Solving Initial Value Problems: A Step-by-Step Guide

In this comprehensive guide, we'll walk through the process of solving an initial value problem (IVP). An initial value problem involves a differential equation along with initial conditions that help us determine a unique solution. Let's consider the following initial value problem as our example:

y'' + 25y = f(t); y(0) = 0, y'(0) = 5
f(t) = { 5, 0 ≤ t < (3π/5) ; 0, (3π/5) ≤ t

Here, we have a second-order linear differential equation with a piecewise-defined forcing function f(t) and two initial conditions. Solving this type of problem requires a systematic approach, which we'll break down into clear steps.

1. Understanding the Problem: Initial Value Problems Explained

Before diving into the solution, let's clarify what an initial value problem is and why it's important. An IVP consists of:

• A differential equation: This equation relates a function to its derivatives. In our case, it's y'' + 25y = f(t), where y'' represents the second derivative of y with respect to t.
• Initial conditions: These are values of the function and its derivatives at a specific point. Here, we have y(0) = 0 and y'(0) = 5, meaning the function's value at t = 0 is 0, and its first derivative's value at t = 0 is 5.
• A forcing function: In many real-world applications, differential equations do not stand alone. They are forced by some outside influence of the surrounding environment. This is represented by f(t), which is 5 for a period of time, and then becomes 0.

The goal of solving an IVP is to find the unique function y(t) that satisfies both the differential equation and the initial conditions. This is crucial in various fields, including physics, engineering, and economics, where we often model systems that evolve over time.

2. Laplace Transforms: A Powerful Tool for Solving IVPs

A popular and effective method for solving linear differential equations with initial conditions is the Laplace transform. The Laplace transform converts a differential equation in the time domain (t) into an algebraic equation in the frequency domain (s), which is often easier to solve. Here's a brief overview of the Laplace transform:

• Definition: The Laplace transform of a function f(t), denoted as F(s) or L{f(t)}, is defined as:

  F(s) = L{f(t)} = ∫₀^∞ e^(-st) f(t) dt
  

  where s is a complex variable.

• Key Properties: The Laplace transform has several useful properties that simplify solving differential equations:

  • Linearity: L{af(t) + bg(t)} = aL{f(t)} + bL{g(t)}

  • Derivative Property: L{f'(t)} = sF(s) - f(0)

  • Second Derivative Property: L{f''(t)} = s²F(s) - sf(0) - f'(0)

  • Laplace Transform of Step Functions: To handle the piecewise function f(t), we need the concept of the unit step function, also known as the Heaviside function. The Heaviside Step Function, often denoted as u(t-c), is used to signify when t is greater than or equal to c, and is key to solving a differential equation that involves piecewise functions.

    u(t-c) = { 0, t<c ; 1, t≥c
    

    L{u(t-c)} = e^(-cs) / s

• Inverse Laplace Transform: After solving for Y(s) in the frequency domain, we need to apply the inverse Laplace transform, denoted as L⁻¹, to get the solution y(t) back in the time domain.

3. Applying the Laplace Transform to Our IVP

Now, let's apply the Laplace transform to our given initial value problem. We'll transform both sides of the differential equation and use the initial conditions.

3.1. Transforming the Left-Hand Side

Taking the Laplace transform of y'' + 25y, we use the derivative properties:

L{y'' + 25y} = L{y''} + 25L{y}
           = [s²Y(s) - sy(0) - y'(0)] + 25Y(s)

where Y(s) = L{y(t)}. Substituting the initial conditions y(0) = 0 and y'(0) = 5, we get:

L{y'' + 25y} = s²Y(s) - 5 + 25Y(s)
           = (s² + 25)Y(s) - 5

3.2. Transforming the Right-Hand Side: Handling the Piecewise Function

To find the Laplace transform of f(t), we first need to express the piecewise function using Heaviside step functions. Recall that f(t) is defined as:

f(t) = { 5, 0 ≤ t < (3π/5) ; 0, (3π/5) ≤ t

We can rewrite this as:

f(t) = 5[u(t) - u(t - (3π/5))]

where u(t) is the unit step function (Heaviside function). Now, we can take the Laplace transform:

L{f(t)} = 5L{u(t) - u(t - (3π/5))}
        = 5[L{u(t)} - L{u(t - (3π/5))}]
        = 5[1/s - (e^(-(3π/5)s) / s)]
        = (5/s)[1 - e^(-(3π/5)s)]

3.3. Combining the Transformed Equation

Now we can combine the Laplace transforms of both sides of the differential equation:

(s² + 25)Y(s) - 5 = (5/s)[1 - e^(-(3π/5)s)]

