question_answer

Let y = y(x) be the solution of the differential equation, \[{{\left( {{x}^{2}}+1 \right)}^{2}}\frac{dy}{dx}+2x\left( {{x}^{2}}+1 \right)y=1\] such that\[y(0)=0.\]If \[\sqrt{a}y(1)=\frac{\pi }{32},\]then the value of 'a' is: [JEE Main 8-4-2019 Morning]

A) \[\frac{1}{2}\]                        

B) \[\frac{1}{16}\]

C) \[\frac{1}{4}\]                        

D) 1

Correct Answer: B

Solution :

we have, \[{{\left. 2y.\frac{dy}{dx}=1\Rightarrow \frac{dy}{dx} \right]}_{P(2+{{t}^{2}},t)}}=\frac{1}{2t}=1\] \[\Rightarrow \]\[t=\frac{1}{2}\] \[\therefore \]    \[P\left( \frac{9}{4},\frac{1}{2} \right)\] So, shortest distance \[=\frac{\left| \frac{9}{4}-\frac{2}{4} \right|}{\sqrt{2}}=\frac{7}{4\sqrt{2}}\] \[\frac{dy}{dx}+\left( \frac{2x}{{{x}^{2}}+1} \right)y=\frac{1}{{{({{x}^{2}}+1)}^{2}}}\] (Linear differential equation) \[\therefore \]\[I.F.={{e}^{\ell n\left( {{x}^{2}}+1 \right)}}=({{x}^{2}}+1)\] So, general solution is \[y.({{x}^{2}}+1)=ta{{n}^{-1}}x+c\] As\[y(0)=0\Rightarrow c=0\] \[\therefore \]\[y(x)=\frac{{{\tan }^{-1}}x}{{{x}^{2}}+1}\] As,\[\sqrt{a}.y\,(1)=\frac{\pi }{32}\]\[\Rightarrow \]\[\sqrt{a}=\frac{1}{4}\Rightarrow a=\frac{1}{16}\]