Question

The solution of the differential equation
$(x^2+y^2)dx-2xydy=0$ is

• A. $\frac{y}{x^2+y^2}=c$
• B. $\frac{x^2+y^2}{x}=c$
• C. $\frac{y^2-x^2}{y}=c$
• D. $\frac{x^2-y^2}{x}=c$

Answer 1

Answer: (D)

Given $\left(x^2+y^2\right)dx=2xydy$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)$
Putting $y=tx$
$\Rightarrow \frac{dy}{dx}=t+x\frac{dt}{dx}$
$\therefore$ Given equation becomes,
$x\frac{dt}{dx}=\frac{1-t^2}{2t}$
$\Rightarrow \frac{2t}{1-t^2}dt=\frac{dx}{x}$
On integrating, we get
$log\left(1-t^2\right)=-logx+logc$
$\Rightarrow 1-t^2=\frac{c}{x}$
$\Rightarrow \frac{x^2-y^2}{x^2}=\frac{c}{x}$
$\Rightarrow \frac{x^2-y^2}{x}=c$