4. Solving for Y(s) in the Frequency Domain

Our next step is to isolate Y(s). First, add 5 to both sides:

(s² + 25)Y(s) = (5/s)[1 - e^(-(3π/5)s)] + 5

Now, divide both sides by (s² + 25):

Y(s) = (5/s)[1 - e^(-(3π/5)s)] / (s² + 25) + 5 / (s² + 25)

We can simplify this expression further:

Y(s) = (5 / [s(s² + 25)])[1 - e^(-(3π/5)s)] + 5 / (s² + 25)
     = 5 / [s(s² + 25)] - (5e^(-(3π/5)s)) / [s(s² + 25)] + 5 / (s² + 25)

5. Inverse Laplace Transform: Returning to the Time Domain

To find the solution y(t), we need to apply the inverse Laplace transform to Y(s). This often involves using partial fraction decomposition and a table of Laplace transforms.

5.1. Partial Fraction Decomposition

We need to decompose the term 5 / [s(s² + 25)] into partial fractions:

5 / [s(s² + 25)] = A/s + (Bs + C) / (s² + 25)

Multiplying both sides by s(s² + 25), we get:

5 = A(s² + 25) + (Bs + C)s
  = As² + 25A + Bs² + Cs

Comparing coefficients, we have:

• A + B = 0
• C = 0
• 25A = 5

Solving this system, we find A = 1/5, B = -1/5, and C = 0. Therefore:

5 / [s(s² + 25)] = (1/5)/s + (-s/5) / (s² + 25)
                 = 1 / (5s) - s / [5(s² + 25)]

5.2. Applying the Inverse Laplace Transform

Now we can rewrite Y(s) using the partial fraction decomposition:

Y(s) = [1 / (5s) - s / [5(s² + 25)]][1 - e^(-(3π/5)s)] + 5 / (s² + 25)
     = 1 / (5s) - s / [5(s² + 25)] - (e^(-(3π/5)s)) / (5s) + (s e^(-(3π/5)s)) / [5(s² + 25)] + 5 / (s² + 25)

Applying the inverse Laplace transform to each term:

y(t) = L⁻¹{Y(s)}
     = L⁻¹{1 / (5s)} - L⁻¹{s / [5(s² + 25)]} - L⁻¹{(e^(-(3π/5)s)) / (5s)} + L⁻¹{(s e^(-(3π/5)s)) / [5(s² + 25)]} + L⁻¹{5 / (s² + 25)}

Using standard Laplace transform pairs, we get:

y(t) = (1/5)L⁻¹{1/s} - (1/5)L⁻¹{s / (s² + 25)} - (1/5)L⁻¹{(e^(-(3π/5)s)) / s} + (1/5)L⁻¹{(s e^(-(3π/5)s)) / (s² + 25)} + L⁻¹{5 / (s² + 25)}
     = (1/5)(1) - (1/5)cos(5t) - (1/5)u(t - (3π/5)) + (1/5)cos(5(t - (3π/5)))u(t - (3π/5)) + sin(5t)

5.3. Simplifying the Solution

Finally, we can simplify the solution:

y(t) = 1/5 - (1/5)cos(5t) + sin(5t) - (1/5)u(t - (3π/5)) + (1/5)cos(5t - 3π)u(t - (3π/5))

Since cos(5t - 3π) = cos(5t - π) = -cos(5t), the final solution is:

y(t) = 1/5 - (1/5)cos(5t) + sin(5t) - (1/5)u(t - (3π/5)) - (1/5)cos(5t)u(t - (3π/5))

6. Understanding the Solution

The solution y(t) is a piecewise function that describes the behavior of the system over time. The terms involving the unit step function u(t - (3π/5)) represent the change in the system's response after t = 3π/5. Specifically, we can write the solution as:

y(t) = { 1/5 - (1/5)cos(5t) + sin(5t), 0 ≤ t < (3π/5) ; 1/5 - (1/5)cos(5t) + sin(5t) - (1/5)[1 + cos(5t)], t ≥ (3π/5)

For 0 ≤ t < (3π/5), the solution is a combination of sinusoidal functions and a constant term. After t = 3π/5, the forcing function f(t) becomes 0, and the solution changes accordingly. This demonstrates how the initial value problem captures the dynamic behavior of the system in response to the forcing function and initial conditions.

Conclusion

Solving initial value problems can seem daunting, but by breaking them down into steps and utilizing powerful tools like the Laplace transform, we can find elegant solutions. This guide demonstrated how to solve a second-order linear differential equation with a piecewise forcing function and initial conditions. Remember to practice these techniques to master them and apply them confidently in various applications. For further information on differential equations and Laplace transforms, visit a trusted resource like Khan Academy's Differential Equations Course.