Solving Differential Equations Using **Laplace Transform** A Step-by-Step Guide

Hey guys! Today, we're diving into the fascinating world of Laplace transforms and how we can wield them to solve initial-value problems. Specifically, we'll be tackling the equation:

$y^{\prime \prime}-y^{\prime}=e^t \cos t, \quad y(0)=0, y^{\prime}(0)=0$

So, buckle up and let's embark on this mathematical journey together!

Understanding Laplace Transforms

Before we jump into solving the problem, let's take a moment to understand what Laplace transforms are and why they're so darn useful. In essence, the Laplace transform is a mathematical tool that converts a function of time, $f(t)$, into a function of a complex variable, $s$. This transformation often simplifies the process of solving differential equations, particularly those with initial conditions.

The Laplace transform of a function $f(t)$, denoted as $F(s)$ or $\mathcal{L}{f(t)}$, is defined by the integral:

$F(s) = \mathcal{L}{f(t)} = \int_{0}^{\infty} e^{-st} f(t) dt$

where $s$ is a complex number. The magic of Laplace transforms lies in their ability to turn differentiation into multiplication. This means that when we apply the Laplace transform to a differential equation, we often end up with an algebraic equation in the $s$-domain, which is much easier to solve. Once we've found the solution in the $s$-domain, we can use the inverse Laplace transform to get back to the time domain and obtain our solution $y(t)$. The initial-value problems, which are commonly encountered in physics and engineering, are particularly well-suited for the Laplace transform method because the initial conditions are naturally incorporated into the transformation process. In control systems, the Laplace transform is used to analyze the stability and response of systems to various inputs. In circuit analysis, it simplifies the solution of circuits with capacitors and inductors by converting differential equations into algebraic equations. The Laplace transform is also widely used in signal processing to analyze and design filters, and in probability theory for analyzing probability distributions. The transform is particularly effective for linear time-invariant systems, where it provides a straightforward way to determine the system's output for a given input. The properties of the Laplace transform, such as linearity, time shifting, and differentiation, make it a powerful tool for solving complex problems. By transforming differential equations into algebraic equations, the Laplace transform allows engineers and scientists to solve problems more efficiently. The inverse Laplace transform is then used to convert the solution back to the original domain, providing the time-domain response of the system.

Applying Laplace Transform to the Given Equation

Now, let’s roll up our sleeves and apply the Laplace transform to our given initial-value problem:

$y^{\prime \prime}-y^{\prime}=e^t \cos t, \quad y(0)=0, y^{\prime}(0)=0$

First, we take the Laplace transform of both sides of the equation. Remember the following key properties of Laplace transforms:

• $\mathcal{L}{y^{\prime}(t)} = sY(s) - y(0)$
• $\mathcal{L}{y^{\prime \prime}(t)} = s^2Y(s) - sy(0) - y^{\prime}(0)$
• $\mathcal{L}{e^{at}\cos(bt)} = \frac{s-a}{(s-a)^2 + b^2}$

where $Y(s)$ is the Laplace transform of $y(t)$.

Applying these properties, we get:

$\mathcal{L}{y^{\prime \prime}-y^{\prime}} = \mathcal{L}{e^t \cos t}$

$s^2Y(s) - sy(0) - y^{\prime}(0) - [sY(s) - y(0)] = \frac{s-1}{(s-1)^2 + 1}$

Given the initial conditions $y(0) = 0$ and $y^{\prime}(0) = 0$, the equation simplifies to:

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

Now, we factor out $Y(s)$:

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

Next, we solve for $Y(s)$ by dividing both sides by $(s^2 - s)$:

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

We can further simplify the denominator by factoring $s$ from $(s^2 - s)$:

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

Now, cancel out the $(s-1)$ terms:

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

This is our solution in the $s$-domain. The equation we've arrived at, $Y(s) = \frac{1}{s[(s-1)^2 + 1]}$, represents the Laplace transform of the solution to the initial-value problem. To find the solution $y(t)$ in the time domain, we need to compute the inverse Laplace transform of $Y(s)$. This involves finding a function $y(t)$ such that when we apply the Laplace transform to it, we get $Y(s)$. The process of finding the inverse Laplace transform often involves techniques such as partial fraction decomposition and the use of Laplace transform tables. These tables provide the Laplace transforms of common functions, which can be matched with the terms in the partial fraction decomposition to find the corresponding time-domain functions. The properties of the Laplace transform, such as linearity and the transforms of derivatives and integrals, are essential in this process. The initial conditions, $y(0) = 0$ and $y^{\prime}(0) = 0$, played a crucial role in simplifying the equation in the $s$-domain. They allowed us to eliminate terms involving $y(0)$ and $y^{\prime}(0)$, which greatly simplified the algebraic manipulations required to solve for $Y(s)$. Without these initial conditions, the equation would have been significantly more complex to solve. The next step, computing the inverse Laplace transform, will bring us back to the time domain, where we will obtain the solution $y(t)$ that satisfies the original differential equation and the given initial conditions. This solution will describe the behavior of the system over time, providing valuable insights into its dynamics.

