Question

For the differential equation $\frac{dy}{dx}=\frac{x{y}^2-x^{2}y}{x^3}$, the solution is $y-Cx=k{x}^{2}y$, here k is the const. of integration, find C?

Answer 1

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