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 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 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 $y ^ { tt } + 8 y ^ { t} + 41 y = 0 , y ( 0 ) = 1 , y ^ { t } ( 0 ) = 4$