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Differential Equations

JEE Mathematics · 179 questions · Page 10 of 18 · Click an option or "Show Solution" to reveal answer

Q91
Let y=y(x)y = y(x) be the solution of the differential equation (x+1)y′−y=e3x(x+1)2(x + 1)y' - y = {e^{3x}}{(x + 1)^2}, with y(0)=13y(0) = {1 \over 3}. Then, the point x=−43x = - {4 \over 3} for the curve y=y(x)y = y(x) is :
A not a critical point
B a point of local minima
C a point of local maxima
D a point of inflection
Correct Answer
Option B
Solution
(x+1)dydx−y=e3x(x+1)2(x + 1){{dy} \over {dx}} - y = {e^{3x}}{(x + 1)^2}
dydx−yx+1=e3x(x+1){{dy} \over {dx}} - {y \over {x + 1}} = {e^{3x}}(x + 1)

If

e−∫1x+1x=e−log⁡(x+1)=1x+1{e^{ - \int {{1 \over {x + 1}}x} }} = {e^{ - \log (x + 1)}} = {1 \over {x + 1}}

∴\therefore

y(1x+1)=∫e3x(x+1)x+1dxy\left( {{1 \over {x + 1}}} \right) = \int {{{{e^{3x}}(x + 1)} \over {x + 1}}dx}
yx+1=∫e3xdx{y \over {x + 1}} = \int {{e^{3x}}dx}
yx+1=e3x3+c{y \over {x + 1}} = {{{e^{3x}}} \over 3} + c

∵\because

y(0)=13y(0) = {1 \over 3}
13=13+c{1 \over 3} = {1 \over 3} + c

∴\therefore

c=0c = 0

So :

y=e3x3(x+1)y = {{{e^{3x}}} \over 3}(x + 1)
y′=e3x(x+1)+e3x3=e3x(x+43)y' = {e^{3x}}(x + 1) + {{{e^{3x}}} \over 3} = {e^{3x}}\left( {x + {4 \over 3}} \right)
y′′=3e3x(x+43)+e3x=e3x(3x+5)y'' = 3{e^{3x}}\left( {x + {4 \over 3}} \right) + {e^{3x}} = {e^{3x}}(3x + 5)
y′=0y' = 0

at

x=−43x = {{ - 4} \over 3}

&

y′′=e−4(1)>0y'' = {e^{ - 4}}(1) > 0

at

x=−43x = {{ - 4} \over 3}
⇒x=−43\Rightarrow x = {{ - 4} \over 3}

is point of local minima

Q92
If the solution curve y=y(x)y = y(x) of the differential equation y2dx+(x2−xy+y2)dy=0{y^2}dx + ({x^2} - xy + {y^2})dy = 0, which passes through the point (1, 1) and intersects the line y=3xy = \sqrt 3 x at the point (α,3α)(\alpha ,\sqrt 3 \alpha ), then value of log⁡e(3α){\log _e}(\sqrt 3 \alpha ) is equal to :
A π3{\pi \over 3}
B π2{\pi \over 2}
C π12{\pi \over 12}
D π6{\pi \over 6}
Correct Answer
Option C
Solution
dydx=y2xy−x2−y2{{dy} \over {dx}} = {{{y^2}} \over {xy - {x^2} - {y^2}}}

Put

y=vxy = vx

we get

v+xdvdx=v2v−1−v2v + x{{dv} \over {dx}} = {{{v^2}} \over {v - 1 - {v^2}}}
⇒xdvdx=v2−v2+v+v3v−1−v2\Rightarrow x{{dv} \over {dx}} = {{{v^2} - {v^2} + v + {v^3}} \over {v - 1 - {v^2}}}
⇒∫v−1−v2v(1+v2)dv=∫dxx\Rightarrow \int {{{v - 1 - {v^2}} \over {v(1 + {v^2})}}dv = \int {{{dx} \over x}} }
tan⁡−1(yx)−ln⁡(yx)=ln⁡x+c{\tan ^{ - 1}}\left( {{y \over x}} \right) - \ln \left( {{y \over x}} \right) = \ln x + c

