Solve the initial-value problem. $$y ^ { tt } + 8 y ^ {t} + 41 y = 0 , y ( 0 ) = 1 , y ^ { t} ( 0 ) = 2$$
A)$y = e ^ { 5 x } \left( \cos ( 5 x ) + \frac { 6 } { 5 } \sin ( 5 x ) \right)$
B)$y = e ^ { - 4 x } \left( \cos ( 5 x ) + \frac { 6 } { 5 } \sin ( 5 x ) \right)$
C)$y = e ^ { 5 x } \left( \cos ( 4 x ) + \frac { 6 } { 5 } \sin ( 4 x ) \right)$
D)$y = e ^ { 4 x } \left( \cos ( 5 x ) - \frac { 2 } { 5 } \sin ( 5 x ) \right)$
E)$y = e ^ { - 5 x } \left( \cos ( 5 x ) + \frac { 2 } { 5 } \sin ( 5 x ) \right)$

Question

Quizplus · Accepted Answer

The characteristic equation of the differential equation $y^{tt} + 8y^t + 41y = 0$ is $r^2 + 8r + 41 = 0$. Solving this quadratic equation, we find the roots to be complex: $r = -4 \pm 5i$, indicating a solution of the form $y = e^{-4x}(A\cos(5x) + B\sin(5x))$. Using the initial conditions $y(0) = 1$ and $y^t(0) = 2$, we can solve for $A$ and $B$, leading to the solution in option B.

Solve the Initial-Value Problem ytt+8yt+41y=0,y(0)=1,yt(0)=2y ^ { tt } + 8 y ^ {t} + 41 y = 0 , y ( 0 ) = 1 , y ^ { t} ( 0 ) = 2ytt+8yt+41y=0,y(0)=1,yt(0)=2

Solve the Initial-Value Problem $y ^ { tt } + 8 y ^ {t} + 41 y = 0 , y ( 0 ) = 1 , y ^ { t} ( 0 ) = 2$