I stuck with these ODE question. solve the ODE $T$ is function of $t$ $T''+T'+\lambda T=0$ $T(0)=0,T'(0)=0,\lambda=n^2 $ when $n\in \Bbb N$ My sulution So far: Characteristic polynomial $r^2+r+\lambda=0$ sol: $r_{1,2}= \frac{-1\pm \sqrt {1-4\lambda} }{2} $ therefor $T(t)=c_1e^{t\frac{-1+ \sqrt {1-4\lambda} }{2}}+c_2e^{t\frac{-1-\sqrt {1-4\lambda}}{2}}$
$T(0)=0 \rightarrow T(0)=c_1 +c_2 =0 \rightarrow c_1=-c_2 $
$T'(0)=c_1(-\frac{1}{2}+\frac{\sqrt{1-4 \lambda }}{2})+c_2(-\frac{1}{2}-\frac{\sqrt{1-4 \lambda }}{2}) =-1$ remember $c_1=-c_2 $ $c_1(-\frac{1}{2}+\frac{\sqrt{1-4 \lambda }}{2})-c_1(-\frac{1}{2}-\frac{\sqrt{1-4 \lambda }}{2}) =-1$
$c_1(\sqrt{1-4 \lambda })=-1$ therefor $c_1=\frac {-1}{\sqrt {1-4 \lambda }}$ and $c_2=\frac {1}{\sqrt {1-4 \lambda }}$ plug in $c_1,c_2$ at$T(t)$:
$T(t)=\frac {-1}{\sqrt {1-4 \lambda }}e^{t\frac{-1+ \sqrt {1-4\lambda} }{2}}+\frac {1}{\sqrt {1-4 \lambda }}e^{t\frac{-1-\sqrt {1-4\lambda}}{2}}$ plug in $\lambda=n^2$ and we got
$$T(t)=\frac {-1}{\sqrt {1-4 n^2 }}e^{t\frac{-1+ \sqrt {1-4n^2} }{2}}+\frac {1}{\sqrt {1-4 n^2 }}e^{t\frac{-1-\sqrt {1-4n^2}}{2}}$$ how do i get to the solution: $e^{-\frac{t}{2}}[Asin(\frac{\sqrt {4n^2-1}}{2})+Bcos(\frac{\sqrt {4n^2-1}}{2})]$ from here thank's
I would say, if
$T(t)=C_1e^{t\frac{-1+ \sqrt {1-4n^2} }{2}}+C_2e^{t\frac{-1-\sqrt {1-4n^2}}{2}}$ then is for $n=1,2,3\cdots$
$T(t)=e^{-t/2}\left(C_1e^{\frac{it\sqrt {4n^2-1 } }{2}}+C_2e^{\frac{-it\sqrt {4n^2-1}}{2}}\right), $ i is the imaginary unit
$=e^{-t/2}\left(C_1\cos \frac{t\sqrt {4n^2-1 } }{2}+iC_1\sin \frac{t\sqrt {4n^2-1 } }{2}+C_2\cos \frac{t\sqrt {4n^2-1 } }{2}-iC_2\sin \frac{t\sqrt {4n^2-1 } }{2}\right)=e^{-t/2}\left((C_1+C_2)\cos \frac{t\sqrt {4n^2-1 } }{2}+i(C_1-C2)\sin \frac{t\sqrt {4n^2-1 } }{2}\right)=e^{-t/2}\left(A\cos \frac{t\sqrt {4n^2-1 } }{2}+B\sin \frac{t\sqrt {4n^2-1 } }{2}\right)$