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Differential Equations

JEE Mathematics · 179 questions · Page 12 of 18 · Click an option or "Show Solution" to reveal answer

Q111
The area enclosed by the closed curve C\mathrm{C} given by the differential equation dydx+x+ay−2=0,y(1)=0\dfrac{d y}{d x}+\dfrac{x+a}{y-2}=0, y(1)=0 is 4π4 \pi. Let PP and QQ be the points of intersection of the curve C\mathrm{C} and the yy-axis. If normals at PP and QQ on the curve C\mathrm{C} intersect xx-axis at points RR and SS respectively, then the length of the line segment RSR S is :
A 433\dfrac{4 \sqrt{3}}{3}
B 232 \sqrt{3}
C 2
D 233\dfrac{2 \sqrt{3}}{3}
Correct Answer
Option A
Solution
dydx+x+ay−2=0dydx=x+a2−y(2−y)dy=(x+a)dx2y−y2=x22+ax+ca+c=−12 as y(1)=0X2+y2+2ax−4y−1−2a=0πr2=4πr2=44=a2+4+1+2a(a+1)2=0\begin{aligned} & \frac{d y}{d x}+\frac{x+a}{y-2}=0 \\\\ & \frac{d y}{d x}=\frac{x+a}{2-y} \\\\ & (2-y) d y=(x+a) d x \\\\ & 2 y \frac{-y}{2}=\frac{x^2}{2}+\mathrm{ax}+\mathrm{c} \\\\ & \mathrm{a}+\mathrm{c}=-\frac{1}{2} \text{ as } \mathrm{y}(1)=0 \\\\ & \mathrm{X}^2+\mathrm{y}^2+2 \mathrm{ax}-4 \mathrm{y}-1-2 \mathrm{a}=0 \\\\ & \pi \mathrm{r}^2=4 \pi \\\\ & \mathrm{r}^2=4 \\\\ & 4=\sqrt{a^2+4+1+2 a} \\\\ & (\mathrm{a}+1)^2=0 \end{aligned}
P,Q=(0,2±3)P, Q=(0,2 \pm \sqrt{3})

Equation of normal at P,Q\mathrm{P}, \mathrm{Q} are y−2=3(x−1)\mathrm{y}-2=\sqrt{3}(\mathrm{x}-1)

y−2=−3(x−1)R=(1−23,0)S=(1+23,0)RS=43=433\begin{aligned} & \mathrm{y}-2=-\sqrt{3}(\mathrm{x}-1) \\\\ & \mathrm{R}=\left(1-\frac{2}{\sqrt{3}}, 0\right) \\\\ & \mathrm{S}=\left(1+\frac{2}{\sqrt{3}}, 0\right) \\\\ & \mathrm{RS}=\frac{4}{\sqrt{3}}=4 \frac{\sqrt{3}}{3} \end{aligned}
Q112
If y=y(x)y=y(x) is the solution curve of the differential equation dydx+ytan⁡x=xsec⁡x,0≤x≤π3,y(0)=1\dfrac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \dfrac{\pi}{3}, y(0)=1, then y(π6)y\left(\dfrac{\pi}{6}\right) is equal to
A π12−32log⁡e(23e)\dfrac{\pi}{12}-\dfrac{\sqrt{3}}{2} \log _{e}\left(\dfrac{2 \sqrt{3}}{e}\right)
B π12+32log⁡e(23e)\dfrac{\pi}{12}+\dfrac{\sqrt{3}}{2} \log _{e}\left(\dfrac{2 \sqrt{3}}{e}\right)
C π12+32log⁡e(2e3)\dfrac{\pi}{12}+\dfrac{\sqrt{3}}{2} \log _{e}\left(\dfrac{2}{e \sqrt{3}}\right)
D π12−32log⁡e(2e3)\dfrac{\pi}{12}-\dfrac{\sqrt{3}}{2} \log _{e}\left(\dfrac{2}{e \sqrt{3}}\right)
Correct Answer
Option D
Solution

dydx+ytan⁡x=xsec⁡x\dfrac{d y}{d x}+y \tan x=x \sec x ∴\therefore I.F =e∫tan⁡xdx=sec⁡x=e^{\int \tan x d x}=\sec x ⇒ysec⁡x=∫xsec⁡2xdx\Rightarrow y \sec x=\int x \sec ^{2} x d x ⇒ysec⁡x=xtan⁡x−ln⁡∣sec⁡x∣+c\Rightarrow y \sec x=x \tan x-\ln |\sec x|+c Given,

