \over {dx}} - 25y = 0

Therefore, λ\lambda = 1, k = −-25; hence,

λ+k=−24\lambda + k = - 24
Q24
If (2+sin⁡x)dydx+(y+1)cos⁡x=0\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0 and y(0) = 1, then y(π2)y\left( {{\pi \over 2}} \right) is equal to :
A −23 - {2 \over 3}
B −13 - {1 \over 3}
C 43{4 \over 3}
D 13{1 \over 3}
Correct Answer
Option D
Solution
(2+sin⁡x)dydx+(y+1)cos⁡x=0\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0

⇒\Rightarrow

ddx(2+sin⁡x)(y+1)=0{d \over {dx}}\left( {2 + \sin x} \right)\left( {y + 1} \right) = 0

On integrating, we get (2 + sin x) (y + 1) = C At x = 0, y = 1 we have (2 + sin 0) (1 + 1) = C ⇒\Rightarrow C = 4 ⇒\Rightarrow

(y+1)=42+sin⁡x\left( {y + 1} \right) = {4 \over {2 + \sin x}}

⇒\Rightarrow y =

42+sin⁡x−1{4 \over {2 + \sin x}} - 1

Now

y(π2)=42+sin⁡π2−1y\left( {{\pi \over 2}} \right) = {4 \over {2 + \sin {\pi \over 2}}} - 1

=

43−1{4 \over 3} - 1

=

13{1 \over 3}
Q25
Let y = y(x) be the solution of the differential equation sin⁡xdydx+ycos⁡x=4x\sin x{{dy} \over {dx}} + y\cos x = 4x, x∈(0,π)x \in \left( {0,\pi } \right). If y(π2)=0y\left( {{\pi \over 2}} \right) = 0, then y(π6)y\left( {{\pi \over 6}} \right) is equal to :
A −49π2 - {4 \over 9}{\pi ^2}
B 493π2{4 \over {9\sqrt 3 }}{\pi ^2}
C −893π2 - {8 \over {9\sqrt 3 }}{\pi ^2}
D −89π2 - {8 \over 9}{\pi ^2}
Correct Answer
Option D
Solution

Given, sin x

dydx+ycos⁡y=4x{{dy} \over {dx}} + y\cos y = 4x
⇒    dydx \Rightarrow \,\,\,\,{{dy} \over {dx}}\,

+ y cot x = 4x cosec x This is a linear differential equation of form,

dydx {{dy} \over {dx}}\,

+ py = Q Where p = cot x and Q = 4x cosec x So, Integrating factor (I. F) =

e∫pdx{e^{\int {pdx} }}

=

e∫cot⁡dx{e^{\int {\cot dx} }}

=

eln⁡∣sin⁡ x∣{e^{\ln \left| {\sin \,x} \right|}}

= sin x as

x∈(0,π)x \in \left( {0,\pi } \right)

Solution of the differential equation is y sin x =

∫\int {}

4x cosecx sinx dx + c

⇒    \Rightarrow \,\,\,\,

y sinx =

∫4x dx +c\int {4x\,dx\, + c}
⇒    \Rightarrow \,\,\,\,

y sinx = 4.

x22+c{{{x^2}} \over 2} + c
⇒    \Rightarrow \,\,\,\,

y sinx = 2x2 + c . . . . . (1) Given that,

y(π2)=0y\left( {{\pi \over 2}} \right) = 0
∴   \therefore\,\,\,

x =

y(π2)=0y\left( {{\pi \over 2}} \right) = 0

and y = 0 Put this x =

π2{\pi \over 2}

and y = 0 at equation (1) 0.1 = 2.

(π2)\left( {{\pi \over 2}} \right)

2 + c

⇒   \Rightarrow \,\,\,

c =

−π22- {{{\pi ^2}} \over 2}

So, differential equation is y sin x = 2x2 −-

π22  .....(2){{{\pi ^2}} \over 2}\,\,.....(2)

Now we have to find y

(π6).\left( {{\pi \over 6}} \right).

So, put x =

π6{{\pi \over 6}}

at equation (2) y . sin

π6{{\pi \over 6}}

= 2

(π6)2−π22{\left( {{\pi \over 6}} \right)^2} - {{{\pi ^2}} \over 2}
⇒       y.12\Rightarrow \,\,\,\,\,\,\,y.{1 \over 2}

= 2.

