Some help with this problem.Find the solution of the equation
v''- 4v'+5v=0,such that v=-1 and v'=1 when x=pi=3.14159
Thank you for your time.
$\displaystyle \displaystyle m^2-4m+5=0\Rightarrow m=\frac{4\pm\sqrt{16-4(5)}}{2}=\frac{4\pm 2i}{2}=2\pm i$
$\displaystyle \alpha=2, \ \ \ \beta=1$
$\displaystyle v=e^{2x}\left(C_1\cos(x)+C_2\sin(x)\right)$
$\displaystyle v(\pi)=e^{2\pi}\left(C_1\cos(\pi)+C_2\sin(\pi)\rig ht)=-1\Rightarrow e^{2\pi}\left(-1\cdot C_1)=-1$
$\displaystyle \displaystyle C_1e^{2\pi}=1\Rightarrow C_1=e^{-2\pi}$
$\displaystyle v'=e^{2x-2\pi}\left((C_2e^{-2\pi}+2)\cos(x)+(2e^{2\pi}C_2-1)\sin(x)\right)$
$\displaystyle \displaystyle v'(\pi)=(C_2e^{-2\pi}+2)\cos(\pi)+(2e^{2\pi}C_2-1)\sin(\pi)=1\Rightarrow -(C_2e^{-2\pi}+2)=1\Rightarrow C_2=-3e^{2\pi}$
$\displaystyle v(x)=e^{2x-2\pi}(\cos(x)-3e^{4\pi}\sin(x))$