# Solving a system of forced, coupled, differential equations

Given a system of forced, coupled, differential equations $$\ddot{x}(t) + \dot{x}(t) + \omega_{z}^{2} x(t) = F_{0} \cos(\omega_{0} t) y(t)$$ $$\ddot{y}(t) + \omega_{y}^{2} y(t) = F_{0} \cos(\omega_{0} t) x(t)$$ which I have solved it with pen and paper.

I'd like to see if Mathematica agrees with me using DSolve. If I follow the same algebraic steps I used to solve the system by hand, but using Mathematica to verify the steps, I get the same result as my pen and paper approach (but that doesn't rule out my method was wrong).

When solving the system by hand I chose solutions $$x(t) = A e^{-i \omega_{x} t} X(t) \quad {\text{and}} \quad y(t) = B e^{-i \omega_{y} t} Y(t)$$ where I make the assumption $$\ddot{X}(t) = \ddot{Y}(t) = 0$$.

My Mathematica approach is

x[t] = A Exp[-I wx t] X[t];
y[t] = B Exp[-I wy t] Y[t];

X''[t] = 0;
Y''[t] = 0;

xSystem = D[x[t], {t, 2}] + D[x[t], t] + wz^2 x[t] == F0 Cos[w0 t] y[t]
ySystem = D[y[t], {t, 2}] + wy^2 y[t] == F0 Cos[w0 t] z[t]

DSolve[
{
xSystem , ySystem ,
X == X0, X' == vX0, Y == Y0, Y' == vY0
},  {X[t], Y[t]}, t
]


This returns the error:

For some branches of the general solution, the given boundary conditions lead to an empty solution


What am I missing? I am also open to other alternatives, DSolve is simply the approach i first reached for.

• What do you think will happen to your ode's when you add X''[t] = 0; Y''[t] = 0; ? Try xSystem /. X''[t] -> 0; ySystem /. Y''[t] -> 0 and you will see they become first order odes's now in X and Y. Basically you deleted the second order derivatives from your odes. I have no idea why you did that, but If you remove X''[t] = 0; Y''[t] = 0; then it will solve them (with some internal warnings from Solve. Jan 10 at 11:45
• If you want to set this condition X''=Y''=0 then first solve the ode's for X and Y as second order odes', then take second derivatives of the solutions found, and only then set up this equation. i.e. do it after solving. not before solving. Jan 10 at 11:54
• Okay removing the X''[t] = 0; Y''[t] = 0;' did indeed improve things, albeit with a strange integral identity in the solution. Could you show me what you mean in your second comment? Jan 10 at 12:32

Could you show me what you mean in your second comment?

This is little too long for comment. But basically solve the system, then set up your equation you wanted. I admit I do not understand why you are doing this. i.e. what is the point of $$X''(t)=0,Y''(t)=0$$ but you know better than me about this.

ClearAll["Globals*"]
x[t_] := A Exp[-I wx t] X[t];
y[t_] := B Exp[-I wy t] Y[t];
xSystem = D[x[t], {t, 2}] + D[x[t], t] + wz^2 x[t] == F0 Cos[w0 t] y[t]
ySystem = D[y[t], {t, 2}] + wy^2 y[t] == F0 Cos[w0 t] z[t]
ode = {xSystem, ySystem}
ic = {X == X0, X' == vX0, Y == Y0, Y' == vY0};
{solX, solY} = DSolveValue[{ode, ic}, {X[t], Y[t]}, t];


eq1 = D[solX, {t, 2}] == 0;