Solve the differential equation. $$y ^ { tt } + 4 y ^ { t } - 5 y = 0$$
A) $y = C _ { 1 } e ^ { x } + C _ { 2 } e ^ { - 5 x }$
B) $y = \frac { C _ { 1 } e ^ { x } } { 3 } + \frac { C _ { 2 } e ^ { - 5 x } } { 2 }$
C) $y = C _ { 1 } e ^ { - 5 x } + C _ { 2 } e ^ { - 5 x }$
D) $y = C _ { 1 } e ^ { x } + C _ { 2 } e ^ { - 7 x }$
E) None of these

Question

Quizplus · Accepted Answer

The characteristic equation is $r^2 + 4r - 5 = 0$, which has roots $r = 1$ and $r = -5$. Therefore, the general solution is $y = C_1 e^x + C_2 e^{-5x}$.

Solve the Differential Equation ytt+4yt−5y=0y ^ { tt } + 4 y ^ { t } - 5 y = 0ytt+4yt−5y=0

Solve the Differential Equation $y ^ { tt } + 4 y ^ { t } - 5 y = 0$