Solving For Y(s) Laplace Transform Of Initial Value Problem

In this article, we will delve into the process of solving for $Y(s)$, which represents the Laplace transform of the solution $y(t)$ to a given initial value problem. Specifically, we will focus on the second-order linear ordinary differential equation with constant coefficients, a common type of problem encountered in various fields of engineering and physics. The initial value problem under consideration is:

$y^{\prime\prime} + 6y = 6t^2 - 7$, with initial conditions $y(0) = 0$ and $y^{\prime}(0) = -9$.

This problem involves finding a function $y(t)$ that satisfies the differential equation and the given initial conditions. To accomplish this, we will employ the powerful technique of Laplace transforms, which transforms the differential equation in the time domain ($t$) into an algebraic equation in the complex frequency domain ($s$). By solving for $Y(s)$ in the $s$-domain, we can then use the inverse Laplace transform to obtain the solution $y(t)$ in the time domain. This method provides a systematic and efficient way to solve linear ordinary differential equations, particularly those with initial conditions.

The Laplace transform is a valuable tool in numerous engineering disciplines, including electrical engineering, mechanical engineering, and control systems. It allows engineers to analyze and design systems by considering their behavior in the frequency domain, which can provide insights that are not readily apparent in the time domain. Understanding the Laplace transform and its applications is crucial for any engineer dealing with dynamic systems and control problems. The ability to solve initial value problems using Laplace transforms is a fundamental skill that enables engineers to predict the behavior of systems and design controllers to meet specific performance requirements. This article aims to provide a comprehensive guide to solving for $Y(s)$ in the given initial value problem, equipping readers with the necessary knowledge and skills to tackle similar problems.

The Laplace transform is a mathematical tool that transforms a function of time, $y(t)$, into a function of a complex variable, $s$. This transformation is particularly useful for solving linear differential equations, as it converts the differential equation into an algebraic equation, which is generally easier to solve. The Laplace transform of a function $y(t)$ is defined as:

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

where $s$ is a complex variable ($s = \sigma + j\omega$). The Laplace transform exists for functions that satisfy certain conditions, such as being piecewise continuous and of exponential order. The inverse Laplace transform, denoted by $\mathcal{L}^{-1}{Y(s)}$, transforms a function in the $s$-domain back to the time domain.

One of the key advantages of using Laplace transforms is that they simplify the process of solving differential equations, especially those with initial conditions. The Laplace transform of derivatives has a particularly useful property:

$\mathcal{L}{y^{\prime}(t)} = sY(s) - y(0)$

$\mathcal{L}{y^{\prime\prime}(t)} = s^2Y(s) - sy(0) - y^{\prime}(0)$

These properties allow us to transform differential equations into algebraic equations in the $s$-domain. For example, consider a second-order linear differential equation with constant coefficients:

$ay^{\prime\prime}(t) + by^{\prime}(t) + cy(t) = f(t)$

Applying the Laplace transform to both sides of the equation, we get:

$a[s^2Y(s) - sy(0) - y^{\prime}(0)] + b[sY(s) - y(0)] + cY(s) = F(s)$

where $F(s)$ is the Laplace transform of $f(t)$. This equation can be rearranged to solve for $Y(s)$:

$Y(s) = \frac{F(s) + a[sy(0) + y^{\prime}(0)] + by(0)}{as^2 + bs + c}$

Once we have $Y(s)$, we can find the solution $y(t)$ by taking the inverse Laplace transform:

$y(t) = \mathcal{L}^{-1}{Y(s)}$

The Laplace transform method is particularly powerful because it incorporates initial conditions directly into the algebraic equation, making it easier to solve initial value problems. It also provides a systematic way to handle various types of forcing functions, $f(t)$, such as step functions, impulse functions, and sinusoidal functions. Understanding the properties and applications of the Laplace transform is essential for engineers and scientists working with dynamic systems and control problems. This article will demonstrate how to apply the Laplace transform to solve the given initial value problem, providing a step-by-step guide to finding $Y(s)$ and, subsequently, $y(t)$.

To effectively apply the Laplace transform method, it is crucial to clearly define the problem statement and identify the initial conditions. In our case, we are given the following second-order linear ordinary differential equation with constant coefficients:

$y^{\prime\prime} + 6y = 6t^2 - 7$

This equation describes the relationship between the unknown function $y(t)$ and its second derivative $y^{\prime\prime}(t)$. The term $6t^2 - 7$ represents the forcing function, which is an external influence on the system. To obtain a unique solution for $y(t)$, we need to specify two initial conditions, as the equation is of second order. The initial conditions provided are:

$y(0) = 0$

$y^{\prime}(0) = -9$

The first initial condition, $y(0) = 0$, specifies the value of the function $y(t)$ at time $t = 0$. In this case, the function starts at zero. The second initial condition, $y^{\prime}(0) = -9$, specifies the value of the first derivative of $y(t)$ at time $t = 0$. This represents the initial rate of change of the function. These initial conditions are essential for determining the particular solution to the differential equation, as they provide specific constraints that the solution must satisfy.

The initial value problem consists of the differential equation and the initial conditions. Solving the initial value problem means finding a function $y(t)$ that satisfies both the differential equation and the initial conditions. This function describes the behavior of the system over time, taking into account the initial state and the external influences. The Laplace transform method is particularly well-suited for solving initial value problems because it incorporates the initial conditions directly into the transformation process. This allows us to solve for the Laplace transform of the solution, $Y(s)$, without having to deal with undetermined constants, as would be the case with traditional methods of solving differential equations. In the following sections, we will apply the Laplace transform to the given differential equation and use the initial conditions to find $Y(s)$. This will involve using the properties of the Laplace transform and algebraic manipulation to solve for $Y(s)$ in the complex frequency domain. The result, $Y(s)$, will then be used to find the solution $y(t)$ in the time domain using the inverse Laplace transform.

