# Solve first order differential equation.

Recently I have met such an equation:

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

First of all what type it has? I can not refer it to any I am aware of(I am a newbie so it is likely that I'm missing something). Trying to solve it gave no result therefore I googled and youtube revealed that I had to use change of variables $y = v(x)x$. However it looks like a cheating. Why use this particular replacement and is there any way to solve it without it?

• It is a perfect approach and is not cheating, let $y = v x \implies y' = v + v' x$, substitute and solve. – Moo Jun 20 '17 at 21:28
• Why do you consider it cheating? It is a homogeneous ODE, so the substitution $y=vx$ is obvious: $$\frac{dy}{dx}=\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}$$ It is in the form $\frac{dy}{dx}=F\left(\frac{y}{x}\right)$, as required. – projectilemotion Jun 20 '17 at 21:34
• One could call it a homogeneous ODE, and you will find such a classification on Wikipedia too. – Triatticus Jun 20 '17 at 21:36
• @projectilemotion Thanks! Till now I was not aware of two types of homogeneity: by argument and by right side. My book in some reason provides information only about homogeneity by right side. Or possibly I somehow just missed it. Going to read it more carefully. – Long Smith Jun 20 '17 at 21:40

$$x\frac { dy }{ dx } =y+\sqrt { x^{ 2 }+y^{ 2 } } \\ x\frac { dy }{ dx } =y+x\sqrt { 1+\frac { { y }^{ 2 } }{ { x }^{ 2 } } } \\ \frac { dy }{ dx } =\frac { y }{ x } +\sqrt { 1+\frac { { y }^{ 2 } }{ { x }^{ 2 } } }$$ and let $y=zx$ $$\frac { dy }{ dx } =x\frac { dz }{ dx } +z\\ x\frac { dz }{ dx } +z=z+\sqrt { 1+{ z }^{ 2 } } \\ x\frac { dz }{ dx } =\sqrt { 1+{ z }^{ 2 } } \\ \int { \frac { dz }{ \sqrt { 1+{ z }^{ 2 } } } } =\int { \frac { dx }{ x } } \\ \\ \\ \\ \\ \\ \\ \\ \\$$ $$z=\tan { t } \\ dz=\frac { dt }{ \cos ^{ 2 }{ t } } \\ \int { \frac { dt }{ \cos { t } } } =\ln { \left| x \right| +C } \\ \int { \frac { \cos { t } }{ \cos ^{ 2 }{ t } } dt } =\ln { \left| x \right| +C } \\ \int { \frac { d\sin { t } }{ 1-\sin ^{ 2 }{ t } } } =\ln { \left| x \right| +C } \\ u=\sin { t } \\ \int { \frac { du }{ 1-{ u }^{ 2 } } } =\ln { \left| x \right| +C } \\ \frac { 1 }{ 2 } \int { \left[ \frac { 1 }{ 1-u } +\frac { 1 }{ 1+u } \right] } =\ln { \left| x \right| +C } \\ \frac { 1 }{ 2 } \ln { \left| \frac { u+1 }{ u-1 } \right| } =\ln { \left| x \right| +C } \\ \frac { 1 }{ 2 } \ln { \left| \frac { \sin { \left( \arctan { \frac { y }{ x } } \right) +1 } }{ \sin { \left( \arctan { \frac { y }{ x } } \right) -1 } } \right| } =\ln { \left| x \right| +C } \\ \frac { 1 }{ 2 } \ln { \left| \frac { y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ y-\sqrt { { x }^{ 2 }+{ y }^{ 2 } } } \right| } =\ln { \left| x \right| +C } \\ \\ \\$$