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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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up vote 2 down vote accepted

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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Yeah, but then I'll have a problem with expressing $y$. –  Jozef Mar 17 '12 at 11:39
    
In the theory of differential equtions it is not always possible to express $y$ in terms of $x$, so it is not required and by convention relation between $x$ and $y$ that you have obtained is called solution of differential equation –  Norbert Mar 17 '12 at 12:02
    
Oh, I didn't know it. Thanks a lot! –  Jozef Mar 17 '12 at 12:12
    
@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 $y=f(x)$ –  Kirthi Raman Mar 17 '12 at 13:17
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