Case with Complex Roots

We continue our discussion of ODEs of the form ${\displaystyle y''+ay'+by=0}$. We illustrate the case where the characteristic polynomial has complex roots using the following example:

Consider the ODE ${\displaystyle y''-2y'+2y=0}$. The characteristic polynomial is ${\displaystyle \lambda ^{2}-2\lambda +2}$, which has no real roots. Using the quadratic formula, we can find that ${\displaystyle 1+i,1-i}$ are the two roots of this polynomial. So then the general solution is ${\displaystyle y(t)=e^{t}(c_{1}e^{it}+c_{2}e^{-it})}$, 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

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

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

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

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 ${\displaystyle c_{1}}$ and ${\displaystyle c_{2}}$ are complex conjugates, and any two choices of ${\displaystyle {\hat {c}}_{1}}$ and ${\displaystyle {\hat {c}}_{2}}$ can be obtained with appopriate choices of ${\displaystyle c_{1},c_{2}}$. An alternative justification could be that if ${\displaystyle {\hat {c}}_{1},{\hat {c}}_{2}}$ are picked to satisfy a problem's initial conditions, ${\displaystyle y(t)={\hat {c}}_{1}e^{t}\cos t+{\hat {c}}_{2}e^{t}\sin t}$ will solve the given IVP and we can know we found the right solution using the existence and uniqueness theorem.