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

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

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

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

$$y''+ 2𝑘y'+ (k^2+ 4)𝑦$$

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

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

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

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

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

$$=0, x\in \Complex$$

Therefore the function

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

is a solution to

$$y''+ 2𝑘y'+ (k^2+ 4)𝑦= 0.$$