As it passes through (1, 1)

c=π4c = {\pi \over 4}
⇒tan⁡−1(yx)ln⁡(yx)=ln⁡x+π4\Rightarrow {\tan ^{ - 1}}\left( {{y \over x}} \right)\ln \left( {{y \over x}} \right) = \ln x + {\pi \over 4}

Put

y=3xy = \sqrt 3 x

we get

⇒π3−ln⁡3=ln⁡x+π4\Rightarrow {\pi \over 3} - \ln \sqrt 3 = \ln x + {\pi \over 4}
⇒ln⁡x=π12−ln⁡3=ln⁡α\Rightarrow \ln x = {\pi \over {12}} - \ln \sqrt 3 = \ln \alpha

∴\therefore

ln⁡(3α)=ln⁡3+ln⁡α\ln \left( {\sqrt 3 \alpha } \right) = \ln \sqrt 3 + \ln \alpha
=ln⁡3+π12−ln⁡3=π12= \ln \sqrt 3 + {\pi \over {12}} - \ln \sqrt 3 = {\pi \over {12}}
Q93
If x = x(y) is the solution of the differential equation ydxdy=2x+y3(y+1)ey, x(1)=0y{{dx} \over {dy}} = 2x + {y^3}(y + 1){e^y},\,x(1) = 0; then x(e) is equal to :
A e3(ee−1){e^3}({e^e} - 1)
B ee(e3−1){e^e}({e^3} - 1)
C e2(ee+1){e^2}({e^e} + 1)
D ee(e2−1){e^e}({e^2} - 1)
Correct Answer
Option A
Solution

dxdy−2xy=y2(y+1)ey\dfrac{d x}{d y}-\dfrac{2 x}{y}=y^{2}(y+1) e^{y}

 If =e∫−2ydy=e−2ln⁡y=1y2\text{ If }=e^{\int-\frac{2}{y} d y}=e^{-2 \ln y}=\frac{1}{y^{2}}

Solution is given by

x⋅1y2=∫y2(y+1)ey⋅1y2dy⇒xy2=∫(y+1)eydy⇒xy2=yey+c\begin{aligned} &x \cdot \frac{1}{y^{2}}=\int y^{2}(y+1) e^{y} \cdot \frac{1}{y^{2}} d y \\\\ \Rightarrow & \frac{x}{y^{2}}=\int(y+1) e^{y} d y \\\\ \Rightarrow & \frac{x}{y^{2}}=y e^{y}+c \end{aligned}

⇒x=y2(yey+c)\Rightarrow x=y^{2}\left(y e^{y}+c\right) at, y=1,x=0y=1, x=0 ⇒0=1(1⋅e1+c)⇒c=−e\Rightarrow 0=1\left(1 \cdot e^{1}+c\right) \Rightarrow c=-e at y=ey=e, x=e2(e.ee−e)x=e^{2}\left(e . e^{e}-e\right)

Q94
Let dydx=ax−by+abx+cy+a, a,b,c∈R{{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}},\,a,b,c \in R, represents a circle with center (α\alpha, β\beta). Then, α\alpha + 2β\beta is equal to :
A −-1
B 0
C 1
D 2
Correct Answer
Option C
Solution

Given,

dydx=ax−by+abx+cy+a{{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}
⇒bxdy+cydy+ady=axdx−bydx+adx\Rightarrow bxdy + cydy + ady = axdx - bydx + adx
⇒bxdy+bydx+(cy+a)dy=(ax+a)dx\Rightarrow bxdy + bydx + (cy + a)dy = (ax + a)dx
⇒b(xdy+ydx)+(cy+a)dy=(ax+a)dx\Rightarrow b(xdy + ydx) + (cy + a)dy = (ax + a)dx
⇒bd(xy)+(cy+a)dy=(ax+a)dx\Rightarrow bd(xy) + (cy + a)dy = (ax + a)dx

Integrating both sides, we get

⇒b∫d(xy)+∫(cy+a)dy=∫(ax+a)dx\Rightarrow b\int {d(xy) + \int {(cy + a)dy = \int {(ax + a)dx} } }
⇒b . xy+c . y22+ay=ax22+ax+k\Rightarrow b\,.\,xy + c\,.\,{{{y^2}} \over 2} + ay = {{a{x^2}} \over 2} + ax + k
⇒ax22−cy22−bxy+ax−ay+k=0\Rightarrow {{a{x^2}} \over 2} - {{c{y^2}} \over 2} - bxy + ax - ay + k = 0

