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.