Question

Evaluate:
$\frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}{x}$

Answer 1

$\frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}{x}$

$x.\frac{dy}{dx}=y+\sqrt{x^2+y^2}$

so Let $y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

$x[v+x\frac{dv}{dx}]=vx+\sqrt{x^2+v^2{x}^2}$

$=vx+x\sqrt{1+v^2}$

$x[v+x.\frac{dv}{dx}]=x[v+\sqrt{1+v^2}]$

$v+x\frac{dv}{dx}=v+\sqrt{1+v^2}$

$\int \frac{dv}{\sqrt{1+v^2}}=\int \frac{dx}{x}+\log c$

$=\log x+\log nc$

$=\log \left(cx\right)$

Let $v=\tan \theta$

$dv={\sec }^2\theta .d\theta$

$\int \frac{{\sec }^2\theta .d\theta }{\sqrt{1+{\tan }^2\theta }}=\int \sec \theta .d\theta$

$\log (\sec \theta +\tan \theta )$

$=\log \left(cx\right)$

$\sec \theta +\tan \theta =cx$

or $\tan \theta +\sqrt{1+{\tan }^2\theta }=cx$

$v+\sqrt{1+v^2}=cx$

$vx+x\sqrt{1+v^2}=c{x}^2$

$y+\sqrt{x^2+v^2{x}^2}=c{x}^2$

$y+\sqrt{x^2+y^2}=c{x}^2$