y(0)=1y(0)=1

⇒1=c\Rightarrow 1=c ∴\therefore ysec⁡x=xtan⁡x−ln⁡∣sec⁡x∣+1 y \sec x=x \tan x-\ln |\sec x|+1 ⇒y=xsin⁡x−cos⁡xln⁡∣sec⁡x∣+cosx\Rightarrow y=x \sin x-\cos x \ln |\sec x|+cosx Now, at x=π/6x=\pi / 6, we have

y=π12−32log⁡e23+32=π12−32[log⁡e23−log⁡ee]=π12−32log⁡e(2e3)\begin{aligned} & y=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e \frac{2}{\sqrt{3}}+\frac{\sqrt{3}}{2} \\\\ & =\frac{\pi}{12}-\frac{\sqrt{3}}{2}\left[\log _e \frac{2}{\sqrt{3}}-\log _e e\right]\\\\ &=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right) \end{aligned}
Q113
Let a differentiable function ff satisfy f(x)+∫3xf(t)tdt=x+1,x≥3f(x)+\int\limits_{3}^{x} \dfrac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3. Then 12f(8)12 f(8) is equal to :
A 19
B 34
C 17
D 1
Correct Answer
Option C
Solution

Differentiating both sides we get,

f′(x)+f(x)x=12x+1⇒dydx+yx=12x+1⇒IF=x⇒yx=12∫xx+1dx+c⇒yx=12((x+1)3232−2(x+1)12)+cxy=13(x+1)32−(x+1)12+cf(3)=2\begin{aligned} & f^{\prime}(x)+\frac{f(x)}{x}=\frac{1}{2 \sqrt{x+1}} \\\\ & \Rightarrow \frac{d y}{d x}+\frac{y}{x}=\frac{1}{2 \sqrt{x+1}} \\\\ & \Rightarrow \mathrm{IF}=x \\\\ & \Rightarrow y x=\frac{1}{2} \int \frac{x}{\sqrt{x+1}} d x+c \\\\ & \Rightarrow y x=\frac{1}{2}\left(\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}-2(x+1)^{\frac{1}{2}}\right)+c \\\\ & x y=\frac{1}{3}(x+1)^{\frac{3}{2}}-(x+1)^{\frac{1}{2}}+c \\\\ & f(3)=2 \end{aligned}

So, x=3,y=2x=3, y=2 ⇒c=163\Rightarrow c=\dfrac{16}{3} Now, x=8x=8 8f(8)=273−3+163=3438 f(8)=\dfrac{27}{3}-3+\dfrac{16}{3}=\dfrac{34}{3} 12f(8)=343×128=1712 f(8)=\dfrac{34}{3} \times \dfrac{12}{8}=17

Q114
The solution of the differential equation dydx=−(x2+3y23x2+y2),y(1)=0\dfrac{d y}{d x}=-\left(\dfrac{x^2+3 y^2}{3 x^2+y^2}\right), y(1)=0 is :
A log⁡e∣x+y∣+xy(x+y)2=0\log _e|x+y|+\dfrac{x y}{(x+y)^2}=0
B log⁡e∣x+y∣−xy(x+y)2=0\log _e|x+y|-\dfrac{x y}{(x+y)^2}=0
C log⁡e∣x+y∣+2xy(x+y)2=0\log _e|x+y|+\dfrac{2 x y}{(x+y)^2}=0
D log⁡e∣x+y∣−2xy(x+y)2=0\log _e|x+y|-\dfrac{2 x y}{(x+y)^2}=0
Correct Answer
Option C
Solution
y=vxy = vx
v+xdvdx=−(1+3v23+v2)v + x{{dv} \over {dx}} = - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}}} \right)
xdvdx=−(1+3v23+v2+v)x{{dv} \over {dx}} = - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}} + v} \right)
dvdx=−((1+v)33+v2){{dv} \over {dx}} = - \left( {{{{{(1 + v)}^3}} \over {3 + {v^2}}}} \right)
⇒3+v2(1+v)3=−dxx\Rightarrow {{3 + {v^2}} \over {{{(1 + v)}^3}}} = {{ - dx} \over x}
⇒ln⁡∣v+1∣+2v(v+1)2=C−ln⁡∣x∣\Rightarrow \ln |v + 1| + {{2v} \over {{{(v + 1)}^2}}} = C - \ln |x|
x=1,v=0⇒C=0x = 1,v = 0 \Rightarrow C = 0
⇒ln⁡∣x+y∣−ln⁡∣x∣+2xy(x+y)2=−ln⁡∣x∣\Rightarrow \ln |x + y| - \ln |x| + {{2xy} \over {{{(x + y)}^2}}} = - \ln |x|
Q115
Let ∫0x1−(y′(t))2dt=∫0xy(t)dt,0≤x≤3,y≥0,y(0)=0\int\limits_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_0^x y(t) d t, 0 \leq x \leq 3, y \geq 0, y(0)=0. Then at x=2,y′′+y+1x=2, y^{\prime \prime}+y+1 is equal to
A 2\sqrt2
B 2
C 1/2
D 1
Correct Answer
Option D
Solution
∫0x1−(y′(t))2dt=∫0xy(t)dt\int\limits_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int\limits_0^x y(t) d t