For equation of circle, Coefficient of x2 = Coefficient of y2 ∴\therefore

a2=−c2{a \over 2} = - {c \over 2}
⇒a=−c\Rightarrow a = - c

And coefficient of

xy=0xy = 0

∴\therefore

−b=0- b = 0
⇒b=0\Rightarrow b = 0

∴\therefore Circle equation becomes,

ax22+ay22+ax−ay+k=0{{a{x^2}} \over 2} + {{a{y^2}} \over 2} + ax - ay + k = 0
⇒x2+y2+2x−2y+2ka=0\Rightarrow {x^2} + {y^2} + 2x - 2y + {{2k} \over a} = 0

∴\therefore Center

=(−g,−f)=(−1,1)=(α,β)= ( - g, - f) = ( - 1,1) = (\alpha ,\beta )

∴\therefore

α=−1\alpha = - 1

and

β=1\beta = 1

∴\therefore

α+2β=−1+2×(1)=1\alpha + 2\beta = - 1 + 2 \times (1) = 1
Q95
The slope of the tangent to a curve C:y=y(x)C: y=y(x) at any point (x,y)(x, y) on it is 2e2x−6e−x+92+9e−2x\dfrac{2 \mathrm{e}^{2 x}-6 \mathrm{e}^{-x}+9}{2+9 \mathrm{e}^{-2 x}}. If CC passes through the points (0,12+π22)\left(0, \dfrac{1}{2}+\dfrac{\pi}{2 \sqrt{2}}\right) and (α,12e2α)\left(\alpha, \dfrac{1}{2} \mathrm{e}^{2 \alpha}\right), then eα\mathrm{e}^{\alpha} is equal to :
A 3+23−2\dfrac{3+\sqrt{2}}{3-\sqrt{2}}
B 32(3+23−2)\dfrac{3}{\sqrt{2}}\left(\dfrac{3+\sqrt{2}}{3-\sqrt{2}}\right)
C 12(2+12−1) \dfrac{1}{\sqrt{2}}\left(\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\right)
D 2+12−1\dfrac{\sqrt{2}+1}{\sqrt{2}-1}
Correct Answer
Option B
Solution

dydx=2e2x−6e−x+92+9e−2x=e2x−6e−x2+9e−2x\dfrac{d y}{d x}=\dfrac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}=e^{2 x}-\dfrac{6 e^{-x}}{2+9 e^{-2 x}}

∫dy=∫e2xdx−3∫1+(3e−x2)2⏟put e−x=tdx=e2x2+3∫dt1+(3t2)2=e2x2+2tan⁡−13t2+C\begin{aligned} &\int d y=\int e^{2 x} d x-3 \int \underbrace{1+\left(\frac{3 e^{-x}}{\sqrt{2}}\right)^{2}}_{\text{put } e^{-x}=t} d x \\\\ &=\frac{e^{2 x}}{2}+3 \int \frac{d t}{1+\left(\frac{3 t}{\sqrt{2}}\right)^{2}} \\\\ &=\frac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1} \frac{3 t}{\sqrt{2}}+C \end{aligned}

y=e2x2+2tan⁡−1(3e−x2)+Cy=\dfrac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1}\left(\dfrac{3 e^{-x}}{\sqrt{2}}\right)+C It is given that the curve passes through

(0,12+π22)\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)
12+π22=12+2tan⁡−1(32)+C\begin{aligned} & \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}=\frac{1}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)+C \end{aligned}

⇒C=π22−2tan⁡−1(32)\Rightarrow \quad C=\dfrac{\pi}{2 \sqrt{2}}-\sqrt{2} \tan ^{-1}\left(\dfrac{3}{\sqrt{2}}\right) Now if (α,12e2α)\left(\alpha, \dfrac{1}{2} e^{2 \alpha}\right) satisfies the curve, then

