Question

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

$x^2\frac{dy}{dx}=x^2-2y^2+xy$

Answer 1

Given, $\frac{dy}{dx}=\frac{x^2-2y^2+xy}{x^2}$ ...(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) becomes $v+x\frac{dv}{dx}=\frac{x^2-2x^2{v}^2+x^{2}v}{x^2}\Rightarrow v+x\frac{dv}{dx}=1-2v^2+v\Rightarrow x\frac{dv}{dx}=1-2v^2\Rightarrow \frac{1}{1-2v^2}dv=\frac{1}{x}dx$
On integrating both sides, we get
$\int \frac{1}{1-2v^2}dv=\int \frac{dx}{x}\Rightarrow \frac{1}{2}\int \frac{1}{{\left(\frac{1}{\sqrt{2}}\right)}^2-v^2}dv=\int \frac{dx}{x}\Rightarrow \frac{1}{2}.\frac{1}{2.\frac{1}{\sqrt{1}}}log\left|\frac{\frac{1}{\sqrt{2}}+v}{\frac{1}{\sqrt{2}}-v}\right|=log|x|+C[\because \int \frac{dx}{a^2-x^2}=\frac{1}{2a}log\left|\frac{a+x}{a-x}\right|+C]\Rightarrow \frac{1}{2\sqrt{2}}log\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right|=log|x|+C\Rightarrow \frac{1}{2\sqrt{2}}log\left|\frac{1+\sqrt{\frac{y}{x}}}{1-\sqrt{2}\frac{y}{x}}\right|=log|x|+C$ $Putv=\frac{y}{x}$
$\Rightarrow \frac{1}{2\sqrt{2}}log\left|\frac{x+\sqrt{2}y}{x-\sqrt{2}y}\right|=log|x|+C$
This is the required solution of the given differential equation.