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 $y^{tt} - 2y^{t} - 24y = 0$ is $r^2 - 2r - 24 = 0$, which factors to $(r - 6)(r + 4) = 0$, giving roots $r = 6$ and $r = -4$. Thus, the general solution is $y(x) = C_1e^{6x} + C_2e^{-4x}$. Using the initial conditions $y(1) = 4$ and $y^{t}(1) = 6$, we solve for $C_1$ and $C_2$ to find the specific solution, which matches the form of choice A when adjusted for the initial condition at $x = 1$.

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 $y ^ { tt } - 2 y ^ { t } - 24 y = 0 , y ( 1 ) = 4 , y ^ { t } ( 1 ) = 6$