Solve the initial-value problem.
$$y ^ { tt } - 2 y ^ { t } - 24 y = 0 , y ( 1 ) = 4 , y ^ { t } ( 1 ) = 6$$
A) $y ( x ) = \frac { 22 } { 10 } e ^ { 6 ( x - 1 ) } + \frac { 18 } { 10 } e ^ { - 4 ( x - 1 ) }$
B) $y ( x ) = e ^ { 6 x } - e ^ { - 4 x }$
C)
$y ( x ) = \frac { 1 } { 10 } e ^ { 4 ( x - 1 ) } + \frac { 1 } { 23 } e ^ { - 6 ( x - 1 ) }$
D)
$y ( x ) = \frac { 1 } { 10 } e ^ { x } - \frac { 1 } { 23 } e ^ { - x }$
E)
$y ( x ) = \frac { 1 } { 10 } e ^ { 4 x } - \frac { 1 } { 23 } e ^ { - 6 x }$

Question

Quizplus · Accepted Answer

The characteristic equation for the differential equation is $r^2 - 2r - 24 = 0$, which factors to $(r - 6)(r + 4) = 0$. Thus, the general solution is $y(x) = C_1 e^{6x} + C_2 e^{-4x}$. Applying the initial conditions $y(1) = 4$ and $y^t(1) = 6$, we solve for $C_1$ and $C_2$ to find the specific solution $y(x) = \frac{22}{10} e^{6(x-1)} + \frac{18}{10} e^{-4(x-1)}$.

Solve the Initial-Value Problem ytt−2yt−24y=0,y(1)=4,yt(1)=6y ^ { tt } - 2 y ^ { t } - 24 y = 0 , y ( 1 ) = 4 , y ^ { t } ( 1 ) = 6ytt−2yt−24y=0,y(1)=4,yt(1)=6

Solve the Initial-Value Problem $y ^ { tt } - 2 y ^ { t } - 24 y = 0 , y ( 1 ) = 4 , y ^ { t } ( 1 ) = 6$