# O D E question

• Jan 19th 2009, 11:58 AM
miepie
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 (Doh)

$y'=\sqrt{4+2y-1}$
• Jan 19th 2009, 12:18 PM
skeeter
Quote:

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 (Doh)

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

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

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

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

integrate ...
• Jan 19th 2009, 09:32 PM
miepie
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!
• Jan 20th 2009, 05:56 AM
Krizalid
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 $\sqrt{4x+2y-1}-2\ln \left( \sqrt{4x+2y-1}+2 \right)=x+k.$