Solving For Y(s) Laplace Transform Of A Differential Equation

#Introduction

In the realm of engineering and mathematics, solving differential equations is a fundamental task. The Laplace transform provides a powerful tool for tackling such problems, particularly those with initial conditions. This article delves into the process of solving for Y(s), the Laplace transform of the solution y(t), to a given initial value problem. We will explore the application of Laplace transforms, inverse Laplace transforms, and partial fraction decomposition to arrive at the solution. Let us consider the following initial value problem:

$\qquad y^{\prime \prime}+6 y=6 t^2-7, \quad y(0)=0, \quad y^{\prime}(0)=-9$

This second-order linear ordinary differential equation (ODE) is accompanied by two initial conditions: y(0) = 0 and y'(0) = -9. These conditions are crucial for determining the unique solution to the problem. Our goal is to find Y(s), the Laplace transform of y(t), which will allow us to ultimately find the solution y(t) in the time domain.

The Laplace transform is a mathematical transformation that converts a function of time, t, into a function of complex frequency, s. This transformation is particularly useful for solving linear differential equations because it converts them into algebraic equations, which are generally easier to solve. Once we have solved for Y(s), we can use the inverse Laplace transform to obtain the solution y(t) in the time domain. This process involves utilizing Laplace transform properties, algebraic manipulation, and techniques such as partial fraction decomposition.

Laplace Transform Properties and Application

The Laplace transform is a powerful tool for solving linear differential equations with constant coefficients. It converts differential equations into algebraic equations, making them easier to solve. The Laplace transform of a function f(t), denoted by F(s), is defined as:

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

where s is a complex variable. The inverse Laplace transform, denoted by $\mathcal{L}^{-1}{F(s)}$, recovers the original function f(t) from its Laplace transform F(s).

To solve the given initial value problem, we will utilize several key properties of the Laplace transform. These include the linearity property, the transform of derivatives, and the transforms of common functions. The linearity property states that the Laplace transform of a linear combination of functions is the linear combination of their Laplace transforms. The transform of derivatives is particularly important for solving differential equations. It states that:

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

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

where Y(s) is the Laplace transform of y(t). These properties allow us to convert the derivatives in the differential equation into algebraic terms involving Y(s). Additionally, we need the Laplace transforms of common functions, such as t^n and constants. The Laplace transform of t^n is given by:

$\qquad \mathcal{L}{t^n} = \frac{n!}{s^{n+1}}$

and the Laplace transform of a constant c is:

$\qquad \mathcal{L}{c} = \frac{c}{s}$

Applying these properties, we can transform the given differential equation and initial conditions into an algebraic equation in terms of Y(s). This algebraic equation can then be solved for Y(s), which is the Laplace transform of the solution y(t). The subsequent step involves finding the inverse Laplace transform of Y(s) to obtain the solution y(t) in the time domain.

Solving the Initial Value Problem

Let's apply the Laplace transform to the given initial value problem:

$\qquad y^{\prime \prime}+6 y=6 t^2-7, \quad y(0)=0, \quad y^{\prime}(0)=-9$

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

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

Using the properties of Laplace transforms, we have:

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

Substituting the initial conditions y(0) = 0 and y'(0) = -9, we get:

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

Simplifying the equation, we have:

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

Now, we solve for Y(s):

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

$\qquad Y(s) = \frac{1}{s^2 + 6} \left(\frac{12}{s^3} - \frac{7}{s} - 9\right)$

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

This expression for Y(s) represents the Laplace transform of the solution y(t). To find the solution y(t) in the time domain, we need to find the inverse Laplace transform of Y(s). This often involves using partial fraction decomposition to break down the complex rational functions into simpler terms that we can find in a table of Laplace transforms.

Partial Fraction Decomposition

To find the inverse Laplace transform of Y(s), we first need to decompose the rational functions using partial fraction decomposition. This technique allows us to express a complex rational function as a sum of simpler fractions, each of which has a known inverse Laplace transform. Let's consider the first term in Y(s):

$\qquad \frac{12}{s^3(s^2 + 6)}$

We can decompose this fraction as follows:

$\qquad \frac{12}{s^3(s^2 + 6)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{Ds + E}{s^2 + 6}$

Multiplying both sides by s^3(s2 + 6), we get:

$\qquad 12 = A s^2(s^2 + 6) + B s(s^2 + 6) + C(s^2 + 6) + (Ds + E)s^3$

Expanding and collecting like terms, we have:

