Separation of Variables for : $$u_{tt} + 2αu_t = c^2u_{xx}$$
$α$ and $c$ are real positive constants such that $α < cπ$, and is subject to boundary conditions: $$u(0, t) = u(1, t) = 0$$
The string has an initial displacement $u(x, 0) = f(x)$ and is initially at rest.
Question: Use separation of variables to show that the displacement of the string is given by $$u(x, t) = \sum_{n=1}^{\infty} c_n ~e^{-αt} ~\sin(nπx)~ \left\{(γ_n/α)~ \cos(γ_n t) +\sin(γ_n t)\right\}$$ where $$γ_n = \sqrt{c^2n^2π^2-α^2~}$$ and give a formula for $~c_n~$.
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I'm quite stuck of the second part of the question. So far my working out has been:
$1)\quad$ Using separation of variable: Letting $u(x,t) = X(x)T(t)$
Eventually separating the PDE into $~2~$ ODE's:
$$X''(x) + λX(X) = 0\qquad \text{and}\qquad T''(t) + 2αT'(t) + c^2λT(t) =0$$
$2)\quad$ Using the boundary conditions I got $~X(0) =0~ $and $~X(1) = 0~$.
Therefore we have a Sturm-Liouville boundary problem $~X''(x) + λX(X) = 0~$ subject to $~X(0) =0~$ and $~X(1) = 0~$.
I found the eigenvalues to be $~λ_n = n^2π^2~$ with corresponding eigen-function $~X_n(x) = \sin(nπx)~$.
Then the $T(t)$ ODE becomes $$T''(t) + 2αT'(t) + c^2n^2π^2T(t) =0$$
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Edit$~1~$: To solve this I found the characteristic equation to be $$r^2 +2αr + c^2n^2π^2=0$$ and by completing the square I got $$(r+α)^2= α^2-c^2n^2π^2$$ since $~α<cπ~$ I got $$ ~r= -α \pm \sqrt{α^2-c^2n^2π^2}$$ Therefore the general solution for $~T~$ is given by
$$T_n(t) = A_n ~e^{-αt}~\cos \left(\sqrt{c^2n^2π^2-α^2~}~ t \right) + B_n~e^{-αt}~\sin \left(\sqrt{c^2n^2π^2-α^2~}~ t \right)$$
Am I correct?
Would really appreciate some help/corrections from this point on. Thanks.
Edit$~2~$: I found $~T_n'(t)~$ and using the condition $~T'(0)=0~$ I got $$T'(0) = B_n ~\sqrt{c^2}~ \sqrt{n^2}~ π =0$$ Therfore $~B_n = 0~$ and the solution for $$T_n = e^{-αt}~ A_n \cos \left(\sqrt{α^2-c^2n^2π^2~}~t \right)$$
Now to solve the entire problem $~u(x,t)~$ we have to consider a linear combination of the product of the solutions.