Question

The solution of $x^2+y^2-2xy\frac{dy}{dx}=0$ is

• A. $x^2-y^2=cx$
• B. $x^2+y^2=cx$
• C. $2(x^2-y^2)=cx$
• D. None of these

Answer 1

The correct option is A $x^2-y^2=cx$

$\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$ is the given equation,

$⟹\frac{dy}{dx}=\frac{x}{2y}+\frac{y}{2x}$

Substituting $\frac{y}{x}=v$

$⟹\frac{dy}{dx}=v+x\frac{dv}{dx}$

Substituting this in the main equation gives,

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

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

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

$⟹\ln \left(\frac{1}{1-v^2}\right)=\ln cx$

Where $c$ is the integration constant,

replacing back $v=\frac{y}{x}$ gives and solving gives,

$\frac{x^2}{x^2-y^2}=cx$

$\therefore x^2-y^2=cx$ where $c$ is a constant.