Good day.
I was working on this problem from my lecture notes.
"Find eigenvalues and corresponding eigenfunctions for the BVP
$\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+\lambda y=0,$ $y(0)=y(1)=1$
and verify their orthogonality by direct calculation"
There's also a solution provided, which I've followed along with. Though solving the whole exercise is the end goal, I'm just concerned with finding the eigenvalues for the time being.
I understand how to set up the auxiliary equation, and how to solve the general system of the ODE for $\lambda < 1$ and $\lambda = 1$, but I don't understand what the solution does to solve it for $\lambda > 1$.
"For $\lambda > 1$, say $\lambda = 1 + \mu^2, \mu = \sqrt{\lambda-1} > 0$, the general solution is $y(x) = (A\cos(\mu x) + B\sin(\mu x))e^{-x},$ $y(0) = 0 \to A=0,$ then $y(1) = 0 \Rightarrow B = 0,$ or $\sin(\mu) = 0,$ $\mu = n\pi,$ $n = 1, 2, 3...,$ giving the result:
$\lambda_{n} = 1 + n^2\pi^2,$ $\phi_n(x) = e^{-x}\sin(n\pi x),$ $n = 1,2,3,...,$"
(It then checks orthogonality)
I don't understand why it substitutes $\mu$. I thought when dealing with complex roots, you need it to be in form $m = p \pm qi$, to substitute into $y = e^{px}(C_{1}\cos(qx)+C_{2}\sin(qx))$. I've thought about it for a bit, but the link isn't clear yet.
After some thinking and reflection, I think I can better articulate, and answer my question.
So, after finding the auxiliary equation, we can complete the square to find $m$ in terms of $\lambda$.
$m^2+2m+\lambda=0$
...
$m=-1 \pm \sqrt{1-\lambda}$
For $\lambda > 1$, we'll have complex roots. I was previously taught that given complex roots in the form $m=p\pm qi$, we can substitute into this complementary function accordingly:
$y = e^{px}(C_{1}cos(qx) + C_{2}sin(qx))$
However, $m=-1 \pm \sqrt{1-\lambda}$ is not in the form $m = p \pm qi$ (since there isn't explicitly any $i$). That was the crux of my question. What do you do from here?
It's simply a case of equating and solving for $q$.
$qi = \sqrt{1-\lambda}$
$-q^2=1-\lambda$
$q^2 = \lambda - 1$
$q = \pm \sqrt{\lambda - 1}$
So
$m = p \pm q i$
$p = -1$
$q = \sqrt{\lambda - 1}$
Hopefully this makes sense and potentially helps someone in the future :)
(Got no idea why $\mu$ was used instead of $q$ in our notes)