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In an ODE class, the differential equation is given

$y' + ky = kq_e(t)$

where the input $q_e(t)$ is given as $cos \ \omega t$. The teacher "complexifies" the problem by using the real part of $e^{i\omega t}$ due to Euler's formula.

Then we have

$y' + ky = k e^{i \omega t}$

But since the solution is also complefied, teacher changes notation from $y$, to $\tilde{y}$, hence we now have

$\tilde{y}' + k\tilde{y} = k e^{i \omega t}$

where the complex solution is $\tilde{y} = y_1 + iy_2$. The claim is we find $\tilde{y}$, then $y_1$ solves the original ODE.

Then teacher goes ahead and solves the problem using exponentials, etc. However what I am looking for is the proof for the statement above, that solving complexified ODE will solve the original ODE.

I tried this

Plug in $\tilde{y} = y_1 + iy_2$ in complexified ODE

$(y_1 + iy_2)' + k(y_1 + iy_2) = k e^{i\omega t}$

$y_1' + iy_2' + ky_1 + kiy_2 = ke^{i\omega t}$

Group real #'s and complex #'s together

$(y_1+ky_1) + i(y_2' + ky_2) = ke^{i\omega t}$

Then the real part of LHS above is $(y_1+ky_1)$, exactly the LHS of primary ODE, and the real part of RHS above is $kcos \ \omega t$ which matches the RHS of original ODE. Is this okay, in terms of language, reasoning, etc.

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Yes, absolutely fine. There is a technical problem arguing in the other direction (when can real solutions be complexified?). – André Nicolas Jul 29 '11 at 14:43
up vote 2 down vote accepted

Yes, that's fine. Two complex quantities are equal iff the real parts are equal and the imaginary parts are equal, so it is OK to separate the real and imaginary components like you do in the last step.

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