Solve the initial-value problem. $$y ^ { tt } + 3 y ^ { t } - 4 y = 0 , y ( 0 ) = 2 , y ^ { t } ( 0 ) = 1$$
A)$y = e ^ { 2 x } + \frac { 1 } { 5 } e ^ { - 2 x }$
B) $y = \frac { 9 } { 5 } e ^ { x } + \frac { 1 } { 5 } e ^ { - 4 x }$
C) $y = \frac { 1 } { 5 } e ^ { 2 x } + \frac { 9 } { 5 } e ^ { - 2 x }$
D) $y = e ^ { x } + \frac { 1 } { 5 } e ^ { - 3 x }$
E) None of these

Question

Quizplus · Accepted Answer

The characteristic equation of the differential equation is $r^2 + 3r - 4 = 0$, which factors to $(r - 1)(r + 4) = 0$. The solutions are $r = 1$ and $r = -4$, leading to the general solution $y = C_1 e^{x} + C_2 e^{-4x}$. Applying the initial conditions $y(0) = 2$ and $y^t(0) = 1$, we find $C_1 = \frac{9}{5}$ and $C_2 = \frac{1}{5}$, resulting in $y = \frac{9}{5} e^{x} + \frac{1}{5} e^{-4x}$.

Solve the Initial-Value Problem ytt+3yt−4y=0,y(0)=2,yt(0)=1y ^ { tt } + 3 y ^ { t } - 4 y = 0 , y ( 0 ) = 2 , y ^ { t } ( 0 ) = 1ytt+3yt−4y=0,y(0)=2,yt(0)=1

Solve the Initial-Value Problem $y ^ { tt } + 3 y ^ { t } - 4 y = 0 , y ( 0 ) = 2 , y ^ { t } ( 0 ) = 1$