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

Question

Solve the differential equation

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

My try:

I used $x=y \tan z$

$$\frac{dx}{dy}=\tan z+\sec^2 z\frac{dz}{dy}$$

So we get:

$$\tan z+\sec^2 z\frac{dz}{dy}+\sin z=y$$

Any clue from here?

Answer 1

Hint:

Let $x=yu$ ,

Then $\dfrac{dx}{dy}=y\dfrac{du}{dy}+u$

$\therefore y\dfrac{du}{dy}+u+\dfrac{yu}{\sqrt{y^2u^2+y^2}}=y$

$y\dfrac{du}{dy}+u+\dfrac{u}{\sqrt{u^2+1}}=y$

$y\dfrac{du}{dy}=y-u-\dfrac{u}{\sqrt{u^2+1}}$

$\left(y-u-\dfrac{u}{\sqrt{u^2+1}}\right)\dfrac{dy}{du}=y$

This belongs to an Abel equation of the second kind.

Let $v=y-u-\dfrac{u}{\sqrt{u^2+1}}$ ,

Then $y=v+u+\dfrac{u}{\sqrt{u^2+1}}$

$\dfrac{dy}{du}=\dfrac{dv}{du}+1+\dfrac{1}{(u^2+1)^\frac{3}{2}}$

$\therefore v\left(\dfrac{dv}{du}+1+\dfrac{1}{(u^2+1)^\frac{3}{2}}\right)=v+u+\dfrac{u}{\sqrt{u^2+1}}$

$v\dfrac{dv}{du}+\left(1+\dfrac{1}{(u^2+1)^\frac{3}{2}}\right)v=v+u+\dfrac{u}{\sqrt{u^2+1}}$

$v\dfrac{dv}{du}=-\dfrac{v}{(u^2+1)^\frac{3}{2}}+u+\dfrac{u}{\sqrt{u^2+1}}$