Thank you for reporting, we will resolve it shortly
Q.
If $x^{2} y-x^{3} \frac{d y}{d x}=y^{4} \cos x$, then $x^{3} y$ is equal to
EAMCETEAMCET 2005
Solution:
Given that
$x^{2} y-x^{3} \frac{d y}{d x}=y^{4} \cos x$
On dividing by $y^{4}$, we get
$\frac{x^{2}}{y^{3}}-\frac{x^{3}}{y^{4}} \frac{d y}{d x}=\cos x $
$\Rightarrow \frac{x^{3}}{y^{4}} \frac{d y}{d x}=\frac{x^{2}}{y^{3}}-\cos x $
$\Rightarrow \frac{1}{y^{4}} \frac{d y}{d x}-\frac{1}{x y^{3}}=-\frac{1}{x^{3}} \cos x $
Let $-\frac{1}{y^{3}}=t $
$\Rightarrow \frac{1}{y^{4}} \frac{d y}{d x}=\frac{1}{3} \cdot \frac{d t}{d x} $
$\Rightarrow \frac{1}{3} \cdot \frac{d t}{d x}+\frac{1 t}{x}=\frac{1}{x^{3}} \cos x $
$\Rightarrow \frac{d t}{d x}+\frac{3}{x} t=\frac{3}{x^{3}} \cos x$
This is a linear differential equation in $t$,
on comparing with $\frac{d t}{d x}+P t=Q$, we get
$P=\frac{3}{x}, Q=\frac{3}{x^{3}} \cos x$
$\therefore $ I.F. $=e^{\int P d x}=e^{\int \frac{3}{x} d x}=e^{\log x^{3}}=x^{3}$
$\therefore $ Complete solution is
$t x^{3}=3 \int \frac{x^{3}}{x^{3}} \cos\, x \,d x+c_{1} $
$\Rightarrow t x^{3}=3 \sin x+c_{1} $
$\Rightarrow -\frac{1}{y^{3}} x^{3}=3 \sin x+c_{1} $
$\Rightarrow y^{-3} x^{3}=-3 \sin x+c$