Differentiating both side

1−(y′(x))2=y(x)(dydx)2+y2=1y′2+y2=12y′y′′+2yy′=0y′′+y=0\begin{aligned} & \sqrt{1-\left(y^{\prime}(x)\right)^2}=y(x) \\ & \left(\frac{d y}{d x}\right)^2+y^2=1 \\ & y^{\prime 2}+y^2=1 \\ & 2 y^{\prime} y^{\prime \prime}+2 y y^{\prime}=0 \\ & y^{\prime \prime}+y=0 \end{aligned}

∴\therefore

Q116
Let the solution curve y=y(x)y=y(x) of the differential equation dy dx−3x5tan⁡−1(x3)(1+x6)3/2y=2xexp⁡{x3−tan⁡−1x3(1+x6)} pass through the origin. Then y(1) is equal to :  \frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{3 x^{5} \tan ^{-1}\left(x^{3}\right)}{\left(1+x^{6}\right)^{3 / 2}} y=2 x \exp \left\{\frac{x^{3}-\tan ^{-1} x^{3}}{\sqrt{\left(1+x^{6}\right)}}\right\} \text{ pass through the origin. Then } y(1) \text{ is equal to : }
A exp⁡(1−π42)\exp \left(\dfrac{1-\pi}{4 \sqrt{2}}\right)
B exp⁡(4−π42)\exp \left(\dfrac{4-\pi}{4 \sqrt{2}}\right)
C exp⁡(4+π42)\exp \left(\dfrac{4+\pi}{4 \sqrt{2}}\right)
D exp⁡(π−442)\exp \left(\dfrac{\pi-4}{4 \sqrt{2}}\right)
Correct Answer
Option B
Solution
dydx−3x5tan⁡−1(x3)(1+x6)32y=2xexp⁡{x3−tan⁡−1x31+x6}{{dy} \over {dx}} - {{3{x^5}{{\tan }^{ - 1}}({x^3})} \over {{{(1 + {x^6})}^{{3 \over 2}}}}}y = 2x\exp \left\{ {{{{x^3} - {{\tan }^{ - 1}}{x^3}} \over {\sqrt {1 + {x^6}} }}} \right\}
IF=e−∫3x5tan⁡−1(x3)(1+x6)32dxIF = {e^{ - \int {{{3{x^5}{{\tan }^{ - 1}}({x^3})} \over {{{(1 + {x^6})}^{{3 \over 2}}}}}dx} }}

Let

tan⁡−1x3=t⇒3x21+x6dx=dt{\tan ^{ - 1}}{x^3} = t \Rightarrow {{3{x^2}} \over {1 + {x^6}}}dx = dt
⇒IF=e−∫tan⁡tsec⁡t.t dt=e−∫sin⁡t.tdt=etcos⁡t−sin⁡t\Rightarrow IF = {e^{ - \int {{{\tan t} \over {\sec t}}.t\,dt} }} = {e^{ - \int {\sin t.tdt} }} = {e^{t\cos t - \sin t}}
⇒IF=etan⁡−1(x3)1+x6−x31+x6\Rightarrow IF = {e^{{{{{\tan }^{ - 1}}({x^3})} \over {\sqrt {1 + {x^6}} }} - {{{x^3}} \over {\sqrt {1 + {x^6}} }}}}

