Answer to Question #264907 in Calculus for tina

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

$𝑦 = 𝑐_1𝑒 ^{(−𝑘+2𝑖)𝑥} + 𝑐_2𝑒 ^{(−𝑘−2𝑖)𝑥} ........(1)\\ solution \space to\\ 𝑦 ^{′′} + 2𝑘𝑦 ^{′} + (𝑘^ 2 + 4)𝑦 = 0.$

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

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

$𝑐_1 (−𝑘+2𝑖)^2 𝑒^{ (−𝑘+2𝑖)𝑥}+ 𝑐_2𝑒 ^{(−𝑘-2𝑖)𝑥)}(-k-2i)^2 + 2k(−𝑘−2𝑖) 𝑐_1𝑒 ^{(−𝑘+2𝑖)𝑥 }+ 2k (−𝑘−2𝑖) 𝑐_2𝑒^{ (−𝑘-2𝑖)𝑥} + (𝑘^2+4) 𝑐_1𝑒^{ (−𝑘+2𝑖)𝑥 }+ (𝑘^2+4) 𝑐_2𝑒^ {(−𝑘+2𝑖)k}$

hence verified