How the does complex part cancel out in complex roots with Euler's formula?

Question

Suppose I have a second order differential equation

I can find the complementary solution $y_c$ by finding the roots of the characteristic equation.

Let those roots be $\lambda \pm\mu$

Thus $y_c(t) = C_1e^{t(\lambda + \mu)} + C_2e^{t(\lambda - \mu)}$ where $C_1$ and $C_2$ are real constants

I was taught that this simplifies to $y_c = e^{\lambda}(Acos(\mu t) + Bsin(\mu t))$

I do not understand how the complex part disappears.

$$y_c = C_1e^{t(\lambda + \mu)} + C_2e^{t(\lambda - \mu)} = e^{\lambda t}(C_1e^{\mu t} + C_2e^{-\mu t})$$

$$= e^{\lambda t}(C_1cos(\mu t) +C_1isin(\mu t)+C_2cos(-\mu t) + C_2isin(-\mu t))$$

Since $cos(\mu t) = cos(-\mu t)$ and $sin(\mu t) = -sin(\mu t)$

$$= e^{\lambda t}(C_1cos(\mu t) +C_1isin(\mu t)+C_2cos(-\mu t) + C_2isin(-\mu t))$$

$$= e^{\lambda t}((C_1 + C_2)cos(\mu t) + (C_1-C_2)isin(\mu t))$$

If we let $A = C_1 + C_2$ and $B = C_1 - C_2$ then $$= e^{\lambda t}(Acos(\mu t) + Bisin(\mu t))$$

Notice that there $i$ is still present in the equation. How is the complex part removed?

Answer 1

We should have $\lambda \pm i\mu$ and

$$y_c(t) = C_1e^{t(\lambda + i\mu)} + C_2e^{t(\lambda - i\mu)}$$

with $C_2 =\overline{C_1}$, since $y_c$ needs to be real, and with this set upwe can show that

$$y_c(t) = e^{\lambda}(A\cos(\mu t) + B\sin(\mu t))$$

indeed we have

$$y_c(t) = C_1e^{t(\lambda + i\mu)} + C_2e^{t(\lambda - i\mu)}=e^{\lambda t}\left(C_1e^{i\mu t} + C_2e^{- i\mu t}\right)=$$

and by $C_1=\frac{A+iB}2$ and $C_2=\frac{A-iB}2$

$$=e^{\lambda t}\left(A\frac{e^{i\mu t}+e^{-i\mu t}}2 + B\frac{e^{i\mu t}-e^{-i\mu t}}{2i}\right)=e^{\lambda}(A\cos(\mu t) + B\sin(\mu t))$$

Answer 2

Suppose our characteristic values are $a\pm ib,$ where $a,b$ are real and $b\ne 0.$ Then it follows that the functions $e^{(a+ib)t}$ and $e^{(a-ib)t}$ are cjaracteristic solutions of the differential equation. Now here's the idea -- since these solutions are independent, any linear combination of these solutions is also a solution, and an arbitrary linear combination of two independent solutions represents the general solution of the system.

Before we proceed we may simplify the functions to be $e^{at}e^{ibt}=e^{at}(\cos bt+i\sin bt)$ and $e^{at}e^{-ibt}=e^{at}(\cos bt-i\sin bt).$ If we now add these solutions, we obtain $2e^{at}\cos bt$ and if we subtract we get $2ie^{at}\sin bt,$ which are also solutions by the above italicised fact. Again, by that fact, an arbitrary linear combination of these solutions represents the general solution of the equation, so that we have the general solution to be $$Ae^{at}\cos bt+Be^{at}\sin bt=e^{at}(A\cos bt+B\sin bt),$$ where $A,B$ are constants (not necessarily real now). Thus, the complex constants have not disappeared. They may appear depending on the initial conditions, thought they rarely do in applications -- that is, usually, the initial conditions always fix $A$ and $B$ to be real.