Question

Solve the differential equation $\frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}{x}$

Answer 1

$y=x\tan t$ ........ $\left(1\right)$

$dy=x{\sec }^{2}tdt+\tan tdx$

Given equation can be written as

$xdy-ydx-\sqrt{x^2+y^2}dx=0$

Substituting $y$ and $dy$ in above equation we get

$x(\tan tdx+x{\sec }^{2}tdt)-(x\tan t+x\sec t)dx=0$

$x^2{\sec }^{2}tdt=x\sec tdx$

$x\sec tdt=dx$

$dx-x\sec tdt=0$

$\frac{dx}{x}-\sec tdt=0$

$ln|x|-ln|\sec t+\tan t|=lnc$

$\frac{x}{\sec t+\tan t}=c$

Rationalizing we get

$\frac{x(\sec t-\tan t)}{{\sec }^{2}t-{\tan }^{2}t}=c$

Since, ${\sec }^{2}t-{\tan }^{2}t=1$

$\Rightarrow x\sec t=\sqrt{x^2(1+{\tan }^{2}t)}$

$=\sqrt{x^2+(x\tan t{)}^2}$

$=\sqrt{x^2+y^2}$ ...... $\left(2\right)$

Hence, $x\sec t-x\tan t=c$

From (1) and (2)

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

which is the solution to given differential equation.