Question

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

• A. $x^2+y^2=cy$
• B. $c(x^2-y^2)=x$
• C. $x^2-y^2=cy$
• D. $x^2+y^2=cx$ (here c is an arbitrary constant)

Answer 1

Answer: (B)

Given differential equation can be rewritten as
$\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$
Put $y=vx$
$\Rightarrow$ $\frac{dy}{dx}=v+x\frac{dv}{dx}$
Then, given differential equation becomes
$v+x\frac{dv}{dx}=\frac{x^2+v^2{x}^2}{2xvx}$
$\Rightarrow$ $x\frac{dv}{dx}=\frac{1+v^2}{2v}-v$
$\Rightarrow$ $x\frac{dv}{dx}=\frac{1-v^2}{2v}$
$\Rightarrow$ $\frac{2v}{1-v^2}dv=\frac{dx}{x}$
On integrating, we get $-\log (1-v^2)=\log x+\log c$
$\Rightarrow$ $\log {(1-v^2)}^{-1}=\log xc$
$\Rightarrow$ ${\left(1-\frac{y^2}{x^2}\right)}^{-1}=xc$
$\Rightarrow$ ${\left(\frac{x^2-y^2}{x^2}\right)}^{-1}=xc$
$\Rightarrow$ $\frac{x^2}{x^2-y^2}=xc$
$\Rightarrow$ $x=c(x^2-y^2)$