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Differential Equations

JEE Mathematics · 179 questions · Page 5 of 18 · Click an option or "Show Solution" to reveal answer

Q41
The solution curve of the differential equation, (1 + e-x)(1 + y2)dydx{{dy} \over {dx}} = y2, which passes through the point (0, 1), is :
A y2 + 1 = y(log⁡e(1+e−x2)+2)\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)
B y2 + 1 = y(log⁡e(1+ex2)+2)\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)
C y2 = 1 + ylog⁡e(1+e−x2){y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}
D y2 = 1 + ylog⁡e(1+ex2){y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}
Correct Answer
Option D
Solution

Given (1 + e-x)(1 + y2)

dydx{{dy} \over {dx}}

= y2 ⇒\Rightarrow

(y2+1y2)dy=exdxex+1\left( {{{{y^2} + 1} \over {{y^2}}}} \right)dy = {{{e^x}dx} \over {{e^x} + 1}}

Integrating both sides,

∫(y2+1y2)dy=∫exdxex+1\int {\left( {{{{y^2} + 1} \over {{y^2}}}} \right)dy = } \int {{{{e^x}dx} \over {{e^x} + 1}}}

⇒\Rightarrow

y−1y=ln∣ex+1∣+cy - {1 \over y} = ln\left| {{e^x} + 1} \right| + c

It passes through (0, 1) ⇒\Rightarrow c =

−ln2- ln2

⇒\Rightarrow

y2−1=yln(ex+12){y^2} - 1 = yln\left( {{{{e^x} + 1} \over 2}} \right)

⇒\Rightarrow

y2=1+yln(ex+12){y^2} = 1 + yln\left( {{{{e^x} + 1} \over 2}} \right)
Q42
If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation, 2x2dy= (2xy + y2)dx, then f(12)f\left( {{1 \over 2}} \right) is equal to :
A 11−log⁡e2{1 \over {1 - {{\log }_e}2}}
B 11+log⁡e2{1 \over {1 + {{\log }_e}2}}
C −11+log⁡e2{{ - 1} \over {1 + {{\log }_e}2}}
D 1+log⁡e2{1 + {{\log }_e}2}
Correct Answer
Option B
Solution
2x2dy=(2xy+y2)dx2{x^2}dy = \left( {2xy + {y^2}} \right)dx
⇒2x2dydx=2xy+y2\Rightarrow 2{x^2}{{dy} \over {dx}} = 2xy + {y^2}
⇒2x22x2y2dydx=2xy2x2y2+y22x2y2\Rightarrow {{2{x^2}} \over {2{x^2}{y^2}}}{{dy} \over {dx}} = {{2xy} \over {2{x^2}{y^2}}} + {{{y^2}} \over {2{x^2}{y^2}}}
⇒1y2dydx−1x1y=12x2\Rightarrow {1 \over {{y^2}}}{{dy} \over {dx}} - {1 \over x}{1 \over y} = {1 \over {2{x^2}}}

Let

−1y=t- {1 \over y} = t
⇒1y2dydx=dtdx\Rightarrow {1 \over {{y^2}}}{{dy} \over {dx}} = {{dt} \over {dx}}
⇒dtdx+t1x=12x2\Rightarrow {{dt} \over {dx}} + t{1 \over x} = {1 \over {2{x^2}}}

This is linear differentiatial equation.

∴I.F=e∫1xdx=eln⁡x\therefore I.F = {e^{\int {{1 \over x}dx} }} = {e^{\ln x}}
∴t.eln⁡x=∫12x2.eln⁡xdx\therefore t.{e^{\ln x}} = \int {{1 \over {2{x^2}}}.{e^{\ln x}}dx}
⇒−1y.x=∫12x2.xdx\Rightarrow - {1 \over y}.x = \int {{1 \over {2{x^2}}}.xdx}
⇒−xy=12∫dxx\Rightarrow - {x \over y} = {1 \over 2}\int {{{dx} \over x}}
⇒−xy=12ln⁡x+c\Rightarrow - {x \over y} = {1 \over 2}\ln x + c

This curve passes through the point (1, 2)

∴−12=0+c\therefore - {1 \over 2} = 0 + c
⇒c=−12\Rightarrow c = - {1 \over 2}
∴−xy=12ln⁡x−12\therefore - {x \over y} = {1 \over 2}\ln x - {1 \over 2}
⇒−2xy=ln⁡x−1\Rightarrow - {{2x} \over y} = \ln x - 1
⇒y=2x1−ln⁡x\Rightarrow y = {{2x} \over {1 - \ln x}}
⇒f(x)=2x1−ln⁡x\Rightarrow f\left( x \right) = {{2x} \over {1 - \ln x}}

