Mathematics - Differential Equation Question with Solution | TestHub

Given, $\left({\sin }^{2}2x\right)\frac{dy}{dx}+\left(8{\sin }^{2}2x+2\sin 4x\right)y=$

$2e^{-4x}\left(2\sin 2x+\cos 2x\right)$ 

Now rewriting the given differential equation we get,

$\Rightarrow \frac{dy}{dx}+\left(8+4\cot 2x\right)y=\frac{2e^{-4x}}{{\sin }^{2}2x}\left(2\sin x+\cos 2x\right)$

Which is a linear differential equation.

Now calculating  $IF={\mathrm{e}}^{\int \left(8+4\cot 2\mathrm{x}\right)\mathrm{dx}}={\mathrm{e}}^{8x+2\ln \left(\sin 2x\right)}$

$=e^{8x}·{\sin }^{2}2x$

Now solution of differential equation is given by,

$y\times IF=\int \frac{2e^{-4x}\left(2\sin 2x+\cos 2x\right)}{{\sin }^{2}2x}\times IFdx$

$\Rightarrow y\left(e^{8x}·{\sin }^{2}2x\right)=\int 2e^{4x}\left(2\sin 2x+\cos 2x\right)dx$

$\Rightarrow y\left(e^{8x}·{\sin }^{2}2x\right)=e^{4x}·\sin 2x+C$

Now given $y\left(\frac{\pi }{4}\right)=e^{-\pi }\Rightarrow C=0$

So, the equation of curve is given by $y=\frac{e^{-4x}}{\sin 2x}$

So, the value of $y\left(\frac{\pi }{6}\right)=\frac{e^{-4\frac{\pi }{6}}}{\sin \left(2·\frac{\pi }{6}\right)}=\frac{2}{\sqrt{3}}{e}^{-\frac{2\pi }{3}}$