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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?

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The "complex version" of a real differential equation of the form $L(y) = f(t)$ (where $L$ is linear, and $f$ is a trigonometric polynomial) is $L(z) = g(t)$, where $g(t)$ is obtained from $f(t)$ by replacing $\cos(\omega t)$ by $\exp(i\omega t)$ and $\sin(\omega t)$ by $-i \exp(i \omega t)$. So in this case it should be $$ z'' + 3 z' + 2 z = -2 i \exp(it)$$ The real part of a solution $z$ will then be a solution of your original equation.

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  • $\begingroup$ Is there a specific name for this? I'd like to look more into it. $\endgroup$ – Zearia Mar 9 '16 at 1:30

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