$\text { Find the general solution of the linear differential equation } y ^ { t t } + 2 y ^ { t } = 0 \text {. }$
A) $y = C _ { 1 } - C _ { 2 } e ^ { - 4 x }$
B) $y = C _ { 1 } + C _ { 2 } e ^ { - 2 x }$
C) $y = C _ { 1 } + 2 C _ { 2 } e ^ { 2 x }$
D) $y = C _ { 1 } - 2 C _ { 2 } e ^ { 2 x }$
E) $y = C _ { 1 } + 4 C _ { 2 } e ^ { - 4 x }$

Question

$\text { Find the general solution of the linear differential equation } y ^ { t t } + 2 y ^ { t } = 0 \text {. }$ 
A) $y = C _ { 1 } - C _ { 2 } e ^ { - 4 x }$
B) $y = C _ { 1 } + C _ { 2 } e ^ { - 2 x }$
C) $y = C _ { 1 } + 2 C _ { 2 } e ^ { 2 x }$
D) $y = C _ { 1 } - 2 C _ { 2 } e ^ { 2 x }$
E) $y = C _ { 1 } + 4 C _ { 2 } e ^ { - 4 x }$

Quizplus · Accepted Answer

The given differential equation is $y'' + 2y' = 0$. This is a homogeneous linear differential equation with constant coefficients. The characteristic equation is $r^2 + 2r = 0$, which factors to $r(r + 2) = 0$. This gives roots $r = 0$ and $r = -2$. Therefore, the general solution is $y = C_1e^{0x} + C_2e^{-2x} = C_1 + C_2e^{-2x}$.

Find the general solution of the linear differential equation ytt+2yt=0. \text { Find the general solution of the linear differential equation } y ^ { t t } + 2 y ^ { t } = 0 \text {. } Find the general solution of the linear differential equation ytt+2yt=0.

$\text { Find the general solution of the linear differential equation } y ^ { t t } + 2 y ^ { t } = 0 \text {. }$