y'= \log(ax+by+c)~ ;\qquad a,b,c \in R
So is this how I am supoosed to ask questions here?
y'= \log(ax+by+c)~ ;\qquad a,b,c \in R
So is this how I am supoosed to ask questions here?
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.