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Solve the following second-order linear ordinary differential equation, with steps please: y''(t) + y(t) = sin(t).
I thank you for any help.
 

Guest Mar 31, 2017
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3+0 Answers

 #1
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Hi the solution is      y = Ae^(-t) - 1/2 sin(t) -1/2 cos(t)

 

You have the auxiliary equation  k^2(k+1) = 0  where  y = Ae^(kt)     and k cannot be zero. So k= -1 and we have y=Ae^(-t)

 

Now for the complementary function ,try CF  = a sin(t) + b cos(t).  To find a  and b

substitute in original equation to get

 

 y= Ae^(-t) + asin(t) + b cos(t)           differentiate to get

y' = -Ae^(-t)  +a cos(t)  - b sin(t)         differentiate again   to get

y'' = Ae^(-t)   -a sin(t)+b cos(t)

 

We want y'' + y'  = sin(t)     so we can equate co-efficients  in sin and cos to get   a= -1/2 and b = -1/2

Guest Mar 31, 2017
 #2
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sorry,slight correction, auxiliary equation is  k(k+1), my typo error.

Guest Mar 31, 2017
 #3
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Solve ( d^2 y(t))/( dt^2) + y(t) = sin(t):
Apply the Laplace transformation ℒ_t[f(t)](s) = integral_0^∞ (f(t))/e^(s t) dt to both sides:
ℒ_t[( d^2 y(t))/( dt^2) + y(t)](s) = ℒ_t[sin(t)](s)

Find the Laplace transformation term by term: :
ℒ_t[( d^2 y(t))/( dt^2)](s) + ℒ_t[y(t)](s) = ℒ_t[sin(t)](s)

Apply ℒ_t[( d^2 y(t))/( dt^2)](s) = s^2 (ℒ_t[y(t)](s)) - s y(0) - y'(0):
ℒ_t[y(t)](s) + -(s y(0)) + s^2 ℒ_t[y(t)](s) - y'(0) = ℒ_t[sin(t)](s)

Apply ℒ_t[sin(t)](s) = 1/(s^2 + 1):
ℒ_t[y(t)](s) + s^2 (ℒ_t[y(t)](s)) - s y(0) - y'(0) = 1/(s^2 + 1)

Simplify:
(s^2 + 1) (ℒ_t[y(t)](s)) - s y(0) - y'(0) = 1/(s^2 + 1)
Solve for ℒ_t[y(t)](s):
ℒ_t[y(t)](s) = (y(0) s^3 + y(0) s + y'(0) + y'(0) s^2 + 1)/(s^2 + 1)^2

Decompose ℒ_t[y(t)](s) via partial fractions:
ℒ_t[y(t)](s) = 1/(s^2 + 1)^2 + (s y(0))/(s^2 + 1) + (y'(0))/(s^2 + 1)
Compute y(t) = ℒ_s^(-1)[1/(s^2 + 1)^2 + (s y(0))/(s^2 + 1) + (y'(0))/(s^2 + 1)](t):

Find the inverse Laplace transformation term by term: :
y(t) = ℒ_s^(-1)[1/(s^2 + 1)^2](t) + ℒ_s^(-1)[(s y(0))/(s^2 + 1)](t) + ℒ_s^(-1)[(y'(0))/(s^2 + 1)](t)

Apply ℒ_s^(-1)[1/(s^2 + 1)^2](t) = 1/2 (sin(t) - t cos(t)):
y(t) = 1/2 (-(t cos(t)) + sin(t)) + ℒ_s^(-1)[(s y(0))/(s^2 + 1)](t) + ℒ_s^(-1)[(y'(0))/(s^2 + 1)](t)

Apply ℒ_s^(-1)[(s y(0))/(s^2 + 1)](t) = y(0) cos(t):
y(t) = 1/2 (-(t cos(t)) + sin(t)) + y(0) cos(t) + ℒ_s^(-1)[(y'(0))/(s^2 + 1)](t)

Apply ℒ_s^(-1)[(y'(0))/(s^2 + 1)](t) = sin(t) y'(0):
y(t) = 1/2 (-(t cos(t)) + sin(t)) + y(0) cos(t) + y'(0) sin(t)

Substitute c_1 = y(0) and c_2 = y'(0):
y(t) = 1/2 (-(t cos(t)) + sin(t)) + c_1 cos(t) + c_2 sin(t)

Simplify the arbitrary constants:
Answer: |y(t) = -1/2 (t cos(t)) + c_1 cos(t) + c_2 sin(t)

Guest Mar 31, 2017

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