Originally Posted by

**lvleph** Like I said I made a mistake and $\displaystyle w = w_1$. Previously, I assumed that the eigenvalues were purely imaginary which leads to a different solutions. This was incorrect. The form of the solution comes from Euler's Formula, i.e.,

$\displaystyle e^{a \pm i b} = e^a \cos b \pm e^a i \sin b$.

So if our solution has complex eigenvalues $\displaystyle w_1,w_2$ the solution

$\displaystyle x(t) = C_1 e^{w_1t} + C_2 e^{w_2t}$

where $\displaystyle w_1 = a+bi,\; w_2=a-bi$ becomes

$\displaystyle x(t) = C_1\left(e^{at} \cos bt + e^{at} i \sin bt\right) + C_2\left(e^{at}\cos bt - e^{at} i \sin bt\right)$.

After further simplification the solution becomes

$\displaystyle x(t) = C_1 e^{at} \cos bt + C_2 e^{at} \sin bt$.

Do you see why my previous solution works if the eigenvalues are purely imaginary?

Again, I am sorry for the confusion.