Solution of differential equation  $x^{2}y-x^3\frac{dy}{dx}=y^4\cos x$  is

$$\begin{gathered} \frac{1}{y^4}\frac{dy}{dx}+(-\frac{1}{x})\frac{1}{y^3}=\frac{-\cos x}{x^3}\mathrm{.....}\left(1\right) \\ Put \frac{1}{y^3}=\vartheta \Rightarrow \frac{1}{y^4}\frac{dy}{dx}=\frac{-1}{3}\frac{d\vartheta }{dx} \\ \therefore \left(1\right)\Rightarrow \frac{d\vartheta }{dx}+\frac{3}{x}\vartheta =3\cos x \end{gathered}$$

   Which is linear is  $\vartheta$ 
IF   $=e^{3\int \frac{1}{x}dx}=e^{3\log x}=x^3$
Solution is  $\vartheta x^3=3\int \cos xdx+c$
$x^3{y}^{-3}=3\sin x+c$