Question

Show that the given differential equation is homogeneous and solve:$x^2\frac{dy}{dx}=x^2-2y^2+xy$

Answer 1

The differential equation is:
$x^2\frac{dy}{dx}=x^2-2y^2+xy$
$\frac{dy}{dx}=\frac{x^2-2y^2+xy}{x^2}...(1)$
Let $F(x,y)=\frac{x^2-2y^2+xy}{x^2}$
$∴F(λx,λy)=\frac{(λx)^2-2(λy)^2+(λx)(λy)}{(λx)^2}=\frac{x^2-2y^2+xy}{x^2}=λ.F(x,y)$
Therefore,the given differential equation is a homogenous equation.
To solve it,we make the substitution as:
$y=vx$
$⇒\frac{dy}{dx}=v+x\frac{dv}{dx}$
Substituting the values of y and $\frac{dy}{dx}$ in equation(1),we get:
$v+x\frac{dv}{dx}=\frac{x^2-2(vx)^2+x.(vx)}{x^2}$
$⇒v+x\frac{dv}{dx}=1-2v^2+v$
$⇒x\frac{dv}{dx}=1-2v^2$
$⇒\frac{dv}{1-2v^2}=\frac{dx}{x}$
Integrating both sides,we get:
$\implies \int\frac{1}{1-2v^2}dv=\int\frac{1}{x}dx$
$\implies \int\frac{1}{(1)^2-(\sqrt{2v})^2}dv=\int\frac{1}{x}dx$
$\implies \frac{1}{2\sqrt2}log\bigg|\frac{1+\sqrt2\frac{y}{x}}{1-\sqrt2\frac{y}{x}}\bigg|=log|x|+c$
$\implies \frac{1}{2\sqrt2}log\bigg|\frac{x+\sqrt2y}{1-\sqrt2y}\bigg|=log|x|+c$
This is the required solution for the given differential equation.