homogenous linear equation

Question

I have an equation that I can't solve. I think it's homogenous.

$$ y' =\frac{y^2 + x \sqrt{4x^2 + y^2}}{xy} $$

Answer 1

Why do you think, it is homogeneous? For the homogeneous differential equations, you should use the change of variation $ y=zx.$ By this, you will have $$ (zxdz+z^{2}dx)-(z^{2}+\sqrt{4+z^{2}})dx=0 .$$ As you you see that it is completely easy. I'm sure with some effort you will solve lots of problems. I am a new user like you!

Answer 2

Yes..its homogenious..

$$ y' =\frac{y^2 + x \sqrt{4x^2 + y^2}}{xy} $$

For this, their is well known method, running by substituting, $$y=vx\implies y'=v+xv'$$

So, it will became as,

\begin{align} v+xv' &=\frac{v^2x^2 + x \sqrt{4x^2 + v^2x^2}}{xvx}\\ &=\frac{v^2 + \sqrt{4 + v^2}}{v} \end{align}

Now, we can write as, $$x v'=\frac{v^2 + \sqrt{4 + v^2}}{v}-v$$ which gives us, $$x v'=\frac{ \sqrt{4 + v^2}}{v}$$

So, $$ \frac v{ \sqrt{4 + v^2}}dv=\frac 1 x dx$$

Can you complete this??