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) $y = C _ { 1 } e ^ { x } + C _ { 2 } e ^ { - 7 x }$

Question

Quizplus · Accepted Answer

The characteristic equation for the differential equation is $r^2 + 4r - 5 = 0$, which factors to $(r - 1)(r + 5) = 0$. The solutions are $r = 1$ and $r = -5$, leading to the general solution $y = C_1 e^{x} + C_2 e^{-5x}$.

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

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