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.