π236−π22{{{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 2}
⇒     y2=π218−π22\Rightarrow \,\,\,\,\,{y \over 2} = {{{\pi ^2}} \over {18}} - {{{\pi ^2}} \over 2}
⇒    y2=π2−9π218\Rightarrow \,\,\,\,{y \over 2} = {{{\pi ^2} - 9{\pi ^2}} \over {18}}
⇒     y=−8π29\Rightarrow \,\,\,\,\,y = - {{8{\pi ^2}} \over 9}
Q26
The general solution of the differential equation (y2 – x3)dx – xydy = 0 (x ≠ \ne 0) is : (where c is a constant of integration)
A y2 + 2x3 + cx2 = 0
B y2 + 2x2 + cx3 = 0
C y2 – 2x + cx3 = 0
D y2 – 2x3 + cx2 = 0
Correct Answer
Option A
Solution
(y2−x3)dx−xydy=0({y^2} - {x^3})dx - xydy = 0
⇒y(ydx−xdy)=x3dx\Rightarrow y(ydx - xdy) = {x^3}dx
⇒yx(ydx−xdyx2)=dx\Rightarrow {y \over x}\left( {{{ydx - xdy} \over {{x^2}}}} \right) = dx
⇒−yxd(yx)=dx\Rightarrow - {y \over x}d\left( {{y \over x}} \right) = dx
⇒−12(yx)2=x+k\Rightarrow - {1 \over 2}{\left( {{y \over x}} \right)^2} = x + k
⇒−y2=2x3+2x2k\Rightarrow - {y^2} = 2{x^3} + 2{x^2}k
⇒y2+2x3+cx2=0\Rightarrow {y^2} + 2{x^3} + c{x^2} = 0
Q27
Consider the differential equation, y2dx+(x−1y)dy=0{y^2}dx + \left( {x - {1 \over y}} \right)dy = 0, If value of y is 1 when x = 1, then the value of x for which y = 2, is :
A 32−1e{3 \over 2} - {1 \over {\sqrt e }}
B 12+1e{1 \over 2} + {1 \over {\sqrt e }}
C 52+1e{5 \over 2} + {1 \over {\sqrt e }}
D 32−e{3 \over 2} - \sqrt e
Correct Answer
Option A
Solution
y2dx+(x−1y)dy=0{y^2}dx + \left( {x - {1 \over y}} \right)dy = 0
⇒dxdy+xy2=1y3\Rightarrow {{dx} \over {dy}} + {x \over {{y^2}}} = {1 \over {{y^3}}}

Integrating factor (I.F) =

e−1y{e^{ - {1 \over y}}}

Now x.

e−1y{e^{ - {1 \over y}}}

=

∫e−1y1y3dy\int {{e^{ - {1 \over y}}}{1 \over {{y^3}}}dy}

by putting

−1y{ - {1 \over y}}

= t x.et =

∫et(−t)dt\int {{e^t}( - t)dt}
⇒xet=−(t.et−et)+c\Rightarrow x{e^t} = - (t.{e^t} - {e^t}) + c
⇒xe−1y=e−1y(1+1y)+c\Rightarrow x{e^{ - {1 \over y}}} = {e^{ - {1 \over y}}}\left( {1 + {1 \over y}} \right) + c
⇒x=1+1y+c.e1y\Rightarrow x = 1 + {1 \over y} + c.e^{1 \over y}

It passes through (1, 1) ∴\therefore c =

−1e-{1 \over e}

Equation of the curve is

x=1+1y−e1y−1x = 1 + {1 \over y} - {e^{{1 \over y} - 1}}

It passes through (k, 2) ∴\therefore k =

1+12−e12−11 + {1 \over 2} - {e^{{1 \over 2} - 1}}

=

323 \over 2

-

1e1 \over{\sqrt e}
Q28
Let y = y(x) be the solution of the differential equation, dydx+ytan⁡x=2x+x2tan⁡x{{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x, x∈(−π2,π2)x \in \left( { - {\pi \over 2},{\pi \over 2}} \right), such that y(0) = 1. Then :
A y(π4)−y(−π4)=2y\left( {{\pi \over 4}} \right) - y\left( { - {\pi \over 4}} \right) = \sqrt 2
B y′(π4)−y′(−π4)=π−2y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2
C y(π4)+y(−π4)=π22+2y\left( {{\pi \over 4}} \right) + y\left( { - {\pi \over 4}} \right) = {{{\pi ^2}} \over 2} + 2
D y′(π4)+y′(−π4)=−2y'\left( {{\pi \over 4}} \right) + y'\left( { - {\pi \over 4}} \right) = - \sqrt 2
Correct Answer
Option B
Solution
dydx+y(tan⁡x)=2x+x2tan⁡x{{dy} \over {dx}} + y(\tan x) = 2x + {x^2}\tan x

