Question

Solve the following:
$x+y\frac{dy}{dx}=\sec (x^2+y^2)$

Answer 1

$x+y\frac{dy}{dx}=\sec (x^2+y^2)-\left(i\right)$

Putting $x^2+y^2=v$ and differentiating we get,

$2x+2y\frac{dy}{dx}=\frac{dv}{dx}\Rightarrow 2(x+y\frac{dy}{dx})=\frac{dy}{dx}x+y\frac{dy}{dx}=\frac{1}{2}\frac{dv}{dx}$

Putting this in $\left(i\right)$

$x+y\frac{dy}{dx}=\sec (x^2+y^2)\frac{1}{2}\frac{dv}{dx}=\sec v\Rightarrow \cos vdv=2dx$

(applying variable separable method)

Integrating both sides

$\int \cos vdv=2\int dx\Rightarrow \sin v=2x+C$

Putting $v=x^2+y^2$

$\sin (x^2+y^2)=2x+C$