Question

Show that the given differential equation is homogeneous and solve:$(x^2-y^2)dx+2xy\,dy=0$

Answer 1

The differential equation is:
$(x^2-y^2)dx+2xy\,dy=0$
$⇒\frac{dy}{dx}=-\frac{(x^2-y^2)}{2xy}...(1)$
Let F(x,y)=-(x2-y2)/2xy.
$∴F(λx,λy)=[\frac{(λx)^2-(λy)^2}{2(λx)(λy)}]=-\frac{(x^2-y^2)}{2xy}=λ.F(x,y)$
Therefore,the given differential equation is a homogenous equation.
To solve it,we make the substitution as:
$y=vx$
$⇒\frac{d}{dx}(y)=\frac{d}{dx}(vx)$
$⇒\frac{dy}{dx}=v+x\frac{dv}{dx}$
Substituting the values of y and $\frac{dy}{dx}$ in equation(1),we get:
$v+x\frac{dv}{dx}=-[\frac{x^2-(vx)^2}{2x.(vx)}]$
$v+x\frac{dv}{dx}=\frac{v^2-1}{2v}$
$⇒x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=\frac{v^2-1-2v^2}{2v}$
$⇒x\frac{dv}{dx}=-\frac{(1+v^2)}{2v}$
$⇒\frac{2v}{1+v^2}dv=-\frac{dx}{x}$
Integrating both sides,we get:
$log(1+v^2)=-logx+logC=log\frac{C}{x}$
$⇒1+v^2=\frac{C}{x}$
$⇒[1+\frac{y^2}{x^2}]=\frac{C}{x}$
$⇒x^2+y^2=Cx$
This is the required solution of the given differential equation.