Consider the differential equation $y'' + 3y' + 2y = 2\sin{t}$. How would I write the complex version of this equation?
I interpreted that question as the general solution to that equation including complex numbers (so BEFORE using Euler's Formula and some magic to remove the complex part). However, as I tried to find the general solution of this differential equation I found that there is no "complex version"?
Here's what I tried: I guessed that $y = e^{\lambda t}$, $y' = \lambda e^{\lambda t}$, $y'' = \lambda^2e^{\lambda t} $
I then substituted this into the original equation and set it to 0.
$$\lambda^2 e^{\lambda t} + 3\lambda e^{\lambda t} + 2e^{\lambda t} = 0 $$ $$\lambda^2 + 3\lambda + 2 = 0 $$
This is where my initial attempts halted, as the solutions to this equation are real and not complex.
Am I misinterpreting the question or is there an error in my method somewhere?