Question

Show that the given differential equation is homogeneous and then solve it.

$(x^2-y^2)dx+2xydy=0$

Answer 1

Given, $(x^2-y^2)dx+2xydy=0$
$\Rightarrow 2xydy=(y^2-x^2)dx\Rightarrow \frac{dy}{dx}=\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$ ...(i)
Here, the numerator and denominator both have polynomial of degree 2. So, the given differential equation is homogeneous.
So, put y=vx $\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx},$then Eq. (i) becaomes
$v+x\frac{dv}{dx}=\frac{v^2{x}^2-x^2}{2x^{2}v}\Rightarrow v+x\frac{dv}{dx}=\frac{v^1-1}{2v}\Rightarrow x\frac{dv}{dx}=\frac{v^2-1}{2v}-v\Rightarrow x\frac{dv}{dx}=\frac{-1-v^2}{2v}\Rightarrow \frac{2v}{v^2+1}dv=-\frac{dx}{x}$
On integrating both sides, we get
$\int \frac{2v}{v^2+1}dv+\int \frac{dx}{x}=0Let{v}^2+1=t\Rightarrow 2v=\frac{dt}{dv}\Rightarrow dv=\frac{dt}{2v}\therefore \int \frac{2v}{t}\times \frac{dt}{2v}+\int \frac{dx}{x}=0\Rightarrow log|t|+log|x|=logC\Rightarrow log|(v^2+1)x|=logC$ $(\because t=1+r^2))$
$\Rightarrow log\left[\left(\frac{y^2+x^2}{x^2}\right)x\right])=logC\Rightarrow \frac{y^2+x^2}{x}=C$ $(\because v=y/x)$
$\Rightarrow x^2+y^2=Cx$
This is the required solution of the given differential equation.