✔ 最佳答案
Method 1
y'' + 4y = 0
by the Laplace transform,
S^2*Y - S*y(0) - y'(0) + 4Y = 0
By y(0) = 1 and y'(0) = 0 , we get
S^2*Y - S + 4Y = 0
Y = S / ( S^2 + 4 )
y = L^ -1 { Y } = L^ -1 { S / ( S^2 + 4 ) } = cos 2t
Ans: y(t) = cos 2t
Method 2
characteristic eq. is λ^2 + 4 = 0
λ = ± 2 i , so
y = C1*cos 2t + C2*sin 2t ... (1)
y' = -2*C1*sin 2t + 2*C2*cos 2t ... (2)
substitute y(0) = 1 into (1), we get C1 = 1
substitute y'(0) = 0 into (2), we get C2 = 0
Hence, y = cos 2t
Checking :
y = cos 2t , y(0) = 1
y' = - 2 sin 2t , y'(0) = 0
y'' = - 4 cos 2t
y'' + 4y = - 4 cos 2t + 4 cos 2t = 0