So the question is to solve
$ y'' + 9y = e^t cos(2t) $
with initial conditions $ y(0) = 0 $ and $ y'(0) = 1 $
This ODE demonstrates dynamical behaviour of particle with unit mass attached to a string (spring constant of k = 9) subject to an external driving force of $F(t) = e^t cos(2t)$ . released from origin with initial velocity one.
ATTEMPT: to solve second order non-homogeneous:
- put in form r
$(r^2 + 9) = 0$
But cant solve for r unless its complex (i've never seen before)
$ ( r- 3i)(r+3i) = 0 $
where $r = 3i, -3i $
so if im on the right track: $y(x) = C_1 e^{3ix} + C_2e^{-3ix} $
particular solution: $ y_p = Ae^tcos(2t) $ ?
Can somebody please help? any hints or answers is appreciated. thanks in advance
Your general solution to the homogeneous equation is correct, but... When dealing with real-valued functions, we often prefer solutions without complex numbers in them. And that's possible even the characteristic equation (with only real coefficients) has complex roots, which are necessarily conjugate. For example, here's how it looks for a second-order equation: if $r=a\pm bi$ are the roots, then the general solution can be written not only as $C_1e^{(a+bi)t}+C_2e^{(a-bi)t}$, but also as $C_1e^{at}\cos(bt)+C_2e^{at}\sin(bt)$. Your form for a particular solution is almost correct, but you also have to include a sine term: $y_p=Ae^t\cos2t+Be^t\sin2t$.