O D E question

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January 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)

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

$<br /> y'=\sqrt{4x+2y-1}<br />$

Thanks for helping!
January 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 $\sqrt{4x+2y-1}-2\ln \left( \sqrt{4x+2y-1}+2 \right)=x+k.$

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