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)

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

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- Jan 19th 2009, 11:58 AMmiepieO 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)

$\displaystyle y'=\sqrt{4+2y-1}$ - Jan 19th 2009, 12:18 PMskeeter
- Jan 19th 2009, 09:32 PMmiepie
I made a mistake, there is also an x on the right side, that's my greatest problem. This is the good one:

$\displaystyle

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

$

Thanks for helping! - Jan 20th 2009, 05:56 AMKrizalid
Put $\displaystyle {{u}^{2}}=4x+2y-1\implies 2uu'=4+2y'\implies uu'=2+y',$ and the ODE becomes $\displaystyle \frac{u}{u+2}\,du=dx$ which after some of integration it's $\displaystyle \sqrt{4x+2y-1}-2\ln \left( \sqrt{4x+2y-1}+2 \right)=x+k.$