Answer to Question #272663 in Differential Equations for Lando

Question

Verify that the function 𝑦 = 𝑐1𝑒

(−𝑘+2𝑖)𝑥 + 𝑐2𝑒

(−𝑘−2𝑖)𝑥 is a

solution to

𝑦

′′ + 2𝑘𝑦

′ + (𝑘

2 + 4)𝑦 = 0

Answer 1

$y=c_1e^{(-k+2i)x}+c_2e^{(-k-2i)x}$

$y'=(-k+2i)c_1e^{(-k+2i)x}+(-k-2i)c_2e^{(-k-2i)x}$

$y''=(-k+2i)^2c_1e^{(-k+2i)x}+(-k-2i)^2c_2e^{(-k-2i)x}$

Substitute

$y''+ 2𝑘y'+ (k^2+ 4)𝑦$

$=(-k+2i)^2c_1e^{(-k+2i)x}+(-k-2i)^2c_2e^{(-k-2i)x}$

$+2k(-k+2i)c_1e^{(-k+2i)x}+2k(-k-2i)c_2e^{(-k-2i)x}$

$+ (k^2+ 4)c_1e^{(-k+2i)x}+ (k^2+ 4)c_2e^{(-k-2i)x}$

$=(k^2-4ik-4-2k^2+4ik+k^2+4)c_1e^{(-k+2i)x}$

$+(k^2+4ik-4-2k^2-4ik+k^2+4)c_2e^{(-k-2i)x}$

$=0, x\in \Complex$

Therefore the function

$y=c_1e^{(-k+2i)x}+c_2e^{(-k-2i)x}$

is a solution to

$y''+ 2𝑘y'+ (k^2+ 4)𝑦= 0.$