Solve the initial-value problem. Select the correct answer.
$$y ^ { tt } + 8 y ^ { t } + 41 y = 0 , y ( 0 ) = 1 , y ^ { t } ( 0 ) = 4$$
A)
$y = e ^ { - 4 x } \left( \cos ( 5 x ) + \frac { 8 } { 5 } \sin ( 5 x ) \right)$
B)
$y = e ^ { - 5 x } \left( \cos ( 5 x ) + \frac { 2 } { 5 } \sin ( 5 x ) \right)$
C)
$y = e ^ { 5 x } \left( \cos ( 4 x ) + \frac { 8 } { 5 } \sin ( 4 x ) \right)$
D)
$y = e ^ { 4 x } \left( \cos ( 5 x ) - \frac { 2 } { 5 } \sin ( 5 x ) \right)$
E)
$y = e ^ { 5 x } \left( \cos ( 5 x ) + \frac { 8 } { 5 } \sin ( 5 x ) \right)$

Question

Quizplus · Accepted Answer

The characteristic equation of the differential equation is $r^2 + 8r + 41 = 0$, which has complex roots $r = -4 \pm 5i$. The general solution is $y = e^{-4x} (C_1 \cos(5x) + C_2 \sin(5x))$. Using the initial conditions $y(0) = 1$ and $y^t(0) = 4$, we find $C_1 = 1$ and $C_2 = \frac{8}{5}$.

Solve the Initial-Value Problem ytt+8yt+41y=0,y(0)=1,yt(0)=4y ^ { tt } + 8 y ^ { t } + 41 y = 0 , y ( 0 ) = 1 , y ^ { t } ( 0 ) = 4ytt+8yt+41y=0,y(0)=1,yt(0)=4

Solve the Initial-Value Problem $y ^ { tt } + 8 y ^ { t } + 41 y = 0 , y ( 0 ) = 1 , y ^ { t } ( 0 ) = 4$