So,

f(12)=2×121−ln⁡(12)=11+log⁡e2f\left( {{1 \over 2}} \right) = {{2 \times {1 \over 2}} \over {1 - \ln \left( {{1 \over 2}} \right)}} = {1 \over {1 + {{\log }_e}2}}
Q43
Let y = y(x) be the solution of the differential equation dydx=2(y+2sin⁡x−5)x−2cos⁡x{{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x such that y(0) = 7. Then y(π\pi) is equal to :
A 2eπ2+52{e^{{\pi ^2}}} + 5
B eπ2+5{e^{{\pi ^2}}} + 5
C 3eπ2+53{e^{{\pi ^2}}} + 5
D 7eπ2+57{e^{{\pi ^2}}} + 5
Correct Answer
Option A
Solution
dydx−2xy=2(2sin⁡x−5)x−2cos⁡x{{dy} \over {dx}} - 2xy = 2(2\sin x - 5)x - 2\cos x

IF =

e−x2{e^{ - {x^2}}}

So,

y.e−x2=∫e−x2(2x(2sin⁡x−5)−2cos⁡x)dxy.{e^{ - {x^2}}} = \int {{e^{ - {x^2}}}(2x(2\sin x - 5) - 2\cos x)dx}
⇒y.e−x2=e−x2(5−2sin⁡x)+c\Rightarrow y.{e^{ - {x^2}}} = {e^{ - {x^2}}}(5 - 2\sin x) + c
⇒y=5−2sin⁡x+c.ex2\Rightarrow y = 5 - 2\sin x + c.{e^{{x^2}}}

Given at x = 0, y = 7 ⇒\Rightarrow 7 = 5 + c ⇒\Rightarrow c = 2 So,

y=5−2sin⁡x+2ex2y = 5 - 2\sin x + 2{e^{{x^2}}}

Now, at x = π\pi, y = 5 + 2

ex2{e^{{x^2}}}
Q44
Let y = y(x) be the solution of the differential equation, 2+sin⁡xy+1.dydx=−cos⁡x{{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x, y > 0,y(0) = 1. If y(π\pi ) = a and dydx{{dy} \over {dx}} at x = π\pi is b, then the ordered pair (a, b) is equal to :
A (2, 1)
B (2,32)\left( {2,{3 \over 2}} \right)
C (1, -1)
D (1, 1)
Correct Answer
Option D
Solution
2+sin⁡xy+1.dydx=−cos⁡x{{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x

⇒\Rightarrow

∫dyy+1=∫(−cos⁡x)dx2+sin⁡x\int {{{dy} \over {y + 1}}} = \int {{{\left( { - \cos x} \right)dx} \over {2 + \sin x}}}

⇒\Rightarrow

ln⁡∣y+1∣=−ln⁡∣2+sin⁡x∣+ln⁡c\ln \left| {y + 1} \right| = - \ln \left| {2 + \sin x} \right| + \ln c

⇒\Rightarrow

ln⁡∣(y+1)(2+sin⁡x)∣=ln⁡c\ln \left| {\left( {y + 1} \right)\left( {2 + \sin x} \right)} \right| = \ln c

As y(0) = 1 ∴\therefore

ln⁡∣(1+1)(2+0)∣=ln⁡c\ln \left| {\left( {1 + 1} \right)\left( {2 + 0} \right)} \right| = \ln c

⇒\Rightarrow ∴\therefore c = 4 ∴\therefore

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

= 4 ⇒\Rightarrow y =

2−sin⁡x2+sin⁡x{{{2 - \sin x} \over {2 + \sin x}}}

Given y(π\pi) = a ∴\therefore y(π\pi) =

2−sin⁡π2+sin⁡π{{{2 - \sin \pi } \over {2 + \sin \pi }}}

= 1 = a

dydx=(2+sin⁡x)(−cos⁡x)−(2−sin⁡x).(cos⁡x)(2+sin⁡x)2{{dy} \over {dx}} = {{\left( {2 + \sin x} \right)\left( { - \cos x} \right) - \left( {2 - \sin x} \right).\left( {\cos x} \right)} \over {{{\left( {2 + \sin x} \right)}^2}}}

⇒\Rightarrow

dydx∣x=π={\left. {{{dy} \over {dx}}} \right|_{x = \pi }} =

1 = b So, (a, b) = (1, 1)

Q45
If dydx=xyx2+y2{{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}; y(1) = 1; then a value of x satisfying y(x) = e is :
A 2e\sqrt 2 e
B 123e{1 \over 2}\sqrt 3 e
C e2{e \over {\sqrt 2 }}
D 3e\sqrt 3 e
Correct Answer
Option D
Solution
dydx=xyx2+y2{{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}

