It's the Euler thing you mentioned.
We know are solutions are in the form:
\(\displaystyle \L\\y=C_{1}e^{({\alpha}+i{\beta})x}+C_{2}e^{({\alpha}-i{\beta})x}\)
But, mostly, we would rather have real functions instead of complex
exponentials, so we use Euler's formula:
\(\displaystyle \L\\e^{i{\theta}}=cos{\theta}+isin{\theta}\)
Therefore,
\(\displaystyle \L\\e^{i{\beta}x}=cos{\beta}x+isin{\beta}x\ and\
e^{-i{\beta}x}=cos{\beta}x-isin{\beta}x\)[1]
we have used \(\displaystyle cos({-\beta}x)=cos({\beta}x)\ and\ sin({-\beta}x)={-
sin({\beta}x)\)
By adding and then subtracting the equations in [1], we get:
\(\displaystyle \L\\e^{i{\beta}x}+e^{-i{\beta}x}=2cos{\beta}x\) and
\(\displaystyle \L\\e^{i{\beta}x}-e^{-i{\beta}x}=2isin{\beta}x\)
Since \(\displaystyle \L\\y=C_{1}e^{({\alpha}+i{\beta})x}+C_{2}e^{({\alpha}-i{\beta})x}\) is a solution of \(\displaystyle \L\\ay''+by'+cy=0\)
for any
C1 and C2, we can choose
C1=C2=1 and C1=1, C2=−1 give two solutions:
y1=e(α+iβ)x+e(α−iβ)x and
y2=e(α+iβ)x−e(α−iβ)x
But, \(\displaystyle \L\\y_{1}=e^{{\alpha}x}(e^{i{\beta}x}+e^{-i{\beta}x})
=2e^{{\alpha}x}cos{\beta}x\ and\ y_{2}=e^{{\alpha}x}(e^{i{\beta}
x}-e^{-i{\beta}x})=2ie^{{\alpha}x}sin{\beta}x\)
The superposition principle says that the real functions \(\displaystyle \L\\e^
{{\alpha}x}cos{\beta}x\ and\ e^{{\alpha}x}sin{\beta}x\) are
solutions of
\(\displaystyle \L\\ay''+by'+cy=0\)
So, the general solution is:
\(\displaystyle \L\\y=C_{1}e^{{\alpha}x}cos{\beta}x+C_{2}e^{{\alpha}x}sin
{\beta}x=e^{{\alpha}x}(C_{1}cos{\beta}x+C_{2}sin{\beta}x)\)
I hope this helps. If you need a more thorough explanation, I'd suggest
seeing your professor. I'm LaTexed out :lol: