𝑄𝑢𝑒𝑠𝑡𝑖𝑜𝑛:𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 𝑦″+4𝑦′+5𝑦=0 y(0)=1 , 𝑦′(0)=0
r^2 + 4r +5 =0
r = (-4±√16-20)/2
r = -2±i
y = C1e^(-2t)sint + C2e^(-2t)cost
initial condition: y(0)=1 , 𝑦′(0)=0
C2 * 1 = 1
C2 = 1
𝑦′ = -2C1e^(-2t)sint + C1e^(-2t)cost + -2C2e^(-2t)cost - C1e^(-2t)sint
C1 -2C2 = 0
C1 = 2
y = 2e^(-2t)sint + e^(-2t)cost
