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$$ y'= \log(ax+by+c)~ ;\qquad a,b,c \in R$$

So is this how I am supoosed to ask questions here?

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Hi panush: welcome to math.SE. I've "texified" your mathematics input. Please consult this discussion on Meta on how to do it yourself in the future. Etiquette wise, it is generally expected that you give a bit more information than just an equation. Is this for school work? What do you want to find? Things like that. And generally just try to be polite. Cheers! –  Willie Wong Mar 5 '11 at 12:59
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This is how you would ask questions to Wolfram alpha. In this forum there are people answering which usually enjoy a bit more interaction... –  Fabian Mar 5 '11 at 13:07
    
I'm sorry about how I asked the question. I've only read the faq, so I didn't know how to ask properly. Thanks for your help. –  panush Mar 5 '11 at 19:58

1 Answer 1

up vote 3 down vote accepted

You can "solve" this equation by separation of variables. First introduce a new function $z(x)= a x + b y(x) + c$. In terms of this function the differential equation reads $$ z'(x) = b \log z(x) +a$$ and is separable (as it does not depend explicitly on $x$). An (implicit) solution reads $$ \int_{z_0}^z dz' \frac{1}{b \log z' +a} = x- x_0$$ with the initial condition $z(x_0) =z_0$ which corresponds to $y(x_0) = (z_0 -a x_0 -c)/b$ in terms of the old variables. The integral can be evaluated and gives (it is rather ugly looking) $$ \frac{e^{-a/b} \text{Ei}\left( \frac{a}{b} + \log z \right)}{b} = x-x_0 + \frac{e^{-a/b} \text{Ei}\left( \frac{a}{b} + \log z_0 \right)}{b},$$ with Ei the exponential integral. Being able to solve explicitly for $z(x)$ or $y(x)$ seems to be unlikely.

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