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 ${d^2 y}\over {dx^2}$?
 A: This is a separable differential equation, as is the case for any equation that can be put into the form $\frac{dy}{dx} = f(x)g(y)$ for some functions $f$ and $g.$ In your case, divide both sides by $e^{by}$ and multiply both sides by $dx,$ then integrate.
A: $$
\frac{d}{dx}\mathrm{e}^{-by} = -b\mathrm{e}^{-by}y' 
$$
Thus we can rewrite your equation as
$$
-\frac{1}{b}\left(\mathrm{e}^{-by}\right)' = ax
$$
A: separate the variables like $$ e^{-by}dy = axdx$$ on integrating, you get $$ -\dfrac{1}{b}e^{-by} = \dfrac{1}{2}ax^2 - \dfrac{C}{2b}$$ which can be solved for $y = \dfrac{1}{b}\ln\left( \dfrac{2}{C - ab x^2}\right)$
for $\dfrac{d^2y}{dx^2},$ differencing $$e^{-by}\dfrac{dy}{dx} = ax$$ gives you 
$$e^{-by}\dfrac{d^2y}{dx^2} -be^{-by}(\dfrac{dy}{dx})^2= a$$ on further simplification you get
$$\dfrac{d^2y}{dx^2} = axe^{by}(1 + a e^{by})$$
A: 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.
