Laplace Transform: Solving Differential Equations Step-by-Step

Hey everyone, let's dive into a cool mathematical tool: the Laplace Transform. We'll explore how it helps us solve those tricky things called differential equations. Specifically, we'll tackle the equation $y' + y = e^t u(t)$, with the initial condition $y(0) = 1$. This guide will break down each step, making the process understandable, even if you're just getting started with Laplace transforms. So, grab your coffee, and let's get started!

Understanding the Laplace Transform

Alright guys, before we jump into the nitty-gritty, let's get a handle on what a Laplace transform actually is. Think of it as a super-smart translator. It takes a function of time, let's say $f(t)$, and transforms it into a function of a complex variable, usually denoted as $s$. This new function, which we'll call $F(s)$, lives in the frequency domain. The beauty of this transformation lies in its ability to turn differential equations (which can be tough to solve directly) into algebraic equations (which are much easier to handle). It's like converting a complex puzzle into a simple equation that can be easily solved. The Laplace transform is defined by the following integral:

$$F(s) = \mathcal{L} \{f(t)\} = \int_0^\infty e^{-st}f(t) dt$$

This integral takes your time-domain function, $f(t)$, multiplies it by a decaying exponential $e^{-st}$, and integrates over all positive time. The result is a function of $s$. Don't let the integral scare you; we usually don't have to calculate it from scratch. We'll use tables and properties to make our lives easier. The Laplace transform is particularly useful for solving linear, constant-coefficient differential equations, which are very common in physics and engineering. Why? Because it turns derivatives into algebraic expressions involving $s$, making it simple to solve for $F(s)$, and then we can use the inverse Laplace transform to get back our solution in the time domain, $y(t)$. This method is especially useful when dealing with discontinuous functions or when the forcing function (the right-hand side of the equation) is complicated. Understanding this translation process is the first key step to mastering the Laplace transform. We are basically moving from one mathematical world to a much easier-to-solve world and coming back.

Applying the Laplace Transform to Our Equation

Now, let's get our hands dirty and apply the Laplace transform to our specific differential equation: $y' + y = e^t u(t)$, with the initial condition $y(0) = 1$. Remember, $u(t)$ is the unit step function, which is 0 for $t < 0$ and 1 for $t \geq 0$. The initial condition, $y(0) = 1$, is super important as it provides the initial value of the function at $t=0$. This information will be needed when applying the Laplace transform to the derivative. Here's the breakdown, step by step:

1. Transforming the equation: We start by taking the Laplace transform of both sides of the equation. Using the linearity property of the Laplace transform, we get:

   $$\mathcal{L} \{y' + y\} = \mathcal{L} \{e^t u(t)\}$$

   $$\mathcal{L} \{y'\} + \mathcal{L} \{y\} = \mathcal{L} \{e^t u(t)\}$$

2. Transforming the derivative: The Laplace transform of a derivative, $y'(t)$, is given by:$\mathcalL} {y'(t)} = sY(s) - y(0)$. Where $Y(s)$ is the Laplace transform of $y(t)$. Using the initial condition $y(0) = 1$, we get $\mathcal{L {y'(t)} = sY(s) - 1$

3. Transforming $y(t)$: The Laplace transform of $y(t)$ is simply $Y(s)$.

4. Transforming the forcing function: The Laplace transform of $e^t u(t)$ is $\frac{1}{s-1}$. This is something we can find in a Laplace transform table. The unit step function, $u(t)$, helps define the function for $t \geq 0$.

5. Putting it all together: Now, we can substitute these results back into the transformed equation:

   $$(sY(s) - 1) + Y(s) = \frac{1}{s-1}$$

See? We've successfully transformed our differential equation into an algebraic one, which is way easier to work with. We have moved the derivative term into a simpler algebraic form which is easy to manage. That's the magic of the Laplace transform.