12e2α=e2α2+2tan⁡−1(3e−α2)+π22−2tan⁡−1(32)\frac{1}{2} e^{2 \alpha}=\frac{e^{2 \alpha}}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3 e^{-\alpha}}{\sqrt{2}}\right)+\frac{\pi}{2 \sqrt{2}}-\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)

tan⁡−1(32)−tan⁡−1(3e−α2)=π22×12=π4\tan ^{-1}\left(\dfrac{3}{\sqrt{2}}\right)-\tan ^{-1}\left(\dfrac{3 e^{-\alpha}}{\sqrt{2}}\right)=\dfrac{\pi}{2 \sqrt{2}} \times \dfrac{1}{\sqrt{2}}=\dfrac{\pi}{4}

32−3e−α21+92e−α=1\frac{\frac{3}{\sqrt{2}}-\frac{3 e^{-\alpha}}{\sqrt{2}}}{1+\frac{9}{2} e^{-\alpha}}=1
32eα−32=eα+92\frac{3}{\sqrt{2}} e^{\alpha}-\frac{3}{\sqrt{2}}=e^{\alpha}+\frac{9}{2}
eα=92+3232−1=32(3+23−2)e^{\alpha}=\frac{\frac{9}{2}+\frac{3}{\sqrt{2}}}{\frac{3}{\sqrt{2}}-1}=\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)
Q96
The general solution of the differential equation (x−y2)dx+y(5x+y2)dy=0\left(x-y^{2}\right) \mathrm{d} x+y\left(5 x+y^{2}\right) \mathrm{d} y=0 is :
A (y2+x)4=C∣(y2+2x)3∣\left(y^{2}+x\right)^{4}=\mathrm{C}\left|\left(y^{2}+2 x\right)^{3}\right|
B (y2+2x)4=C∣(y2+x)3∣\left(y^{2}+2 x\right)^{4}=C\left|\left(y^{2}+x\right)^{3}\right|
C ∣(y2+x)3∣=C(2y2+x)4\left|\left(y^{2}+x\right)^{3}\right|=\mathrm{C}\left(2 y^{2}+x\right)^{4}
D ∣(y2+2x)3∣=C(2y2+x)4\left|\left(y^{2}+2 x\right)^{3}\right|=C\left(2 y^{2}+x\right)^{4}
Correct Answer
Option A
Solution

(x−y2)dx+y(5x+y2)dy=0\left(x-y^{2}\right) d x+y\left(5 x+y^{2}\right) d y=0

ydydx=y2−x5x+y2y \frac{d y}{d x}=\frac{y^{2}-x}{5 x+y^{2}}

Let y2=ty^{2}=t

12⋅dtdx=t−x5x+t\frac{1}{2} \cdot \frac{d t}{d x}=\frac{t-x}{5 x+t}

Now substitute, t=vxt=v x

dtdx=v+xdvdx12{v+xdvdx}=v−15+vxdvdx=2v−25+v−v=−3v−v2−25+v∫5+vv2+3v+2dv=∫−dxx∫4v+1dv−∫3v+2dv=−∫dxx4ln⁡∣v+1∣−3ln⁡∣v+2∣=−ln⁡x+ln⁡C∣(v+1)4(v+2)3∣=cx∣(y2x+1)4(y2x+2)3∣=cx∣(y2+x)4∣=C∣(y2+2x)3∣\begin{aligned} & \frac{d t}{d x}=v+x \frac{d v}{d x} \\\\ & \frac{1}{2}\left\{v+x \frac{d v}{d x}\right\}=\frac{v-1}{5+v} \\\\ & x \frac{d v}{d x}=\frac{2 v-2}{5+v}-v=\frac{-3 v-v^{2}-2}{5+v} \\\\ & \int \frac{5+v}{v^{2}+3 v+2} d v=\int-\frac{d x}{x} \\\\ & \int \frac{4}{v+1} d v-\int \frac{3}{v+2} d v=-\int \frac{d x}{x} \\\\ & 4 \ln |v+1|-3 \ln |v+2|=-\ln x+\ln C \\\\ & \left|\frac{(v+1)^{4}}{(v+2)^{3}}\right|=\frac{c}{x} \\\\ & \left| \frac{\left(\frac{y^{2}}{x}+1\right)^{4}}{\left(\frac{y^{2}}{x}+2\right)^{3}}\right|=\frac{c}{x} \\\\ & \left|\left(y^{2}+x\right)^{4}\right|=C\left|\left(y^{2}+2 x\right)^{3}\right| \end{aligned}
Q97
If $${{dy} \over {dx}} + 2y\tan x = \sin x,\,0
A 18{1 \over 8}
B 34{3 \over 4}
C 14{1 \over 4}
D 38{3 \over 8}
Correct Answer
Option A
Solution
dydx+2ytan⁡x=sin⁡x{{dy} \over {dx}} + 2y\tan x = \sin x

which is a first order linear differential equation. Integrating factor (I. F.)

