Answer
$y(x)=e^{-2x}cos(5x)+Ce^{-2x}sin(5x)$
where $C$ is an arbitrary constant.
Work Step by Step
$y''+4y'+29y=0$, $y(0)=1$ and $y(\pi)=-e^{-2\pi}$
Roots are complex numbers: $\alpha \pm i \beta$, thus the
General solution: $y=e^{\alpha x}[c_1cos(\beta x)+c_2sin(\beta x)]$
Plug in the given values: $y(0)=1$ and $y(\pi)=-e^{-2\pi}$
Hence, $y(0)=1$
$y(\pi)=-e^{-2\pi}$
The solution of the boundary value problem is
$y(x)=e^{-2x}cos(5x)+Ce^{-2x}sin(5x)$
where $C$ is an arbitrary constant.
