Mathematics - Differential Equation Question with Solution | TestHub

Given:

$f\left(x\right)\sin 2x+\sin x-\left(1+{\cos }^{2}x\right)f^{\text{'}}\left(x\right)=0$

Now, let $f\left(x\right)=y$ we get,

$\Rightarrow y\sin 2x+\sin x-\left(1+{\cos }^{2}x\right)\frac{dy}{dx}=0$

$\Rightarrow \frac{y\sin 2x}{\left(1+{\cos }^{2}x\right)}+\frac{\sin x}{\left(1+{\cos }^{2}x\right)}-\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}-\left(\frac{\sin 2x}{1+{\cos }^{2}x}\right)y=\frac{\sin x}{1+{\cos }^{2}x}$

$\Rightarrow \mathrm{I}.\mathrm{F}.=e^{\int -\left(\frac{\sin 2x}{1+{\cos }^{2}x}\right)dx}$

$\Rightarrow \mathrm{I}.\mathrm{F}.=e^{\log \left(1+{\cos }^{2}x\right)}$

$\Rightarrow \mathrm{I}.\mathrm{F}.=1+{\cos }^{2}x$

So, solution of the equation is given by,

$y\left(1+{\cos }^{2}x\right)=\int \frac{\sin x}{\left(1+{\cos }^{2}x\right)}\left(1+{\cos }^{2}x\right)dx$

$\Rightarrow y\left(1+{\cos }^{2}x\right)=\int \sin xdx$

$\Rightarrow y\left(1+{\cos }^{2}x\right)=-\cos x+C$

It is given that, $f\left(0\right)=0$.

$\Rightarrow 0\left(1+1\right)=-\cos 0+C$

$\Rightarrow C=1$

$\Rightarrow y\left(1+{\cos }^{2}x\right)=1-\cos x$

$\Rightarrow y=\frac{1-\cos x}{1+{\cos }^{2}x}$

$\Rightarrow y\left(\frac{\pi }{2}\right)=\frac{1-\cos \frac{\pi }{2}}{1+{\cos }^2\frac{\pi }{2}}$

$\Rightarrow y\left(\frac{\pi }{2}\right)=1$