Question

For the given differential equation find the particular solution satisfying the given conditions.

$2xy+y^2-2x^2\frac{dy}{dx}=0,y=2whenx=1$

Answer 1

Given, $2x^2\frac{dy}{dx}=2xy+y^2\Rightarrow \frac{dy}{dx}=\frac{2xy+y^2}{2x^2}$ ...(i)
Thus, the given differential equation is homogeneous.
So, put $\frac{y}{x}=vi.e.,y=vx\Rightarrow \frac{dv}{dx}=v+x\frac{dv}{dx}$
Then, Eq. (i) becomes $v+x\frac{dv}{dx}=\frac{2x^{2}v+x^2{v}^2}{2x^2}\Rightarrow v+x\frac{dy}{dx}=\frac{x^2(2v+v^2)}{2x^2}\Rightarrow v+x\frac{dv}{dx}=v+\frac{1}{2}{v}^2\Rightarrow x\frac{dv}{dx}=\frac{1}{2}{v}^2\Rightarrow \frac{2}{v^2}dv=\frac{1}{x}dx$
On integrating both sides, we get
$2\int v^{-2}dv=\int \frac{dx}{x}\Rightarrow 2\left(\frac{v^{-2+1}}{-2+1}\right)=log|x|+C\Rightarrow 2\left(\frac{v^{-1}}{-1}\right)=log|x|+C\Rightarrow -\frac{2}{v}=log|x|+C\Rightarrow -\frac{2x}{y}=log|x|+C$ $(\because v=\frac{y}{x})...\left(ii\right)$
When x=1, then y=2, therefore $\frac{-2\times 1}{2}=log1+C\Rightarrow -1=0+C$
On, putting the value of C in Eq. (ii), we get
$-\frac{2x}{y}=log|x|-1\Rightarrow y=\frac{2x}{1-log|x|}$