# Complex Numbers and Constants of Integration

1. Dec 6, 2015

### squelch

1. The problem statement, all variables and given/known data

Suppose that the characteristic equation to a second order, linear, homogeneous differential equation with constant coefficients yielded two complex roots:
$$\begin{array}{l} {\lambda _1} = a + bi\\ {\lambda _2} = a - bi \end{array}$$
This would yield a general solution of:
$$y = {C_1}{e^{(a + bi)x}} + {C_2}{e^{(a - bi)x}}$$

I would like to prove that this is equal to the expression:
$$y = {C_1}{e^{ax}}\sin (bx) + {C_2}{e^{ax}}\cos (bx)$$

2. Relevant equations

Euler's identity:
$${e^{ix}} = \cos (x) + i\sin (x)$$

3. The attempt at a solution

At the end of the proof, I am left with the expression:
$$y = i{C_1}{e^{ax}}\sin (bx) + {C_2}{e^{ax}}\cos (bx)$$

Can $i$ be "rolled up" into the constant of integration $C_1$ and the whole thing just defined as a single, undetermined constant?

2. Dec 6, 2015

### Samy_A

If $C_1$ and $C_2$ are the same constants as in $y = {C_1}{e^{(a + bi)x}} + {C_2}{e^{(a - bi)x}}$, I don't think this is correct...
... and if they are differrent constants, why not "roll up" the $i$ into $C_1$?

3. Dec 6, 2015

### squelch

I'm just trying to derive a textbook definition, in case the derivation is required on an exam.
At a point in the derivation, coming from the original equation, ${e^{ax}}\cos (bx)[{C_1} + {C_2}] + i{e^{ax}}\sin (bx)[{C_1} - {C_2}]$.
I combined the constants into a new $C_1$ and $C_2$, mostly to match the textbook equation. I suppose it'd be more clear (and proper) to call them $C_1 '$ and $C_2 '$
I don't see a reason why I wouldn't be able to, but if it's a legal operation or not is my question. Call it a sanity check.

4. Dec 6, 2015

### Samy_A

Yes, you can do that. As the $C$'s and $C'$'s are arbitrary complex numbers anyway, there is absolutely no reason why you couldn't do it.