Let y = vx ∴\therefore

dydx{{dy} \over {dx}}

= v + x.

dvdx{{dv} \over {dx}}

⇒\Rightarrow v + x.

dvdx{{dv} \over {dx}}

=

x(vx)x2+v2x2{{x\left( {vx} \right)} \over {{x^2} + {v^2}{x^2}}}

=

v1+v2{v \over {1 + {v^2}}}

⇒\Rightarrow x.

dvdx{{dv} \over {dx}}

=

v1+v2{v \over {1 + {v^2}}}

- v =

v−v−v31+v2{{v - v - {v^3}} \over {1 + {v^2}}}

=

−v31+v2- {{{v^3}} \over {1 + {v^2}}}

⇒\Rightarrow

∫1+v2v3dv=−∫dxx\int {{{1 + {v^2}} \over {{v^3}}}} dv = - \int {{{dx} \over x}}

⇒\Rightarrow

−12v2+log⁡v- {1 \over {2{v^2}}} + \log v

=

−log⁡x+C- \log x + C

⇒\Rightarrow

−12x2y2+log⁡(yx)- {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)

=

−log⁡x+C- \log x + C

......(1) putting x = 1, y = 1 we get ⇒\Rightarrow C =

−12- {1 \over 2}

From eq. (1)

−12x2y2+log⁡(yx)- {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)

=

−log⁡x−12- \log x - {1 \over 2}

Put y = e

−12x2e2+log⁡(ex)- {1 \over 2}{{{x^2}} \over {{e^2}}} + \log \left( {{e \over x}} \right)

=

−log⁡x−12- \log x - {1 \over 2}

⇒\Rightarrow x2 = 3e2 ⇒\Rightarrow x = ±\pm3

e\sqrt e
Q46
Let y = y(x) be a solution of the differential equation, 1−x2dydx+1−y2=0\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0, |x| < 1. If y(12)=32y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}, then y(−12)y\left( { - {1 \over {\sqrt 2 }}} \right) is equal to :
A −32 - {{\sqrt 3 } \over 2}
B None of those
C 12{{1 \over {\sqrt 2 }}}
D −12-{{1 \over {\sqrt 2 }}}
Correct Answer
Option C
Solution

Given

1−x2dydx+1−y2=0\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0

⇒\Rightarrow

dydx=−1−y21−x2{{dy} \over {dx}} = - {{\sqrt {1 - {y^2}} } \over {\sqrt {1 - {x^2}} }}

⇒\Rightarrow

dy1−y2+dx1−x2=0{{dy} \over {\sqrt {1 - {y^2}} }} + {{dx} \over {\sqrt {1 - {x^2}} }} = 0

⇒\Rightarrow sin-1 y + sin-1 x = c Given that

y(12)=32y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}

So when x =

12{1 \over 2}

then y =

32{{\sqrt 3 } \over 2}

∴\therefore sin-1

32{{\sqrt 3 } \over 2}

+ sin-1

12{1 \over 2}

= c ⇒\Rightarrow c =

π3{\pi \over 3}

+

π6{\pi \over 6}

=

π2{\pi \over 2}

So sin-1 y + sin-1 x =

π2{\pi \over 2}

⇒\Rightarrow sin-1 y =

π2{\pi \over 2}

- sin-1 x ⇒\Rightarrow sin-1 y = cos-1 x Now

y(−12)y\left( { - {1 \over {\sqrt 2 }}} \right)

means x =

−12- {1 \over {\sqrt 2 }}

and find y. Putting x =

−12- {1 \over {\sqrt 2 }}

sin-1 y = cos-1

(−12)\left( { - {1 \over {\sqrt 2 }}} \right)

=

3π4{{3\pi } \over 4}

y (at x =

−12- {1 \over {\sqrt 2 }}

) = sin(

3π4{{3\pi } \over 4}

) =

12{{1 \over {\sqrt 2 }}}
Q47
Let y = y(x) be the solution curve of the differential equation, (y2−x)dydx=1\left( {{y^2} - x} \right){{dy} \over {dx}} = 1, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :
A 2 + e
B -e
C 2
D 2 - e
Correct Answer
Option D
Solution
(y2−x)dydx=1\left( {{y^2} - x} \right){{dy} \over {dx}} = 1

