# Trigonometric Input to an First Order Differential Equation, Exponentials

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.

-
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