Question

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

• A. $x^2+y^2=xC$
• B. $x^2-y^2=xC$
• C. $x^2+y^2=C$
• D. $x^2-y^2=C$

Answer 1

Answer: (B)

$x^2-y^2=xC$

Given differential equation is
$(x^2+y^2)-2xy\frac{dy}{dx}=0$
which is homogeneous.
($\because$ degree of each term is same i.e., $2$)
It can be rewritten as
$\frac{dy}{dx}=\frac{x^2+y^2}{2xy}=\frac{1}{2}[\frac{x}{y}+\frac{y}{x}]$
Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$, so that the differential equation becomes
$v+x\frac{dv}{dx}=\frac{1}{2}(\frac{1}{v}+v)\Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{2v}-v$
$\Rightarrow x\frac{dv}{dx}=\frac{1-v^2}{2v}$
$\Rightarrow \int \frac{2v}{1-v^2}dv=\int \frac{1}{x}dx$
$\Rightarrow -\log (1-v^2)=\log \left|x\right|-\log \left|C\right|$
$\Rightarrow x(1-v^2)=C$
$\Rightarrow \frac{x^2-y^2}{x}=C$ $(\because v=\frac{y}{x})$
Hence, $x^2-y^2=xC$ is the required solution.