=e∫2tan⁡x dx= {e^{\int {2\tan x\,dx} }}
=e2ln⁡∣sec⁡x∣=sec⁡2x= {e^{2\ln |\sec x|}} = {\sec ^2}x

Solution of differential equation can be written as

y . sec⁡2x=∫sin⁡x . sec⁡2x dx=∫sec⁡ x . tan⁡x dxy\,.\,{\sec ^2}x = \int {\sin x\,.\,{{\sec }^2}x\,dx = \int {\sec \,x\,.\,\tan x\,dx} }
y sec⁡2x=sec⁡x+Cy~{\sec ^2}x = \sec x + C
y(π3)=0,0=sec⁡π3+C⇒    C=−2y\left( {{\pi \over 3}} \right) = 0,0 = \sec {\pi \over 3} + C \Rightarrow \,\,\,\,C = - 2
y=sec⁡x−2sec⁡2x=cos⁡x−2cos⁡2xy = {{\sec x - 2} \over {{{\sec }^2}x}} = \cos x - 2{\cos ^2}x
=18−2(cos⁡x−14)2= {1 \over 8} - 2{\left( {\cos x - {1 \over 4}} \right)^2}
ymax⁡=18{y_{\max }} = {1 \over 8}
Q98
Let the solution curve y=f(x)y=f(x) of the differential equation dydx+xyx2−1=x4+2x1−x2 \dfrac{d y}{d x}+\dfrac{x y}{x^{2}-1}=\dfrac{x^{4}+2 x}{\sqrt{1-x^{2}}}, x∈(−1,1)x\in(-1,1) pass through the origin. Then ∫−3232f(x)dx\int\limits_{-\dfrac{\sqrt{3}}{2}}^{\dfrac{\sqrt{3}}{2}} f(x) d x is equal to
A π3−14\dfrac{\pi}{3}-\dfrac{1}{4}
B π3−34\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4}
C π6−34\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4}
D π6−32\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{2}
Correct Answer
Option B
Solution
dydx+xyx2−1=x4+2x1−x2{{dy} \over {dx}} + {{xy} \over {{x^2} - 1}} = {{{x^4} + 2x} \over {\sqrt {1 - {x^2}} }}

which is first order linear differential equation. Integrating factor

(I.F.)=e∫xx2−1dx(I.F.) = {e^{\int {{x \over {{x^2} - 1}}dx} }}
=e12ln⁡∣x2−1∣=∣x2−1∣= {e^{{1 \over 2}\ln |{x^2} - 1|}} = \sqrt {|{x^2} - 1|}
=1−x2= \sqrt {1 - {x^2}}

∵\because

x∈(−1,1)x \in ( - 1,1)

Solution of differential equation

y1−x2=∫(x4+2x)dx=x55+x2+cy\sqrt {1 - {x^2}} = \int {({x^4} + 2x)dx = {{{x^5}} \over 5} + {x^2} + c}

Curve is passing through origin,

c=0c = 0
y=x5+5x251−x2y = {{{x^5} + 5{x^2}} \over {5\sqrt {1 - {x^2}} }}
∫−3232x5+5x251−x2dx=0+2∫032x21−x2dx\int\limits_{{{ - \sqrt 3 } \over 2}}^{{{\sqrt 3 } \over 2}} {{{{x^5} + 5{x^2}} \over {5\sqrt {1 - {x^2}} }}dx = 0 + 2\int\limits_0^{{{\sqrt 3 } \over 2}} {{{{x^2}} \over {\sqrt {1 - {x^2}} }}dx} }

