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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](http://meta.math.stackexchange.com/questions/934/where-is-the-latex-reference-please) 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!2011-03-05
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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...2011-03-05
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    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.2011-03-05

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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.