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I'd love your help with finding the function $y$ from the following differential equation:

y'=\sqrt{5x+2y-3}.

I tried to use $z=5x+2-3$, so z'=5+2y' , and y'=\frac{z'}{2}-2.5

and from the equation \frac{z'}{2}-2.5=y'=\sqrt z, and then z'=2 \sqrt z+5, so $\frac{dz}{2 \sqrt z +5} = dx$, but using integration here is difficult and won't lead me to $y$.

I tries to use substitution in other ways like $z=\sqrt{5x+2y-3}$, or $z^2= 5x+2y-3$ but then again I got stuck in the middle.. Any suggestion?

Thanks a lot

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Integration here is simple $ \int\frac{dz}{2\sqrt{z}+5}=\{t=\sqrt{z}\}=\int\frac{2t dt}{2t+5}=\int\left(1-\frac{5}{2t+5}\right)dt=t-\frac{5}{2}\int\frac{d(2t+5)}{2t+5}= $ $ t-\frac{5}{2}\ln(2t+5)+C=\sqrt{z}-\frac{5}{2}\ln(2\sqrt{z}+5)+C $

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    @Jozef the last expression can then be converted to $x$ terms like $\sqrt{5x-1}-\frac{5}{2}\ln(2\sqrt{5x-1}+5)+C$, so you are getting the final result in the form $y=f(x)$2012-03-17