Consider the following 2nd order homogeneous equation:
\(\displaystyle \displaystyle \frac{d^2y}{dx^2}+a\frac{dy}{dx}+by=0\)
Let the two roots of the auxiliary equation be:
\(\displaystyle \displaystyle p=\frac{-a+\sqrt{a^2-4b}}{2},\,q=\frac{-a-\sqrt{a^2-4b}}{2}\)
In the case where \(\displaystyle a^2-4b<0\), we have complex conjugate roots, which we may write as:
\(\displaystyle \displaystyle p=\frac{-a+i\sqrt{4b-a^2}}{2},\,q=\frac{-a-i\sqrt{4b-a^2}}{2}\)
Let's let:
\(\displaystyle \displaystyle r=-\frac{a}{2},\,\omega=\frac{\sqrt{4b-a^2}}{2}\)
Hence, the general solution may be expressed as:
\(\displaystyle y(x)=c_1e^{(r+\omega i)x}+c_2e^{(r-\omega i)x}=e^{rx}\left(c_1e^{\omega ix}+c_1e^{-\omega ix} \right)\)
Recall that Euler's formula states:
\(\displaystyle e^{ix}=\cos(x)+i\sin(x)\)
and so, we may express the solution to the ODE as:
\(\displaystyle y(x)=e^{rx}\left(c_1(\cos(\omega x)+i\sin(\omega x))+c_2(\cos(\omega x)-i\sin(\omega x)) \right)\)
If we regroup on like trigonometric terms and redefine the parameters, we have:
\(\displaystyle y(x)=e^{rx}\left(c_1\cos(\omega x)+ic_2\sin(\omega x) \right)\)
Let's now define:
\(\displaystyle y_1(x)=c_1e^{rx}\cos(\omega x)\)
\(\displaystyle y_2(x)=c_2e^{rx}\sin(\omega x)\)
And we may now write the solution as:
\(\displaystyle y(x)=y_1(x)+iy_2(x)\)
Since \(\displaystyle y(x)\) is a solution to the ODE, we may now write:
\(\displaystyle \displaystyle \left(\frac{d^2y_1}{dx^2}+a\frac{dy_1}{dx}+by_1 \right)+i\left(\frac{d^2y_2}{dx^2}+a\frac{dy_2}{dx}+by_2 \right)=0\)
Now, a complex value \(\displaystyle a+bi\) is equal to zero only when \(\displaystyle a=0\) and \(\displaystyle b=0\) hence we must have:
\(\displaystyle \displaystyle \frac{d^2y_1}{dx^2}+a\frac{dy_1}{dx}+by_1=0\)
\(\displaystyle \displaystyle \frac{d^2y_2}{dx^2}+a\frac{dy_2}{dx}+by_2=0\)
which implies both \(\displaystyle y_1(x)\) and \(\displaystyle y_2(x)\) must be solutions to the ODE, and so the general solution may be expressed as:
\(\displaystyle y(x)=y_1(x)+y_2(x)=e^{rx}\left(c_1\cos(\omega x)+c_2\sin(\omega x) \right)\)
edit: Only now do I notice this topic has been resurrected...
