Is there a solution for y for ${{dy}\over dx} = axe^{by}$ I have come up with the equation in the form $${{dy}\over dx} = axe^{by}$$, where a and b are arbitrary real numbers, for a project I am working on. I want to be able to find its integral and differentiation if possible. Does anyone know of a possible solution for $y$ and/or ${d^2 y}\over {dx^2}$?

I think there are some scenarios to consider if $a$ and or $b$ is equal to zero.
Case 1: $a=0$
Then $$\frac{dy}{dx} = 0 \implies y = C$$
Case 2: $a \neq 0, b=0$.
$$\frac{dy}{dx} = ax \implies y = \frac{ax^2}{2}+C$$
Case 3: $a \neq 0 \neq b$.
Then $$\frac{dy}{dx} = axe^{by} \implies e^{-by}dy = axdx \\ \implies \int e^{-by}dy = \int axdx \\ \implies \frac{-1}{b}e^{-by} = \frac{ax^2}{2}+C \\ \implies e^{-by} = \frac{-ax^2}{2b}+\tilde{C} \\ \implies y = \frac{-1}{b}\ln\left(\frac{-ax^2}{2b}+\tilde{C} \right)$$
It would probably be wise to look at the solution of $y$ in case $3$ to figure out if there are other constraints on the  constants $a,b,\tilde{C}$. For example, if $a>0, b<0, \tilde{C}<0$ then you would be taking the natural log of a negative number.
