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) =
To solve the initial value problem (IVP):
y” + y = f(t), with initial conditions y(0) = 0, y'(0) = 1,
and the forcing function defined as
f(t) = 1 for t > 1, and f(t) = 0 for 0 < t < 1,
we express f(t) using the unit step function (Heaviside function):
Step 1: Express f(t) using unit step function
f(t) = u₁(t)
This is because the unit step function u₁(t) equals 0 for t < 1 and 1 for t ≥ 1.
Step 2: Take Laplace Transform of both sides
Take Laplace transform of the differential equation:
L{y” + y} = L{u₁(t)}
Recall the Laplace transforms:
- L{y”} = s²Y(s) – s y(0) – y'(0)
- L{y} = Y(s)
- L{u₁(t)} = e^(-s)/s
Substitute into the equation:
(s²Y(s) – s * 0 – 1) + Y(s) = e^(-s)/s
(s²Y(s) + Y(s)) – 1 = e^(-s)/s
Factor Y(s):
Y(s)(s² + 1) = e^(-s)/s + 1
Now solve for Y(s):
Y(s) = [e^(-s)/s + 1] / (s² + 1)
Step 3: Take Inverse Laplace Transform
Break it into two parts:
Y(s) = e^(-s)/[s(s² + 1)] + 1/(s² + 1)
The second term is straightforward:
L⁻¹{1/(s² + 1)} = sin(t)
Now handle the first term using shifting theorem:
L⁻¹{e^(-s)F(s)} = u₁(t) * f(t – 1)
So we find L⁻¹{1/[s(s² + 1)]}
We can use partial fractions to evaluate it:
1 / [s(s² + 1)] = A/s + (Bs + C)/(s² + 1)
Solve:
1 = A(s² + 1) + (Bs + C)(s)
Expand and equate coefficients:
1 = A s² + A + Bs² + Cs
Group terms:
(s²)(A + B) + (Cs) + A = 1
So:
A + B = 0
C = 0
A = 1
Then B = -1
So:
1 / [s(s² + 1)] = 1/s – s/(s² + 1)
Thus, the inverse transform is:
L⁻¹{1/s – s/(s² + 1)} = 1 – cos(t)
Now apply the time shift:
L⁻¹{e^(-s)/[s(s² + 1)]} = u₁(t)(1 – cos(t – 1))
Step 4: Final Solution
So the solution is:
y(t) = u₁(t)(1 – cos(t – 1)) + sin(t)
Explanation
This problem involves solving a second-order linear differential equation with a piecewise forcing function using Laplace transforms. Since f(t) = 0 for t < 1 and f(t) = 1 for t ≥ 1, it fits the unit step function model: f(t) = u₁(t). The Laplace transform helps handle such discontinuous inputs effectively.
We begin by transforming the differential equation using the Laplace properties for derivatives, applying the given initial conditions. This converts the differential equation into an algebraic equation in terms of Y(s), which is the Laplace transform of y(t). By isolating Y(s), we express it as the sum of two rational functions.
The inverse Laplace of the second term, 1/(s² + 1), directly gives sin(t). The first term is more complex and requires using the shifting theorem, which is helpful when an exponential factor appears in the Laplace domain. Before applying the shift, we simplify 1 / [s(s² + 1)] using partial fractions, which breaks it into simpler terms whose inverse transforms are known.
The time-shifting theorem then converts the result into a function multiplied by a unit step: this represents how the system only starts responding at t = 1 due to the delayed input. The full solution, therefore, combines the natural response of the system (sin(t)) with the delayed forced response from the input (1 – cos(t – 1) after t = 1).
This method provides a clear and structured way to solve complex initial value problems with piecewise or delayed inputs.
