Question

For $ 'c' $ is the arbitrary constant, the solution of the differential equation $ (x^2 + 2y^2)dx - xy\,dy= 0 $ is

• A. $ x^{2} +y^{2} = x^{4} c^{2} $
• B. $ x^{2} -y^{2} = x^{2} c^{2} $
• C. $ x + y = x^{4} c^{2} $
• D. $ x^{2} +y^{2} = c^{2} $

Answer 1

The Correct Option is A

We have $\left(x^{2}+2y^{2}\right)dx-xy dy =0$
$\Rightarrow \frac{dy}{dx}=\frac{x^{2}+2y^{2}}{xy}=\frac{x}{y}+\frac{2y}{x} \ldots\left(i\right)$
put $y=vx$ so that $\frac{dy}{dx}=v+x\frac{dv}{dx}$
$\therefore \left(i\right)$ becomes, $v+x \frac{dv}{dx}=\frac{1}{v}+2v$
$\Rightarrow x \frac{dv}{dx}=\frac{1+2v^{2}-v^{2}}{v}$
$=\frac{1+v^{2}}{v}$
$\Rightarrow \frac{v}{1+v^{2}} dv $
$=\frac{dx}{x}$
Integrating both sides, we get
$\int \frac{v}{1+v^{2}} dv $
$=\int\frac{dx}{x}+C_{1}$
$\Rightarrow \frac{1}{2}log \left|1+v^{2}\right|$
$=log \left|x\right|+log c$
$\Rightarrow log \left|1+\frac{y^{2}}{x^{2}}\right|=2\,log\,cx $
$\Rightarrow \frac{x^{2}+y^{2}}{x^{2}}=c^{2}x^{2}$
$\Rightarrow x^{2}+y^{2}=c^{2}x^{4}$, is the required solution