put

x=sin⁡θx = \sin \theta
dx=cos⁡θ dθdx = \cos \theta \,d\theta
I=2∫0π3sin⁡2θ . cos⁡θdθcos⁡θI = 2\int\limits_0^{{\pi \over 3}} {{{{{\sin }^2}\theta \,.\,\cos \theta d\theta } \over {\cos \theta }}}
=∫0π3(1−cos⁡2θ)dθ= \int\limits_0^{{\pi \over 3}} {(1 - \cos 2\theta )d\theta }
=(θ−sin⁡2θ2)∣0π3= \left. {\left( {\theta - {{\sin 2\theta } \over 2}} \right)} \right|_0^{{\pi \over 3}}
=π3−34= {\pi \over 3} - {{\sqrt 3 } \over 4}
Q99
Let y=y1(x)y=y_{1}(x) and y=y2(x)y=y_{2}(x) be two distinct solutions of the differential equation dydx=x+y\dfrac{d y}{d x}=x+y, with y1(0)=0y_{1}(0)=0 and y2(0)=1y_{2}(0)=1 respectively. Then, the number of points of intersection of y=y1(x)y=y_{1}(x) and y=y2(x)y=y_{2}(x) is
A 0
B 1
C 2
D 3
Correct Answer
Option A
Solution
dydx=x+y{{dy} \over {dx}} = x + y

Let

x+y=tx + y = t
1+dydx=dtdx1 + {{dy} \over {dx}} = {{dt} \over {dx}}
dtdx−1=t⇒∫dtt+1=∫dx{{dt} \over {dx}} - 1 = t \Rightarrow \int {{{dt} \over {t + 1}} = \int {dx} }
ln⁡∣t+1∣=x+C′\ln |t + 1| = x + C'
∣t+1∣=Cex|t + 1| = C{e^x}
∣x+y+1∣=Cex|x + y + 1| = C{e^x}

For

y1(x), y1(0)=0⇒C=1{y_1}(x),\,{y_1}(0) = 0 \Rightarrow C = 1

For

y2(x), y2(0)=1⇒C=2{y_2}(x),\,{y_2}(0) = 1 \Rightarrow C = 2
y1(x){y_1}(x)

is given by

∣x+y+1∣=ex|x + y + 1| = {e^x}
y2(x){y_2}(x)

is given by

∣x+y+1∣=2ex|x + y + 1| = 2{e^x}

At point of intersection

ex=2ex{e^x} = 2{e^x}

No solution So, there is no point of intersection of

y1(x){y_1}(x)

and

y2(x){y_2}(x)

.

Q100
Let the solution curve of the differential equation x dy=(x2+y2+y)dx,x>0x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0, intersect the line x=1x=1 at y=0y=0 and the line x=2x=2 at y=αy=\alpha. Then the value of α\alpha is :
A 12\dfrac{1}{2}
B 32\dfrac{3}{2}
C −-32\dfrac{3}{2}
D 52\dfrac{5}{2}
Correct Answer
Option B
Solution
xdy−ydxx2+y2=dx{{xdy - ydx} \over {\sqrt {{x^2} + {y^2}} }} = dx
⇒dydx=x2+y2x+yx\Rightarrow {{dy} \over {dx}} = {{\sqrt {{x^2} + {y^2}} } \over x} + {y \over x}
⇒dydx=1+y2x2+yx\Rightarrow {{dy} \over {dx}} = \sqrt {1 + {{{y^2}} \over {{x^2}}}} + {y \over x}

Let

yx=v{y \over x} = v
⇒v+xdvdx=1+v2+v\Rightarrow v + x{{dv} \over {dx}} = \sqrt {1 + {v^2}} + v
⇒dv1+v2=dxx\Rightarrow {{dv} \over {\sqrt {1 + {v^2}} }} = {{dx} \over x}

OR

ln⁡(v+1+v2)=ln⁡x+C\ln \left( {v + \sqrt {1 + {v^2}} } \right) = \ln x + C

at

x=1, y=0x = 1,\,y = 0
⇒C=0\Rightarrow C = 0
yx+1+y2x2=x{y \over x} + \sqrt {1 + {{{y^2}} \over {{x^2}}}} = x

At

x=2x = 2

,

y2+1+y24=2{y \over 2} + \sqrt {1 + {{{y^2}} \over 4}} = 2
⇒1+y24=4+y24−2y\Rightarrow 1 + {{{y^2}} \over 4} = 4 + {{{y^2}} \over 4} - 2y

OR

y=32y = {3 \over 2}
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