⇒\Rightarrow

dxdy{{dx} \over {dy}}

+ x = y2 I.F =

e∫dy{e^{\int {dy} }}

= ey Solution is given by xey =

∫y2eydy{\int {{y^2}{e^y}dy} }

⇒\Rightarrow xey = (y2 – 2y + 2)ey + C y(0) = 1 means x = 0, y = 1 ∴\therefore C = -e ∴\therefore xey = (y2 – 2y + 2)ey - e put y = 0 ∴\therefore x = 0 – 0 + 2 – e ⇒\Rightarrow x = 2 - e

Q48
If y = y(x) is the solution of the differential equation, ey(dydx−1)=ex{e^y}\left( {{{dy} \over {dx}} - 1} \right) = {e^x} such that y(0) = 0, then y(1) is equal to:
A 2 + loge2
B loge2
C 1 + loge2
D 2e
Correct Answer
Option C
Solution

Given,

ey(dydx−1)=ex{e^y}\left( {{{dy} \over {dx}} - 1} \right) = {e^x}

⇒\Rightarrow

eydydx−ey=ex{e^y}{{dy} \over {dx}} - {e^y} = {e^x}

....(1) Let

ey=t{e^y} = t

⇒\Rightarrow

eydydx=dtdx{e^y}{{dy} \over {dx}} = {{dt} \over {dx}}

∴\therefore

dtdx−t=ex{{dt} \over {dx}} - t = {e^x}

So here p = -1 and q = ex We know, IF =

e∫pdx{e^{\int {pdx} }}

=

e∫−1dx{e^{\int { - 1dx} }}

=

e−x{e^{ - x}}

∴\therefore t.

e−x{e^{ - x}}

=

∫ex.e−xdx\int {{e^x}.{e^{ - x}}dx}

⇒\Rightarrow t.

e−x{e^{ - x}}

= x + c Putting value of t, we get

eye−x{e^y}{e^{ - x}}

= x + c ⇒\Rightarrow

ey−x=x+c{e^{y - x}} = x + c

.....(

2) Given y(0) = 0 means y = 0 when x = 0.

Putting in equation (2), we get e0 = 0 + c ⇒\Rightarrow c = 1 ∴\therefore

ey−x=x+1{e^{y - x}} = x + 1

....(

3) Now we have to find y(1), which means when x = 1 find the value of y in the above equation.

Putting x = 1 in equation (3)

ey−1=1+1{e^{y - 1}} = 1 + 1

⇒\Rightarrow y - 1 = loge2 ⇒\Rightarrow y = 1 + loge2

Q49
The population P = P(t) at time 't' of a certain species follows the differential equation dPdt{{dP} \over {dt}} = 0.5P – 450. If P(0) = 850, then the time at which population becomes zero is :
A log⁡e18{\log _e}18
B 12log⁡e18{1 \over 2}{\log _e}18
C 2log⁡e18{\log _e}18
D log⁡e9{\log _e}9
Correct Answer
Option C
Solution
dpdt=p−9002{{dp} \over {dt}} = {{p - 900} \over 2}
∫8500dpp−900=∫0tdt2\int\limits_{850}^0 {{{dp} \over {p - 900}} = \int\limits_0^t {{{dt} \over 2}} }

⇒\Rightarrow

ln⁡∣p−900∣8500=t2\ln |p - 900|_{850}^0 = {t \over 2}

⇒\Rightarrow

ln⁡900−ln⁡50=t2\ln 900 - \ln 50 = {t \over 2}

⇒\Rightarrow

t2=ln⁡18{t \over 2} = \ln 18
⇒t=2ln⁡18\Rightarrow t = 2\ln 18
Q50
If a curve y = f(x) passes through the point (1, 2) and satisfies xdydx+y=bx4x {{dy} \over {dx}} + y = b{x^4}, then for what value of b, ∫12f(x)dx=625\int\limits_1^2 {f(x)dx = {{62} \over 5}} ?
A 315{{31} \over 5}
B 10
C 5
D 625{{62} \over 5}
Correct Answer
Option B
Solution
dydx+yx=bx3{{dy} \over {dx}} + {y \over x} = b{x^3}

,

I.F.=e∫dxx=xI.F. = {e^{\int {{{dx} \over x}} }} = x

∴\therefore

yx=∫bx4dx=bx55+Cyx = \int {b{x^4}dx} = {{b{x^5}} \over 5} + C

Passes through (1, 2), we get

2=b5+C2 = {b \over 5} + C

........ (i) Also,

∫12(bx45+cx)dx=625\int\limits_1^2 {\left( {{{b{x^4}} \over 5} + {c \over x}} \right)dx = {{62} \over 5}}
⇒b25×32+Cln⁡2−b25=625⇒C=0\Rightarrow {b \over {25}} \times 32 + C\ln 2 - {b \over {25}} = {{62} \over 5} \Rightarrow C = 0

&

b=10b = 10
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