Partial Fraction Decomposition

To find the inverse Laplace transform, we'll use partial fraction decomposition. We want to express $Y(s)$ as a sum of simpler fractions:

$Y(s) = \frac{1}{s[(s-1)^2 + 1]} = \frac{A}{s} + \frac{Bs + C}{(s-1)^2 + 1}$

To find the constants $A$, $B$, and $C$, we multiply both sides by the denominator $s[(s-1)^2 + 1]$:

$1 = A[(s-1)^2 + 1] + (Bs + C)s$

Expanding and collecting terms, we get:

$1 = A(s^2 - 2s + 2) + Bs^2 + Cs$

$1 = (A + B)s^2 + (-2A + C)s + 2A$

Now, we equate the coefficients of the powers of $s$:

• $s^2: A + B = 0$
• $s^1: -2A + C = 0$
• $s^0: 2A = 1$

From the third equation, we find $A = \frac{1}{2}$. Substituting $A$ into the first equation, we get $B = -\frac{1}{2}$. Substituting $A$ into the second equation, we find $C = 1$. Thus, our partial fraction decomposition is:

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

We can rewrite the second term to make it easier to find the inverse Laplace transform:

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

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

Partial fraction decomposition is a crucial step in finding the inverse Laplace transform when dealing with complex rational functions. By breaking down the original fraction into simpler terms, we can easily identify the corresponding time-domain functions using Laplace transform tables. The method involves expressing the denominator of the rational function as a product of linear and irreducible quadratic factors, and then writing the fraction as a sum of simpler fractions, each with one of these factors as its denominator. The coefficients of these fractions are then determined by solving a system of linear equations, which is obtained by equating the coefficients of the powers of $s$ in the numerators. In our case, the denominator was $s[(s-1)^2 + 1]$, which factors into a linear term $s$ and an irreducible quadratic term $(s-1)^2 + 1$. This allowed us to express $Y(s)$ as a sum of three simpler fractions, each of which has a known inverse Laplace transform. The accuracy of the partial fraction decomposition is critical for obtaining the correct solution. Any error in the decomposition will propagate through the remaining steps and lead to an incorrect time-domain solution. Therefore, it is essential to carefully check the decomposition by recombining the fractions and verifying that the result matches the original fraction. The decomposition not only simplifies the inverse Laplace transform process but also provides insights into the system's behavior. Each term in the decomposition corresponds to a particular mode or component of the system's response. By examining these terms, we can understand the contributions of different factors to the overall behavior of the system.

Inverse Laplace Transform

Now, we find the inverse Laplace transform of each term. We'll use the following transforms:

• $\mathcal{L}^{-1}{\frac{1}{s}} = 1$
• $\mathcal{L}^{-1}{\frac{s-a}{(s-a)^2 + b^2}} = e^{at}\cos(bt)$
• $\mathcal{L}^{-1}{\frac{b}{(s-a)^2 + b^2}} = e^{at}\sin(bt)$

Applying these, we get:

$y(t) = \mathcal{L}^{-1}{Y(s)} = \mathcal{L}^{-1}{\left[\frac{1}{2s} - \frac{1}{2} \cdot \frac{s-1}{(s-1)^2 + 1} + \frac{1}{2} \cdot \frac{1}{(s-1)^2 + 1}\right]}$

$y(t) = \frac{1}{2}\mathcal{L}^{-1}{\frac{1}{s}} - \frac{1}{2}\mathcal{L}^{-1}{\frac{s-1}{(s-1)^2 + 1}} + \frac{1}{2}\mathcal{L}^{-1}{\frac{1}{(s-1)^2 + 1}}$

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

So, our final solution is:

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

The inverse Laplace transform is the final step in solving an initial-value problem using the Laplace transform method. It involves converting the solution in the $s$-domain, $Y(s)$, back to the time domain, $y(t)$. This is done by finding a function $y(t)$ whose Laplace transform is $Y(s)$. The process often involves the use of Laplace transform tables, which list the Laplace transforms of common functions, and the properties of the Laplace transform, such as linearity, time shifting, and frequency shifting. In our case, we used the Laplace transform table to find the inverse Laplace transforms of $\frac{1}{s}$, $\frac{s-1}{(s-1)^2 + 1}$, and $\frac{1}{(s-1)^2 + 1}$. These correspond to the time-domain functions 1, $e^t\cos(t)$, and $e^t\sin(t)$, respectively. The linearity property of the Laplace transform allowed us to find the inverse Laplace transform of a sum of terms by simply summing the inverse Laplace transforms of the individual terms. This greatly simplified the process of finding the inverse Laplace transform of $Y(s)$. The final solution, $y(t) = \frac{1}{2} - \frac{1}{2}e^t\cos(t) + \frac{1}{2}e^t\sin(t)$, represents the time-domain solution to the initial-value problem. It describes the behavior of the system over time and satisfies both the differential equation and the initial conditions. This solution can be used to analyze the system's response to different inputs and to design control systems that achieve desired performance characteristics. The inverse Laplace transform is a powerful tool for solving linear differential equations, particularly those with constant coefficients. It provides a systematic approach for finding solutions and is widely used in engineering and physics.

Conclusion

And there you have it, guys! We've successfully used the Laplace transform to solve the given initial-value problem. Remember, the Laplace transform is a powerful tool that can simplify the process of solving differential equations. By transforming the problem into the $s$-domain, we can often solve algebraic equations instead of differential equations, making the whole process much more manageable. Keep practicing, and you'll become a Laplace transform pro in no time!