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}} $