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# Differential equation......

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Solve the following second-order linear ordinary differential equation, with steps of the solution:

y''(t) + y(t) = sin(t). I thank you for any help.

Guest Apr 5, 2017
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### 1+0 Answers

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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 Apr 5, 2017

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