Case with Complex Roots

We continue our discussion of ODEs of the form $y^{″} + a y^{'} + b y = 0$ . We illustrate the case where the characteristic polynomial has complex roots using the following example:

Consider the ODE $y^{″} - 2 y^{'} + 2 y = 0$ . The characteristic polynomial is $λ^{2} - 2 λ + 2$ , which has no real roots. Using the quadratic formula, we can find that $1 + i, 1 - i$ are the two roots of this polynomial. So then the general solution is $y (t) = e^{t} (c_{1} e^{i t} + c_{2} e^{- i t})$ , and we're done. But wait! There's a problem with this. Our differential equation is for a real valued function, but as written, our solution could be complex. What's going on? To try and resolve this, let's first use Euler's Formula and expand these exponentials. Write

$y (t) = e^{t} (c_{1} \cos t + i c_{1} \sin t + c_{2} \cos t - i c_{2} \sin t) = e^{t} ((c_{1} + c_{2}) \cos t + i (c_{1} - c_{2}) \sin t$

Now, we define ${\hat{c}}_{1} = c_{1} + c_{2}$ and ${\hat{c}}_{2} = i (c_{1} - c_{2})$ , so we have

$y (t) = {\hat{c}}_{1} e^{t} \cos t + {\hat{c}}_{2} e^{t} \sin t,$

so this is the general solution for problems with complex roots. The one thing we need to be careful about is ensuring the constants are real. They will be real if $c_{1}$ and $c_{2}$ are complex conjugates, and any two choices of ${\hat{c}}_{1}$ and ${\hat{c}}_{2}$ can be obtained with appopriate choices of $c_{1}, c_{2}$ . An alternative justification could be that if ${\hat{c}}_{1}, {\hat{c}}_{2}$ are picked to satisfy a problem's initial conditions, $y (t) = {\hat{c}}_{1} e^{t} \cos t + {\hat{c}}_{2} e^{t} \sin t$ will solve the given IVP and we can know we found the right solution using the existence and uniqueness theorem.