Solving for Y(s)

Alright, now that we've transformed our differential equation into an algebraic one, it's time to solve for $Y(s)$. This is where we isolate $Y(s)$ and express it as a function of $s$. This is the crucial step that prepares us to move back to the time domain. Let's rearrange the equation from the previous step:

$$(sY(s) - 1) + Y(s) = \frac{1}{s-1}$$

First, group all the terms involving $Y(s)$ on one side and move the other terms to the other side:

$$sY(s) + Y(s) = \frac{1}{s-1} + 1$$

Next, factor out $Y(s)$ from the left side:

$$Y(s)(s + 1) = \frac{1}{s-1} + 1$$

Now, to isolate $Y(s)$, divide both sides by $(s + 1)$:

$$Y(s) = \frac{\frac{1}{s-1} + 1}{s + 1}$$

Let's simplify the expression on the right-hand side. First, let's combine the terms in the numerator:

$$Y(s) = \frac{\frac{1 + (s-1)}{s-1}}{s + 1}$$

Which simplifies to:

$$Y(s) = \frac{s}{(s-1)(s+1)}$$

So, we have found the Laplace transform of our solution, $Y(s) = \frac{s}{(s-1)(s+1)}$. The next step is to bring this expression back to the time domain, we are almost there, keep going!

Inverse Laplace Transform and the Solution

We have transformed our differential equation, solved for $Y(s)$, and now it's time to return to the time domain to find our solution, $y(t)$. To do this, we need to apply the inverse Laplace transform, denoted as $\mathcal{L}^{-1}$. The goal is to find a function of time, $y(t)$, whose Laplace transform is $Y(s)$. Looking at $Y(s) = \frac{s}{(s-1)(s+1)}$, we can use partial fraction decomposition to make it easier to find the inverse Laplace transform. Partial fraction decomposition breaks down a complex fraction into a sum of simpler fractions. So we decompose our expression like this:

$$\frac{s}{(s-1)(s+1)} = \frac{A}{s-1} + \frac{B}{s+1}$$

To find the constants $A$ and $B$, we multiply both sides by $(s-1)(s+1)$:

$$s = A(s+1) + B(s-1)$$

Now, we can solve for $A$ and $B$. Let's substitute values for $s$:

• If $s = 1$, we get: $1 = A(1+1) + B(1-1) \implies 1 = 2A \implies A = \frac{1}{2}$.
• If $s = -1$, we get: $-1 = A(-1+1) + B(-1-1) \implies -1 = -2B \implies B = \frac{1}{2}$.

Now, substitute the values of $A$ and $B$ back into our decomposed fraction:

$$Y(s) = \frac{\frac{1}{2}}{s-1} + \frac{\frac{1}{2}}{s+1}$$

Now, we can take the inverse Laplace transform of each term. From our Laplace transform tables, we know that:

• $\mathcal{L}^{-1} \{\frac{1}{s-1}\} = e^t$
• $\mathcal{L}^{-1} \{\frac{1}{s+1}\} = e^{-t}$

So, applying the inverse Laplace transform to $Y(s)$, we get:

$$y(t) = \frac{1}{2}e^t + \frac{1}{2}e^{-t}$$

And there you have it! We have successfully solved the differential equation using the Laplace transform. The solution is a combination of an increasing and a decaying exponential function. This is the power of the Laplace transform: turning a complex differential equation into an easy-to-solve algebraic problem and then getting the solution back in the time domain.

Conclusion

So, we went through the entire process, from understanding the basics of the Laplace transform, applying it to our differential equation, solving for $Y(s)$, and finally, using the inverse Laplace transform to find the solution $y(t)$. This method provides a powerful tool for solving various differential equations, especially those with complex forcing functions or initial conditions. The Laplace transform simplifies the solution process by converting differential equations into algebraic equations, which are much easier to manipulate. Keep practicing, and you'll become a pro at solving differential equations using the Laplace transform in no time! Now you know how it works, guys, happy solving!