Explanation

(b)
Given that, $x^{2}y - x^{3}\ \frac{dy}{dx} = y^{4}\cos x$
i.e., $x^{3}\frac{dy}{dx} - x^{2}y = - y^{4}\cos x$
on dividing by $–y^{4}x^{3}$, we get
$- \frac{1}{y^{4}}\ \ \frac{dy}{dx} + \frac{1}{y^{3}}.\frac{1}{x} = \frac{1}{x^{3}}\cos x$
Put $\frac{1}{y^{3}} = V$
$\Rightarrow$ $- \frac{1}{y^{4}}\ \frac{dy}{dx} = \frac{1}{3}\frac{dV}{dx}$
$\therefore$ $\frac{1}{3}\ \ \frac{dV}{dx} + \frac{1}{x}V = \frac{1}{x^{3}}\cos x$
$\Rightarrow$ $\frac{dV}{dx} + \frac{3}{x}\ V = \frac{3}{x^{3}}\cos x$
Which is linear in $V$.
$\therefore$ $IF = e^{\int_{}^{}{\frac{3}{x}\ dx}} = e^{3\log x} = x^{3}$
So, the solution is
$x^{3}V = \int_{}^{}x^{3}.\ \frac{3}{x^{3}}\cos x\ dx + c$
$= 3\sin x + c$
$\Rightarrow$ $\frac{x^{3}}{y^{3}} = 3\sin x + c$
Putting $x = 0,\ y = 1,$ we get $c = 0$
Hence, the solution is $x^{3} = 3y^{3}\sin x$