JEE Main On 08 April 2018 Question 24
Question: Let $ y=y(x) $ be the solution of the differential equation $ \sin x\frac{dy}{dx}+y\cos x=4x,x\in (0,\pi ). $ If $ y( \frac{\pi }{2} )=0 $ , then $ y( \frac{\pi }{6} ) $ is equal to: [JEE Main Online 08-04-2018]
Options:
A) $ -\frac{8}{9}{{\pi }^{2}} $
B) $ -\frac{4}{9}{{\pi }^{2}} $
C) $ -\frac{4}{9\sqrt{3}}{{\pi }^{2}} $
D) $ \frac{-8}{9\sqrt{3}}{{\pi }^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{dy}{dx}+y\cot x=\frac{4x}{\sin x} $ Integrating factor
$ \text{=}{e^{\int _{{}}^{{}}{\cot xdx}}} $ $ ={e^{+In|\sin x|}} $ $ =\sin x $
$ (\because x\in (0,\pi )) $
$ \therefore y\cdot (\sin x)=\int _{{}}^{{}}{\frac{4x}{sinx}(\sin x)\cdot dx} $
$ y\cdot (\sin x)=2x^{2}+c $
$ y( \frac{\pi }{2} )=0\Rightarrow c=-\frac{{{\pi }^{2}}}{2} $ $ y( \frac{\pi }{6} ) $
$ \therefore y.( \sin \frac{\pi }{6} )=2{{( \frac{\pi }{6} )}^{2}}-\frac{{{\pi }^{2}}}{2} $
$ y=\frac{-8}{9}{{\pi }^{2}} $