To solve the given initial value problem using the Laplace transform, the first step is to apply the Laplace transform to both sides of the differential equation:

$y^{\prime\prime} + 6y = 6t^2 - 7$

Using the linearity property of the Laplace transform, we can transform each term separately:

$\mathcal{L}{y^{\prime\prime}} + 6\mathcal{L}{y} = 6\mathcal{L}{t^2} - 7\mathcal{L}{1}$

Now, we need to use the Laplace transform properties for derivatives and basic functions. Recall the following properties:

• $\mathcal{L}{y^{\prime\prime}(t)} = s^2Y(s) - sy(0) - y^{\prime}(0)$
• $\mathcal{L}{y(t)} = Y(s)$
• $\mathcal{L}{t^n} = \frac{n!}{s^{n+1}}$ (for $n = 0, 1, 2, ...$)
• $\mathcal{L}{1} = \frac{1}{s}$

Applying these properties, we get:

$[s^2Y(s) - sy(0) - y^{\prime}(0)] + 6Y(s) = 6\frac{2!}{s^3} - 7\frac{1}{s}$

Substituting the given initial conditions $y(0) = 0$ and $y^{\prime}(0) = -9$, we have:

$[s^2Y(s) - s(0) - (-9)] + 6Y(s) = \frac{12}{s^3} - \frac{7}{s}$

Simplifying the equation, we get:

$s^2Y(s) + 9 + 6Y(s) = \frac{12}{s^3} - \frac{7}{s}$

Now, we need to isolate $Y(s)$ to solve for it in the $s$-domain. This involves algebraic manipulation of the equation. The next step is to collect the terms containing $Y(s)$ on one side of the equation and move the other terms to the other side. This will allow us to factor out $Y(s)$ and solve for it explicitly. This process is a crucial part of using Laplace transforms to solve differential equations, as it transforms the differential equation into an algebraic equation in the $s$-domain, which is much easier to solve.

Continuing from the previous step, we have the equation:

$s^2Y(s) + 9 + 6Y(s) = \frac{12}{s^3} - \frac{7}{s}$

Now, we want to isolate $Y(s)$. First, move the constant term 9 to the right side of the equation:

$s^2Y(s) + 6Y(s) = \frac{12}{s^3} - \frac{7}{s} - 9$

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

$Y(s)(s^2 + 6) = \frac{12}{s^3} - \frac{7}{s} - 9$

Now, divide both sides by $(s^2 + 6)$ to solve for $Y(s)$:

$Y(s) = \frac{\frac{12}{s^3} - \frac{7}{s} - 9}{s^2 + 6}$

To simplify this expression, multiply the numerator and denominator by $s^3$ to clear the fractions in the numerator:

$Y(s) = \frac{12 - 7s^2 - 9s^3}{s^3(s^2 + 6)}$

Rearrange the terms in the numerator to write the polynomial in descending order of powers of $s$:

$Y(s) = \frac{-9s^3 - 7s^2 + 12}{s^3(s^2 + 6)}$

This is the expression for $Y(s)$, the Laplace transform of the solution $y(t)$ to the given initial value problem. This expression is a rational function in $s$, and it represents the solution in the complex frequency domain. To find the solution $y(t)$ in the time domain, we would need to take the inverse Laplace transform of $Y(s)$. This often involves using partial fraction decomposition to break down the rational function into simpler terms that can be easily inverted using a table of Laplace transforms or other techniques. The process of finding $Y(s)$ is a crucial step in solving differential equations using Laplace transforms, as it transforms the problem from the time domain to the frequency domain, where algebraic techniques can be used to find the solution. The resulting expression for $Y(s)$ contains all the information needed to reconstruct the solution $y(t)$ in the time domain, including the effects of the initial conditions and the forcing function.

In conclusion, we have successfully solved for $Y(s)$, the Laplace transform of the solution $y(t)$ to the given initial value problem:

$y^{\prime\prime} + 6y = 6t^2 - 7$, with initial conditions $y(0) = 0$ and $y^{\prime}(0) = -9$.

By applying the Laplace transform to both sides of the differential equation, using the properties of Laplace transforms for derivatives and basic functions, and incorporating the initial conditions, we were able to transform the differential equation into an algebraic equation in the $s$-domain. We then solved for $Y(s)$ by isolating it and simplifying the resulting expression. The final expression for $Y(s)$ is:

$Y(s) = \frac{-9s^3 - 7s^2 + 12}{s^3(s^2 + 6)}$

This expression represents the solution to the initial value problem in the complex frequency domain. To obtain the solution $y(t)$ in the time domain, we would need to take the inverse Laplace transform of $Y(s)$. This typically involves using partial fraction decomposition to break down the rational function into simpler terms that can be easily inverted using a table of Laplace transforms or other techniques. While finding $y(t)$ is beyond the scope of this article, the determination of $Y(s)$ is a crucial step in the process of solving differential equations using Laplace transforms. The method demonstrated here provides a systematic approach to solving linear ordinary differential equations with constant coefficients, particularly those with initial conditions. The Laplace transform is a powerful tool in engineering and mathematics, allowing us to transform differential equations into algebraic equations, which are often easier to solve. This technique is widely used in various fields, including electrical engineering, mechanical engineering, and control systems, for analyzing and designing dynamic systems.