Solve the following differential equation {d^2 Y/dt^2 + 4Y (t)=0 {Y(0)=1 Y'(0)=0 They're in the same bracket Detailed answers please?

2015-09-09 5:45 pm

回答 (2)

2015-09-09 6:22 pm
✔ 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
2015-09-09 6:27 pm
y = A cos 2t + B sin 2t
y = cos 2t

and if you haven't learned out why that is...
would it make sense if I said

y'' - k^2 y = 0
r^2 - k^2 = 0
(r-k)(r+k) = 0
y = Ae^kt + B e^-kt?

but now instead of a -k you have a + k

y'' + k^2 y = 0
r^2 + k^2 = 0
(r + ik)(r-ik)
y = A e^ikt + B e^-ikt
e^ikt = cos kt + i sin kt
e^-ikt = cos kt - i sin kt

and if that doesn't work for you
dy/dt = u(y)
d^2y/dt^2 = du/dy dy/dt = du/dy u
u du/dy = -4y
this is a separable diff eq.
u^2 = - 4y^2 + C^2
dy/dt = (C^2 - 4y^2)^1/2
1/(C^2 - 4y^2)^1/2 dy = dt
(1/2) sin^-1 (2/C y) = t + D
y = C sin (2t + D)
which is equivalent to:
y = A sin 2t + B sin 2t


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