# Math Help - O D E question

1. ## O D E question

Hi there,

Can anybody show me how to solve an O D E with a square root in it?
I just don't get it

$y'=\sqrt{4+2y-1}$

2. Originally Posted by miepie
Hi there,

Can anybody show me how to solve an O D E with a square root in it?
I just don't get it

$y'=\sqrt{4+2y-1}$
$\frac{dy}{dx} = \sqrt{3 + 2y}$

$\frac{dy}{\sqrt{3+2y}} = dx$

$\frac{2}{\sqrt{3+2y}} \, dy = 2 \, dx$

integrate ...

3. I made a mistake, there is also an x on the right side, that's my greatest problem. This is the good one:

$
y'=\sqrt{4x+2y-1}
$

Thanks for helping!

4. Put ${{u}^{2}}=4x+2y-1\implies 2uu'=4+2y'\implies uu'=2+y',$ and the ODE becomes $\frac{u}{u+2}\,du=dx$ which after some of integration it's $\sqrt{4x+2y-1}-2\ln \left( \sqrt{4x+2y-1}+2 \right)=x+k.$