Question

The solution of the differential equation
$(x^2 +y^2)dx - 2xy\, dy = 0$ 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 $\left(x^{2}+y^{2}\right)dx=2xy\,dy$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)$
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
$log\left(1-t^{2}\right)=-log\,x+log\,c$
$\Rightarrow 1-t^{2}=\frac{c}{x}$
$\Rightarrow \frac{x^{2}-y^{2}}{x^{2}}=\frac{c}{x}$
$\Rightarrow \frac{x^{2}-y^{2}}{x}=c$