Question

Solve the differential equatiion $(x^2+y^2)dx=2xydy$.

Answer 1

$(x^2+y^2)dx=2xydy$

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

Let $y=vx$

$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

Now the equation becomes-

$v+x\frac{dv}{dx}=\frac{x^2+v^2{x}^2}{2v{x}^2}$

$\Rightarrow v+x\frac{dv}{dx}=\frac{1+v^2}{2v}$

$\Rightarrow x\frac{dv}{dx}=\frac{1+v^2-2v^2}{2v}$

$\Rightarrow \frac{dx}{x}=\frac{2vdv}{1-v^2}$

Integrating both sides, we have

$\int \frac{dx}{x}=\int \frac{2vdv}{1-v^2}$

$\log x=-\log (1-v^2)+\log c$

$\log x+\log (1-v^2)=\log c$

$\log \left(x(1-v^2)\right)=\log c$

$x(1-v^2)=C$

$x(1-v^2)=C$

$x(1-\frac{y^2}{x^2})=C$ where $y=vx\Rightarrow v=\frac{y}{x}$

$\Rightarrow \frac{x^2-y^2}{x}=C$ where $C$ is the constant of integration.

$\Rightarrow x^2-y^2=Cx$ is the solution of the given differential equation.