∴\therefore Solution is

y . etan⁡−1x31+x6−x31+x6=∫2x dx+cy\,.\,{e^{{{{{\tan }^{ - 1}}{x^3}} \over {\sqrt {1 + {x^6}} }} - {{{x^3}} \over {\sqrt {1 + {x^6}} }}}} = \int {2x\,dx + c}
⇒y . etan⁡−1x3−x31+x6=x2+c\Rightarrow y\,.\,{e^{{{{{\tan }^{ - 1}}{x^3} - {x^3}} \over {\sqrt {1 + {x^6}} }}}} = {x^2} + c
y(0)=0⇒c=0y(0) = 0 \Rightarrow c = 0
x=1x = 1
y . eπ4−12=1y\,.\,{e^{{{{\pi \over 4} - 1} \over {\sqrt 2 }}}} = 1
⇒y=e1−π42\Rightarrow y = {e^{{{1 - {\pi \over 4}} \over {\sqrt 2 }}}}
⇒y=e4−π42\Rightarrow y = {e^{{{4 - \pi } \over {4\sqrt 2 }}}}
Q117
Let y=y(x)y=y(x) be the solution of the differential equation xlog⁡exdydx+y=x2log⁡ex,(x>1)x{\log _e}x{{dy} \over {dx}} + y = {x^2}{\log _e}x,(x > 1). If y(2)=2y(2) = 2, then y(e)y(e) is equal to
A 1+e22{{1 + {e^2}} \over 2}
B 1+e24{{1 + {e^2}} \over 4}
C 2+e22{{2 + {e^2}} \over 2}
D 4+e24{{4 + {e^2}} \over 4}
Correct Answer
Option D
Solution
xln⁡xdydx+y=x2ln⁡xx\ln x{{dy} \over {dx}} + y = {x^2}\ln x
dydx+1xln⁡x . y=x{{dy} \over {dx}} + {1 \over {x\ln x}}\,.\,y = x

If

=e∫1xln⁡xdx=e∫1tdt= {e^{\int {{1 \over {x\ln x}}dx} }} = {e^{\int {{1 \over t}dt} }}

, where

t=ln⁡xt = \ln x
=eln⁡t=t=ln⁡x= {e^{\ln t}} = t = \ln x
y.ln⁡x=∫xln⁡x=x22ln⁡x−∫x22 . 1xy.\ln x = \int {x\ln x = {{{x^2}} \over 2}\ln x - \int {{{{x^2}} \over 2}\,.\,{1 \over x}} }
yln⁡x=x22ln⁡x−x24+Cy\ln x = {{{x^2}} \over 2}\ln x - {{{x^2}} \over 4} + C

..... (i)

y(2)=2⇒C=1y(2) = 2 \Rightarrow C = 1

Putting

x=ex = e

in (i),

y=e24+1=4+e24y = {{{e^2}} \over 4} + 1 = {{4 + {e^2}} \over 4}
Q118
Let y=f(x)y=f(x) be the solution of the differential equation y(x+1)dx−x2dy=0,y(1)=ey(x+1)dx-x^2dy=0,y(1)=e. Then lim⁡x→0+f(x)\mathop {\lim }\limits_{x \to {0^ + }} f(x) is equal to
A e2{e^2}
B 0
C 1e2{1 \over {{e^2}}}
D 1e{1 \over e}
Correct Answer
Option B
Solution

Given,

y(x+1)dx−x2dy=0y(x + 1)dx - {x^2}dy = 0
⇒(x+1x2)dx=dyy\Rightarrow \left( {{{x + 1} \over {{x^2}}}} \right)dx = {{dy} \over y}
⇒1xdx+dxx2=dyy\Rightarrow {1 \over x}dx + {{dx} \over {{x^2}}} = {{dy} \over y}

Integrating both sides, we get

∫dxx+∫dxx2=∫dyy\int {{{dx} \over x} + \int {{{dx} \over {{x^2}}} = \int {{{dy} \over y}} } }
⇒ln⁡∣x∣−1x=ln⁡∣y∣+C\Rightarrow \ln |x| - {1 \over x} = \ln |y| + C

..... (1) Given

y(1)=ey(1) = e

∴\therefore

x=1x = 1

and

y=ey = e

Putting value of x and y in equation (1), we get

ln⁡∣1∣−11=ln⁡∣e∣+C\ln |1| - {1 \over 1} = \ln |e| + C
⇒0−1=1+C\Rightarrow 0 - 1 = 1 + C
⇒C=−2\Rightarrow C = - 2

