17.5. Solved Example

Problem: Solve the system using Laplace Transform:

dx/dt = 3x + 4y,

dy/dt = -4x + 3y,

x(0) = 1, y(0) = 0

In this section, we are presented with a problem that involves solving a system of simultaneous linear differential equations using the Laplace Transform method. The equations given are:

1. dx/dt = 3x + 4y
2. dy/dt = -4x + 3y

Additionally, we are provided with initial conditions: x(0) = 1 and y(0) = 0. These initial conditions will be crucial when applying the Laplace Transform because they allow us to find the specific solution to our equations.

Imagine you're trying to figure out how two interconnected water tanks behave over time. One tank fills up faster based on the levels in both tanks, while another tank drains water based on the water level in the first tank. The equations given are similar to the rules that govern how the water levels change, and the initial conditions tell us how much water is in each tank at the start.

Step 1: Take Laplace Transforms

L{dx/dt} = sX(s) - x(0), L{dy/dt} = sY(s) - y(0)

So the equations become:

sX(s) - 1 = 3X(s) + 4Y(s)

sY(s) - 0 = -4X(s) + 3Y(s)

In this step, we apply the Laplace Transform to both of the differential equations we formulated. By doing so, we convert the differential equations into algebraic equations in the s-domain. Applying the Laplace Transform involves using the property that transforms the derivative of a function into a function in the s-domain, adjusted for initial conditions:

• For dx/dt, we have L{dx/dt} = sX(s) - x(0)
• For dy/dt, we have L{dy/dt} = sY(s) - y(0)

This yields two new equations we can work with.

Think of the Laplace Transform as a way to map a changing situation, like a fluctuating water level in our tanks, into a simpler, static snapshot. Instead of tracking the chaos of changing water levels over time, we're reorganizing things so we can analyze them more easily while keeping track of the starting amounts.

Step 2: Rearranging the Equations

sX(s) - 3X(s) - 4Y(s) = 1 ⇒ (s - 3)X(s) - 4Y(s) = 1 (Equation 1) sY(s) + 4X(s) - 3Y(s) = 0 ⇒ 4X(s) + (s - 3)Y(s) = 0 (Equation 2)

Next, we rearrange the transformed equations to isolate terms involving X(s) and Y(s). For our first equation, after substituting in x(0) = 1, we isolate the term for X(s). In the second equation, we isolate Y(s). This gives us two algebraic equations (Equation 1 and Equation 2) that can be solved simultaneously.

Consider a recipe that gives different ingredients needed to prepare a dish. By rearranging the recipe and isolating certain values, you can determine how much of each ingredient you require. Here, we rearranged our equations similar to tweaking a recipe to make it easier to understand how much of each 'variable' we need.

Step 3: Solve Simultaneously

From equation (2): (s - 3) 4X(s) = - (s - 3)Y(s) ⇒ X(s) = - Y(s) / 4

Substituting into equation (1): (s - 3) [-((s - 3)/4)Y(s)] - 4Y(s) = 1

Now, we need to solve the two equations simultaneously. From Equation 2, we expressed X(s) in terms of Y(s). By substituting this expression for X(s) back into Equation 1, we can reduce the equation to just Y(s). This manipulation helps us find a specific form for Y(s) before we evaluate back to find X(s).

Think about trying to balance ingredients in a dish. If you know the ratio of one ingredient to another, you can substitute the known quantities to find out how much of another ingredient you should use. Here, we are substituting Y(s) value to simplify and find X(s) similarly.

Step 4: Take Inverse Laplace Transforms

Using standard transforms: L^{-1}{(s - a)/((s - a)^2 + b^2)} = e^{at}cos(bt) L^{-1}{b/((s - a)^2 + b^2)} = e^{at}sin(bt)

So: X(s) = (s - 3)/((s - 3)^2 + 4^2) ⇒ x(t) = e^{3t}cos(4t) Y(s) = -4/((s - 3)^2 + 4^2) ⇒ y(t) = -e^{3t}sin(4t)

After finding X(s) and Y(s), we now take the Inverse Laplace Transform to convert these algebraic expressions back into functions of time, x(t) and y(t). We've utilized known inverse transforms to achieve this, leading to our final solutions for x(t) and y(t). The transforming process allows us to interpret the outcome in real-world time-dependent behaviors.

Imagine flipping a pancake – the batter transforms from a liquid state into a solid form as it cooks. Similarly, while we worked with equations in the s-domain, we now transform them back into meaningful time functions, which allows us to see how the system behaves over time.

Final Answer: x(t) = e^{3t}cos(4t), y(t) = -e^{3t}sin(4t)

The final step provides the solution to the original system of differential equations. We have successfully found that x(t) = e^{3t}cos(4t) and y(t) = -e^{3t}sin(4t). This step demonstrates the effectiveness of the Laplace Transform technique in solving complex systems and showcases how we can express interactions of x and y in the time domain after going through a systematic approach.

Think about predicting the behavior of a pendulum as it swings over time. After going through the whole calculation and transformation, we now have precise predictions (x(t) and y(t)) for how the pendulum would behave at any moment in time, giving us valuable insights into the dynamics of the system.

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