# 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!

• can you say something about the variables? – Dr. Sonnhard Graubner Nov 22 '14 at 23:18
• i think you can only plot a solution if you have values for the parameters and the constants – Dr. Sonnhard Graubner Nov 22 '14 at 23:37
• WolframAlpha solves this without difficulty. The open source program Maxima also solves it without difficulty. Sage tries to call Maxima but chokes due to the poor interface. There are many purely numerical solvers that can deal with the equation as well. – Mark McClure Nov 28 '14 at 15:05

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 $v'(0)$.