∴\therefore Equation (1) becomes,

ln⁡∣x∣=−1x=ln⁡∣y∣−2\ln |x| = - {1 \over x} = \ln |y| - 2
⇒ln⁡∣y∣=ln⁡∣x∣−1x+2\Rightarrow \ln |y| = \ln |x| - {1 \over x} + 2
⇒y=eln⁡∣x∣ . e2−1x\Rightarrow y = {e^{\ln |x|}}\,.\,{e^{2 - {1 \over x}}}
⇒y=x . e2−1x\Rightarrow y = x\,.\,{e^{2 - {1 \over x}}}

Now,

lim⁡x→0+y\mathop {\lim }\limits_{x \to {0^ + }} y
=lim⁡x→0+(x . e2−1x)= \mathop {\lim }\limits_{x \to {0^ + }} \left( {x\,.\,{e^{2 - {1 \over x}}}} \right)
=0 . e−α= 0\,.\,{e^{ - \alpha }}
=0= 0
Q119
Let y=y(t)y=y(t) be a solution of the differential equation dydt+αy=γe−βt{{dy} \over {dt}} + \alpha y = \gamma {e^{ - \beta t}} where, α>0,β>0\alpha > 0,\beta > 0 and γ>0\gamma > 0. Then lim⁡t→∞y(t)\mathop {\lim }\limits_{t \to \infty } y(t)
A is 0
B is 1
C is −1-1
D does not exist
Correct Answer
Option A
Solution

dydt+αy=γe−βt\dfrac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

 I.F. =e∫αdt=eαt⇒y⋅eαt=γ∫e(α−β)tdt=γe(α−β)t(α−β)+C⇒y=γ(α−β)e−βt+Ce−αtlim⁡x→∞y(t)=lim⁡x→∞[γ(α−β)e−βt+Ce−αt]=0\begin{aligned} & \text{ I.F. }=e^{\int \alpha d t}=e^{\alpha t} \\\\ & \Rightarrow y \cdot e^{\alpha t}=\gamma \int e^{(\alpha-\beta) t} d t=\gamma \frac{e^{(\alpha-\beta) t}}{(\alpha-\beta)}+C \\\\ & \Rightarrow y=\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t} \\\\ & \lim _{x \rightarrow \infty} y(t)=\lim _{x \rightarrow \infty}\left[\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t}\right]=0 \end{aligned}
Q120
Let y=y(x)y = y(x) be the solution curve of the differential equation dydx=yx(1+xy2(1+log⁡ex)),x>0,y(1)=3{{dy} \over {dx}} = {y \over x}\left( {1 + x{y^2}(1 + {{\log }_e}x)} \right),x > 0,y(1) = 3. Then y2(x)9{{{y^2}(x)} \over 9} is equal to :
A x25−2x3(2+log⁡ex3){{{x^2}} \over {5 - 2{x^3}(2 + {{\log }_e}{x^3})}}
B x23x3(1+log⁡ex2)−2{{{x^2}} \over {3{x^3}(1 + {{\log }_e}{x^2}) - 2}}
C x27−3x3(2+log⁡ex2){{{x^2}} \over {7 - 3{x^3}(2 + {{\log }_e}{x^2})}}
D x22x3(2+log⁡ex3)−3{{{x^2}} \over {2{x^3}(2 + {{\log }_e}{x^3}) - 3}}
Correct Answer
Option A
Solution
dydx−yx=y3(1+log⁡ex)1y3dydx−1xy2=1+log⁡ex Let −1y2=t⇒2y3dydx=dtdx∴dtdx+2tx=2(1+log⁡ex) I.F. =e∫2xdx=x2−x2y2=23((1+log⁡ex)x3−x33)+Cy(1)=3y29=C5−2x3(2+log⁡ex3)\begin{aligned} & \frac{d y}{d x}-\frac{y}{x}=y^3\left(1+\log _e x\right) \\\\ & \frac{1}{y^3} \frac{d y}{d x}-\frac{1}{x y^2}=1+\log _e x \\\\ & \text{ Let }-\frac{1}{y^2}=t \Rightarrow \frac{2}{y^3} \frac{d y}{d x}=\frac{d t}{d x} \\\\ & \therefore \frac{d t}{d x}+\frac{2 t}{x}=2\left(1+\log _e x\right) \\\\ & \text{ I.F. }=e^{\int \frac{2}{x} d x}=x^2 \\\\ & \frac{-x^2}{y^2}=\frac{2}{3}\left(\left(1+\log _e x\right) x^3-\frac{x^3}{3}\right)+C \\\\ & y(1)=3 \\\\ & \frac{y^2}{9}=\frac{C}{5-2 x^3\left(2+\log _e x^3\right)} \end{aligned}
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