I.F. =

e∫tan⁡x dx{e^{\int {\tan x\,dx} }}

=

eln⁡sec⁡x{e^{\ln \sec x}}

= sec x y. sec x =

∫(2x+x2tan⁡x)sec⁡ x dx\int {(2x + {x^2}\tan x)\sec \,x\,dx}
⇒ysec⁡x=x2sec⁡x+λ\Rightarrow y\sec x = {x^2}\sec x + \lambda

⇒\Rightarrow y(0) = 0 + λ\lambda = 1 ⇒\Rightarrow λ\lambda = 1 ⇒\Rightarrow y = x2 + cos x ⇒\Rightarrow y' = 2x – sinx

⇒y′(π4)=π2−12\Rightarrow y'\left( {{\pi \over 4}} \right) = {\pi \over 2} - {1 \over {\sqrt 2 }}
⇒y′(−π4)=−π2+12\Rightarrow y'\left( { - {\pi \over 4}} \right) = - {\pi \over 2} + {1 \over {\sqrt 2 }}

∴\therefore

y′(π4)−y′(−π4)=π−2y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2
Q29
If y = y(x) is the solution of the differential equation 5+ex2+y.dydx+ex=0{{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0 satisfying y(0) = 1, then a value of y(loge13) is :
A -1
B 1
C 0
D 2
Correct Answer
Option A
Solution
5+ex2+y.dydx+ex=0{{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0

⇒\Rightarrow

5+ex2+y.dydx=−ex{{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} = -{e^x}

Integrating both sides, ⇒\Rightarrow

∫dy2+y=∫−exex+5dx\int {{{dy} \over {2 + y}}} = \int {{{ - {e^x}} \over {{e^x} + 5}}} dx

⇒\Rightarrow ln (y + 2) = – ln(ex + 5) + k ⇒\Rightarrow (y + 2) (ex + 5) = C ∵\because y(0) = 1 ⇒\Rightarrow C = 18 ∴\therefore y + 2 =

18ex+5{{{18} \over {{e^x} + 5}}}

At x = loge13 y + 2 =

1813+5{{{18} \over {13 + 5}}}

= 1 ⇒\Rightarrow y = -1

Q30
If y = y(x) is the solution of the differential equation dydx=(tan⁡x−y)sec⁡2x{{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}x, x∈(−π2,π2)x \in \left( { - {\pi \over 2},{\pi \over 2}} \right), such that y (0) = 0, then y(−π4)y\left( { - {\pi \over 4}} \right) is equal to :
A 12−e{1 \over 2} - e
B e−2e - 2
C 2+1e2 + {1 \over e}
D 1e−2{1 \over e} - 2
Correct Answer
Option B
Solution
dydx+ysec⁡2x=sec2x tanx{{dy} \over {dx}} + y{\sec ^2}x = se{c^2}x\,tanx

→\to This is linear differential equation

IF=e∫sec⁡2xdx=etan⁡xIF = {e^{\int {{{\sec }^2}xdx} }} = {e^{\tan x}}

Now solution is

y.etan⁡x=∫etan⁡xsec⁡2xtan⁡xdxy.{e^{\tan x}} = \int {{e^{\tan x}}} {\sec ^2}x\tan xdx

∴\therefore Let tanx = t sec2xdx = dt

y.etan⁡x=∫et∣∣t∣dty.{e^{\tan x}} = \int {\mathop {{e^t}}\limits_{||} } \mathop t\limits_| dt

yetanx = tet – et + c yetanx = (tanx – 1)etanx + c y = (tanx – 1) + c · e–tanx Given y(0) = 0 ⇒\Rightarrow 0 = –1 + c ⇒\Rightarrow c = 1

y(−π4)=−1−1+e=−2+ey\left( { - {\pi \over 4}} \right) = - 1 - 1 + e = - 2 + e
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