Solve the initial-value problem.
$y ^ { tt } + 8 y ^ {t } + 41 y = 0 , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 5$
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

We start by finding the characteristic equation of the differential equation:
$$r^2+8r+41=0$$
Solving for the roots using the quadratic formula yields:
$$r=-4\pm5i$$
Thus, the general solution is:
$$y(x)=c_1e^{(-4+5i)x}+c_2e^{(-4-5i)x}$$
Using Euler's formula, we can rewrite this as:
\begin{align*}
y(x)&=c_1e^{-4x}(\cos(5x)+i\sin(5x))+c_2e^{-4x}(\cos(-5x)+i\sin(-5x))\
&=c_1e^{-4x}(\cos(5x)+i\sin(5x))+c_2e^{-4x}(\cos(5x)-i\sin(5x))\
&=e^{-4x}[(c_1+c_2)\cos(5x)+i(c_1-c_2)\sin(5x)]
\end{align*}
Since the initial conditions are $y(0)=1$ and $y'(0)=5$, we can write:
\begin{align*}
y(0)&=c_1+c_2=1\
y'(0)&=-4c_1+5c_2=5\
\end{align*}
Solving the system of equations yields $c_1=-\frac{9}{5}$ and $c_2=\frac{14}{5}$. Thus, the solution is:
$$y(x)=e^{-4x}\left[-\frac{9}{5}\cos(5x)+\frac{14}{5}\sin(5x)ight]$$
which matches choice (C).

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

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