15.14. Laplace Transform in Solving Differential Equations

Consider a second-order linear ODE with constant coefficients:

d²y/dt² + b dy/dt + cy = f(t), where y(0) = y0, y'(0) = y1.

This equation represents a second-order linear ordinary differential equation (ODE) with constant coefficients. The terms represent the second derivative of y with respect to t, the first derivative of y, and the function f(t) which is the input to the system. The constants a, b, and c define the dynamic characteristics of the system, while y(0) and y'(0) are the initial conditions of the function and its first derivative, respectively.

Think of this equation as a model for a car's motion where y(t) is the position of the car over time. The coefficients relate to factors like acceleration (a), friction (b), and potential energy (c) that influence how the car moves, while the initial conditions indicate the car's starting position and speed.

Apply Laplace Transform:

a[s²Y(s)−sy0−y1] + b[sY(s)−y0] + cY(s) = F(s)

To solve the ODE using the Laplace transform, we take the transformation of each term. The terms involving derivatives transform according to specific properties of the Laplace transform: the second derivative leads to s²Y(s) with the initial conditions factored in. The result is an algebraic equation in the Laplace domain (Y(s)), which can be solved for Y(s). Here, F(s) is the Laplace transform of the input function f(t).

Imagine translating the dynamics of a car into a language that only talks about speed and position without directly mentioning time—this transition to Laplace transform does just that, allowing us to manipulate the relationships in a clearer, often simpler, way.

Solve algebraically for Y(s), then find the inverse Laplace transform to obtain y(t).

Once we have expressed the left side as a function of Y(s), we can rearrange the equation to isolate Y(s). This step is typically followed by algebraic manipulations, like factoring, to make the expression simpler. After obtaining Y(s), the next step is to take the inverse Laplace transform, which allows us to revert from the Laplace domain back to the time domain, revealing y(t).

It’s similar to first solving an equation that is in a different system (like Celsius to Fahrenheit) and then converting it back to the system we understand (like reconverting Fahrenheit back to Celsius). We first manipulate it comfortably in the new system (Laplace) before interpreting results back in the original terms (time).

Example Solve: y′′ + 3y′ + 2y = e^−t, y(0) = 0, y′(0) = 0 Step 1: Take Laplace on both sides:

1/(s² + 3s + 2) = (s + 1)/(s + 1) Step 2: Factor and solve:

Y(s) = 1/[(s + 1)(s + 1)(s + 2)] Apply partial fractions, then take inverse Laplace to get y(t).

This example illustrates the procedure using a specific second-order linear ODE. After taking the Laplace transform of each term, we obtain a rational function in Y(s). The next step is to factor this expression appropriately and use partial fraction decomposition to make taking the inverse transform feasible. Finally, applying the inverse Laplace transform gives the original function in time.

Think of it as assembling a complex LEGO structure. First, you sort out the pieces (Laplace transform), then you figure out how to fit them together in a smart way (solving the equation), and finally, you step back to see the whole completed structure (inverse Laplace transform). This breakdown makes it easier to understand how each individual part connects to create the whole.

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