How do you solve a 2nd order differential equation of the form $v = v' - v'' +C^t +D^{t+E}$ I've been working on an economic simulator for a game I've been making and in order to simulate the velocity of money, I created the differential equation of the form $v = v' -v'' + C^t + D^{t+E}$. However, after creating it, I realized that I didn't actually know how to solve it (the most advanced course in mathematics I have taken is Calculus BC), and neither Wolfram Alpha nor Sage seem to be able to use a method to either solve or approximate the solution. Thanks for any help in advance!
 A: Try a solution
$$v = aC^t + b D^{t+E}$$
where $a,b$ are constants. The motivation for this is that if $v_1,v_2$ solves 
$$v_1 = v_1'-v_1'' + C^t$$
$$v_2 = v_2'-v_2'' + D^{t+E}$$
then $v = v_1 + v_2$ solves the equation
$$v = v'-v'' + C^t + D^{t+E}$$
plus the fact that for any constant $Y$ we have $(Y^t)' = \log Y \cdot Y^t$. Plugging our guess above into the equation gives us
$$C^t(1 + \ldots) + D^{t+e}(1+\ldots) = 0$$
where $1+\ldots$ will be some constants. In order for this to hold for all $t$ we need both $1+\ldots = 0$. This gives you two equations for $a,b$ and the particular solution.
In order to get the most general solution note that for any $v_3$ that solves
$$v_3 = v_3' - v_3''$$
then $v= v_1 + v_2 + v_3$ will also be a solution to your equation.
I leave solving this to you (this is a homogeneous linear ODE and can be solved with the standard method of trying $v_3 = e^{\lambda t}$ and solving a quadratic equation for $\lambda$ to give you the two solutions, see for example this page for an explanation). If you do it correctly (and if I have done it correctly) you will get
$$v = \frac{C^t}{1 - \log C + \log^2 C} + \frac{D^{E+t}}{1 -  \log D  + \log^2D} + Fe^{t/2}\cos(\sqrt{3}t/2) + Ge^{t/2}\sin(\sqrt{3}t/2)$$
where $F,G$ are free constants determined by the initial conditions. For example if you want to set the initial conditions at $t=0$ you plug in $t=0$ above to get:
$$v(0) = \frac{1}{1 - \log C + \log^2 C} + \frac{D^{E}}{1 -  \log D  + \log^2D} + F$$
This fixes $F$ in terms of $v(0)$. Taking the derivative of $v$ and plugging in $t=0$ gives:
$$v'(0) = \frac{\log C}{1 - \log C + \log^2 C} + \frac{D^{E}\log D}{1 -  \log D  + \log^2D} + \frac{F}{2} + \frac{G\sqrt{3}}{2}$$
which fixes $G$ in term of $v'(0)$.
