Solve the initial-value problem. $y ^ { tt } + 8 y ^ { t } + 41 y = 0 , y ( 0 ) = 1 , y ^ { t } ( 0 ) = 5$
A)
$y = e ^ { 5 x } \left( \cos ( 5 x ) + \frac { 9 } { 5 } \sin ( 5 x ) \right)$
B)
$y = e ^ { 4 x } \left( \cos ( 5 x ) - \frac { 2 } { 5 } \sin ( 5 x ) \right)$
C)
$y = e ^ { - 4 x } \left( \cos ( 5 x ) + \frac { 9 } { 5 } \sin ( 5 x ) \right)$
D)
$y = e ^ { - 5 x } \left( \cos ( 5 x ) + \frac { 2 } { 5 } \sin ( 5 x ) \right)$
E)
$y = e ^ { 5 x } \left( \cos ( 4 x ) + \frac { 9 } { 5 } \sin ( 4 x ) \right)$

Question

Quizplus · Accepted Answer

The characteristic equation for this differential equation is $r^2+8r+41=0$, which has roots $r=-4+5i$ and $r=-4-5i$. Therefore, the general solution is 
$$y=e^{-4x}(c_1\cos 5x+c_2\sin 5x)$$
Using the initial conditions, we can solve for the constants $c_1$ and $c_2$. We have $y(0)=1$, which gives us $c_1=1$. We also have $y'(0)=5$, which gives us 
$$y'(x)=-4e^{-4x}(\cos 5x+5\sin 5x)+5e^{-4x}(\sin 5x)$$
Substituting in $x=0$ and using $y'(0)=5$, we get $c_2=9/5$. 
Therefore, the solution to the initial-value problem is 
$$y=e^{-4x}(\cos 5x+\frac{9}{5}\sin 5x)$$
which matches choice (C).

Solve the Initial-Value Problem ytt+8yt+41y=0,y(0)=1,yt(0)=5y ^ { tt } + 8 y ^ { t } + 41 y = 0 , y ( 0 ) = 1 , y ^ { t } ( 0 ) = 5ytt+8yt+41y=0,y(0)=1,yt(0)=5

Solve the Initial-Value Problem $y ^ { tt } + 8 y ^ { t } + 41 y = 0 , y ( 0 ) = 1 , y ^ { t } ( 0 ) = 5$