I'm a little stuck on solving this ODE $$T''(t) + 2αT'(t) + c^2n^2π^2T(t) =0$$
Note: We are given that $~α<cπ~$ and the initial condition $~T'(0) = 0~$. I need to find the general solution and using the conditions the specific solution to $~T_n$.
This is what I've managed to do so far :
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 $~ r= -α +/- \sqrt(α^2-c^2n^2π^2)~$. Therefore the general solution for $~T~$ is given by
$$T_n(t) = A_n ~e^{-αt}~\cos(\sqrt{c^2n^2π^2-α^2~}~t~) + B_ne^{-αt}~\sin(\sqrt{c^2n^2π^2-α^2~}~t~)$$ Am I correct?
Would really appreciate the help.
Edit: Corrected $~T(t)~$.
Let $X(x)=e^{mx}$ the, you get the characteristic equation for $m$ as $$ m^2+ 2 \alpha m+ c^2 n^2 \pi^2=0.$$ This gives $$m_1,m_2=-\alpha\pm \sqrt{\alpha^2-c^2n^2\pi^2}.$$ So the general solution is $$X(x)=C_1 e^{m_1 x}+ C_2 e^{m_2 x}~~~(1).$$ If $\alpha< cn \pi$.Then $m_1,m_2=-\alpha\pm i \beta.$ Then $$X(x)= e^{-\alpha x}[ D_1 \sin \beta x+ D_2 \cos \beta x]~~~(2)$$ By putting the condition that $X'(0)=0,$ we get $$X(x)=De^{-\alpha x}[\alpha \sin \beta x + \beta \cos \beta x],~~ \beta =\sqrt{c^2n^2\pi^2-\alpha^2}~~~(3).$$ In your solution $t$ is missing in the arguments of sin and cos.Also the condition that $X'(0=0$ has not been caried out..