Question

For the differential equation $x^{2}y-x^2\frac{dy}{dx}=y^4\cos x$, the solution is $\frac{1}{y^3}{e}^{3x}=\int \frac{k\cos x{e}^{3x}}{x^2}dx+c$, find $k=$

Answer 1

$x^{2}y-x^2\frac{dy}{dx}=y^4\cos x\Rightarrow -\frac{1}{y^4}\frac{dy}{dx}+\frac{1}{y^3}=\frac{\cos x}{x^2}$
Put $\frac{1}{y^3}=v\Rightarrow -\frac{3}{y^4}dy=dv$
$\therefore \frac{dv}{dx}+3v=\frac{\cos x}{x^2}$ ...(1)
Here $P=3\Rightarrow \int Pdx=\int 3dx=3x$
$\therefore I.F.=e^{3x}$
Multiplying (1) by $I.F.$ we get
$e^{3x}\frac{dv}{dx}+3e^{3x}v=\frac{3\cos x{e}^{3x}}{x^2}$
Integrating both sides we get
$v{e}^{3x}=\int \frac{3\cos x{e}^{3x}}{x^2}dx+c\Rightarrow \frac{1}{y^3}{e}^{3x}=\int \frac{k\cos x{e}^{3x}}{x^2}dx+c$
$\therefore k=3$