Question

If $x^{2}y-x^3\frac{dy}{dx}=y^4\cos x$, then $x^{3}y$ is equal to

• A. $\sin x$
• B. $2\sin x+c$
• C. $-3\sin x+c$
• D. $3\cos x+c$

Answer 1

Answer: (C)

Given that
$x^{2}y-x^3\frac{dy}{dx}=y^4\cos x$
On dividing by $y^4$, we get
$\frac{x^2}{y^3}-\frac{x^3}{y^4}\frac{dy}{dx}=\cos x$
$\Rightarrow \frac{x^3}{y^4}\frac{dy}{dx}=\frac{x^2}{y^3}-\cos x$
$\Rightarrow \frac{1}{y^4}\frac{dy}{dx}-\frac{1}{x{y}^3}=-\frac{1}{x^3}\cos x$
Let $-\frac{1}{y^3}=t$
$\Rightarrow \frac{1}{y^4}\frac{dy}{dx}=\frac{1}{3}\cdot \frac{dt}{dx}$
$\Rightarrow \frac{1}{3}\cdot \frac{dt}{dx}+\frac{1t}{x}=\frac{1}{x^3}\cos x$
$\Rightarrow \frac{dt}{dx}+\frac{3}{x}t=\frac{3}{x^3}\cos x$
This is a linear differential equation in $t$,
on comparing with $\frac{dt}{dx}+Pt=Q$, we get
$P=\frac{3}{x},Q=\frac{3}{x^3}\cos x$
$\therefore$ I.F. $=e^{\int Pdx}=e^{\int \frac{3}{x}dx}=e^{\log x^3}=x^3$
$\therefore$ Complete solution is
$t{x}^3=3\int \frac{x^3}{x^3}\cos xdx+c_1$
$\Rightarrow t{x}^3=3\sin x+c_1$
$\Rightarrow -\frac{1}{y^3}{x}^3=3\sin x+c_1$
$\Rightarrow y^{-3}{x}^3=-3\sin x+c$