Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let the solution curve   of the differential equation <img alt="\math

Let the solution curve \mathrm{y=y(x)}  of the differential equation
\mathrm{\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] x \frac{d y}{d x}=x+\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] y}
pass through the points \mathrm{(1,0)} and \mathrm{(2 \alpha, \alpha), \alpha>0}. Then \mathrm{ \alpha} is equal to

Option: 1

\mathrm{\frac{1}{2} \exp \left(\frac{\pi}{6}+\sqrt{e}-1\right)}


Option: 2

\mathrm{\frac{1}{2} \exp \left(\frac{\pi}{3}+e-1\right)}


Option: 3

\mathrm{\exp \left(\frac{\pi}{6}+\sqrt{e}+1\right)}


Option: 4

\mathrm{2\exp \left(\frac{\pi}{3}+\sqrt{e}-1\right)}


Answers (1)

best_answer

\mathrm{Let \: \frac{y}{x}= t\: \Rightarrow \: \frac{dy}{dx}=t+\frac{x\, dt}{dx}. }


\mathrm{\left [ \frac{1}{\sqrt{1-t^{2}}} +e^{t}\right ]x\left ( t+\frac{x\, dt}{dx} \right )= x+\left [ \frac{1}{\sqrt{1-t^{2}}}+e^{t} \right ]t^{x}}

\mathrm{\Rightarrow\: \frac{t}{\sqrt{1-t^{2}}} +\frac{x}{\sqrt{1-t^{2}}}\, \frac{dt}{dx}+t\, e^{t}+x\, e^{t}\frac{dt}{dx}= 1+\frac{t}{\sqrt{1-t^{2}}}+t\, e^{t}}
\mathrm{\Rightarrow\: \frac{x}{\sqrt{1-t^{2}}}\, \frac{dt}{dx}+x\, e^{t}\frac{dt}{dx}= 1}

\mathrm{\Rightarrow\: \int \frac{dt}{\sqrt{1-t^{2}}}+\int e^{t}\, dt= \int \frac{dx}{x}}

\mathrm{\Rightarrow \sin^{-1}t+e^{t}= ln\, x+c}
\mathrm{\Rightarrow \sin^{-1}\left ( \frac{y}{x} \right )+e^{y/x}= ln\, x+c,\: \: at\: \: x= 1,y= 0}
\mathrm{\Rightarrow c= 1, \: \: Also\: \: at\: \: x= 2\, \alpha ,y= \alpha }
\mathrm{\Rightarrow \sin^{-1}\left ( \frac{\alpha }{2\, \alpha } \right ) +e^{\alpha /2\, \alpha }= ln\, 2\, \alpha +1}
\mathrm{\Rightarrow \frac{\pi}{6}+e^{1/2}= 1+ln\, 2\alpha \Rightarrow \, ln\, 2\alpha = \frac{\pi}{6}+\sqrt{e}-1}
\mathrm{\Rightarrow \alpha = \frac{1}{2}\, exp\left ( \frac{\pi}{6}+\sqrt{e}-1 \right )}

Option (A)



 

Posted by

Shailly goel

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question