Question

Solve the differential equation: $\frac{dy}{dx}-x{\sin }^{2}x=\frac{1}{x\log x}$

Answer 1

$\frac{dy}{dx}-x{\sin }^{2}x=\frac{\log x}{x}$

$\Rightarrow \frac{dy}{dx}.dx=\frac{\log x}{x}.dx+x{\sin }^{2}x.dx$

Integrate on both sides, we get

$\Rightarrow \int dy=\int \frac{\log x}{x}.dx+\int x{\sin }^{2}x.dx$

$\Rightarrow y=I_1+I_2$ ........$\left(1\right)$

where $I_1=\int \frac{\log x}{x}.dx$

Let $t=\log x\Rightarrow dt=\frac{1}{x}dx$

So,$I_1=\int t.dt$

$=\frac{t^2}{2}+c_1$

$=\frac{{\left(\log x\right)}^2}{2}+c_1$ where $t=\log x$

And, $I_2=\int x{\sin }^{2}x.dx$

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

$=\frac{1}{2}\int x(1-\cos 2x).dx$

$=\frac{1}{2}\int x.dx-\frac{1}{2}\int x\cos 2x.dx$

$=\frac{1}{2}\frac{x^2}{2}-\frac{1}{2}\int x\cos 2x.dx$

consider $\int x\cos 2x.dx$

$u=x\Rightarrow du=dx$

$dv=\cos 2xdx\Rightarrow v=\frac{\sin 2x}{2}$

$\therefore \int x\cos 2x.dx=\frac{x\sin 2x}{2}-\int \frac{\sin 2x}{2}dx$

$=\frac{x\sin 2x}{2}-\frac{1}{2}\int \sin 2xdx$

$=\frac{x\sin 2x}{2}+\frac{1}{2}\frac{\cos 2x}{2}$

$\therefore I_2=\frac{1}{2}\frac{x^2}{2}-\frac{1}{2}(\frac{x\sin 2x}{2}+\frac{1}{2}\frac{\cos 2x}{2})$

$=\frac{x^2}{4}-\frac{x\sin 2x}{4}-\frac{\cos 2x}{8}+c_2$

$\therefore I=I_1+I_2$

$I=\frac{{\left(\log x\right)}^2}{2}+c_1+\frac{x^2}{4}-\frac{x\sin 2x}{4}-\frac{\cos 2x}{8}+c_2$

$I=\frac{{\left(\log x\right)}^2}{2}+\frac{x^2}{4}-\frac{x\sin 2x}{4}-\frac{\cos 2x}{8}+c$ where $c=c_1+c_2$