Use the Laplace Transform to solve IVP: y” + y = f(t). y(0) = 0, y'(0) = 1, where 0 < t < 6, 1 < t. f(t) =
The Correct Answer and Explanation is:
To solve the initial value problem (IVP) using the Laplace Transform, we are given:
- Differential equation:
y” + y = f(t) - Initial conditions:
y(0) = 0, y'(0) = 1 - For t > 1, the forcing function f(t) = 1, and for 0 < t < 1, f(t) = 0
This means:
f(t) = u₁(t) where u₁(t) is the unit step function starting at t = 1
Step 1: Take the Laplace Transform of both sides
Use the Laplace transform properties:
- L{y”} = s²Y(s) – sy(0) – y'(0) = s²Y(s) – 1
- L{y} = Y(s)
- L{f(t)} = L{u₁(t)} = e^{-s}/s
Now apply the Laplace transform to the differential equation:
s²Y(s) – 1 + Y(s) = e^{-s}/s
Step 2: Solve for Y(s)
Combine like terms:
(s² + 1)Y(s) = 1 + e^{-s}/s
Solve for Y(s):
Y(s) = [1 + e^{-s}/s] / (s² + 1)
Step 3: Take the inverse Laplace Transform
Break it into two terms:
Y(s) = 1/(s² + 1) + e^{-s}/[s(s² + 1)]
- L⁻¹{1/(s² + 1)} = sin(t)
- L⁻¹{e^{-s}F(s)} = u₁(t) * f(t – 1) (time-shifting property)
Let F(s) = 1/[s(s² + 1)]
Use partial fractions:
1/[s(s² + 1)] = A/s + (Bs + C)/(s² + 1)
Solve:
1 = A(s² + 1) + (Bs + C)(s)
→ A = 1, B = -1, C = 0
So:
F(s) = 1/s – s/(s² + 1)
Thus:
L⁻¹{F(s)} = 1 – cos(t)
Now apply the time shift:
L⁻¹{e^{-s}F(s)} = u₁(t)(1 – cos(t – 1))
Final Answer:
y(t) = sin(t) + u₁(t)(1 – cos(t – 1))
Explanation
We solved the second-order linear differential equation using the Laplace transform method. This method is powerful because it turns differential equations into algebraic equations, which are easier to handle.
We began by transforming the original equation y” + y = f(t). Using Laplace transform rules, we converted the left-hand side into terms of Y(s), the transform of y(t). For the right-hand side, we recognized f(t) as the unit step function u₁(t) because the forcing term is zero before t = 1 and one afterward. The Laplace transform of u₁(t) is e^{-s}/s.
After applying the Laplace transform to both sides and substituting initial conditions y(0) = 0 and y'(0) = 1, we solved for Y(s). The solution was a sum of two terms. The first, 1/(s² + 1), corresponds to the inverse transform sin(t). The second term, involving e^{-s}, required the second shifting theorem (also called the time-delay property).
We rewrote 1/[s(s² + 1)] using partial fractions and took its inverse Laplace transform to get 1 – cos(t). Applying the shift gave us u₁(t)(1 – cos(t – 1)).
Combining both parts, the solution to the original IVP is:
y(t) = sin(t) + u₁(t)(1 – cos(t – 1))
This piecewise behavior reflects the fact that the forcing function f(t) turns on at t = 1, affecting the system starting from that time.
