Question

Solve the differential equation $xdy-ydx=(x^2+y^2)dx$.

• A. $y=x\tan (x+c)$
• B. $y=x\tan \left(x\right)+c$
• C. $y=x\tan (-x+c)$
• D. $y=x\tan (-x)+c$

Answer 1

The correct option is A $y=x\tan (x+c)$

$xdy-ydx=(x^2+y^2)dx\Rightarrow x\frac{dy}{dx}-y=x^2+y^2$
Substitute $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\therefore x(x\frac{dv}{dx}+v)-xv=x^2+x^2{v}^2\Rightarrow \frac{dv}{dx}=v^2+1\Rightarrow \frac{1}{v^2+1}\frac{dv}{dx}=1$
Integrating both sides
$\int \frac{1}{v^2+1}\frac{dv}{dx}dx=\int dx\Rightarrow {\tan }^{-1}v=x+c\Rightarrow v=\tan (x+c)\Rightarrow y=x\tan (x+c)$