$\qquad 12 = (A + D)s^4 + (B + E)s^3 + (6A + C)s^2 + 6Bs + 6C$

Equating coefficients, we obtain the following system of equations:

$\qquad A + D = 0$ $\qquad B + E = 0$ $\qquad 6A + C = 0$ $\qquad 6B = 0$ $\qquad 6C = 12$

Solving this system, we find:

$\qquad C = 2$ $\qquad B = 0$ $\qquad A = -\frac{1}{3}$ $\qquad D = \frac{1}{3}$ $\qquad E = 0$

Thus, the partial fraction decomposition of the first term is:

$\qquad \frac{12}{s^3(s^2 + 6)} = -\frac{1}{3s} + \frac{2}{s^3} + \frac{s}{3(s^2 + 6)}$

Similarly, we decompose the second term in Y(s):

$\qquad \frac{7}{s(s^2 + 6)} = \frac{F}{s} + \frac{Gs + H}{s^2 + 6}$

$\qquad 7 = F(s^2 + 6) + (Gs + H)s$

$\qquad 7 = (F + G)s^2 + Hs + 6F$

Equating coefficients, we get:

$\qquad F + G = 0$ $\qquad H = 0$ $\qquad 6F = 7$

Solving this system, we find:

$\qquad F = \frac{7}{6}$ $\qquad G = -\frac{7}{6}$ $\qquad H = 0$

Thus, the partial fraction decomposition of the second term is:

$\qquad \frac{7}{s(s^2 + 6)} = \frac{7}{6s} - \frac{7s}{6(s^2 + 6)}$

Now we can rewrite Y(s) using these partial fraction decompositions.

Finding the Inverse Laplace Transform

Now that we have performed the partial fraction decomposition, we can rewrite Y(s) as:

$\qquad Y(s) = \left(-\frac{1}{3s} + \frac{2}{s^3} + \frac{s}{3(s^2 + 6)}\right) - \left(\frac{7}{6s} - \frac{7s}{6(s^2 + 6)}\right) - \frac{9}{s^2 + 6}$

Combining terms, we get:

$\qquad Y(s) = -\frac{1}{3s} + \frac{2}{s^3} + \frac{s}{3(s^2 + 6)} - \frac{7}{6s} + \frac{7s}{6(s^2 + 6)} - \frac{9}{s^2 + 6}$

$\qquad Y(s) = \left(-\frac{1}{3} - \frac{7}{6}\right)\frac{1}{s} + \frac{2}{s^3} + \left(\frac{1}{3} + \frac{7}{6}\right)\frac{s}{s^2 + 6} - \frac{9}{s^2 + 6}$

$\qquad Y(s) = -\frac{3}{2}\frac{1}{s} + \frac{2}{s^3} + \frac{3}{2}\frac{s}{s^2 + 6} - \frac{9}{s^2 + 6}$

Now, we find the inverse Laplace transform of each term. Recall the following Laplace transforms:

\qquad \mathcal{L}^{-1}\left{\frac{1}{s}\right} = 1 \qquad \mathcal{L}^{-1}\left{\frac{n!}{s^{n+1}}\right} = t^n \qquad \mathcal{L}^{-1}\left{\frac{s}{s^2 + a^2}\right} = \cos(at) \qquad \mathcal{L}^{-1}\left{\frac{a}{s^2 + a^2}\right} = \sin(at)

Applying these, we get:

\qquad y(t) = \mathcal{L}^{-1}{Y(s)} = -\frac{3}{2}\mathcal{L}^{-1}\left{\frac{1}{s}\right} + \mathcal{L}^{-1}\left{\frac{2}{s^3}\right} + \frac{3}{2}\mathcal{L}^{-1}\left{\frac{s}{s^2 + 6}\right} - 9\mathcal{L}^{-1}\left{\frac{1}{s^2 + 6}\right}

$\qquad y(t) = -\frac{3}{2}(1) + t^2 + \frac{3}{2}\cos(\sqrt{6}t) - 9\left(\frac{1}{\sqrt{6}}\sin(\sqrt{6}t)\right)$

$\qquad y(t) = -\frac{3}{2} + t^2 + \frac{3}{2}\cos(\sqrt{6}t) - \frac{9}{\sqrt{6}}\sin(\sqrt{6}t)$

Thus, the solution to the initial value problem is:

$\qquad y(t) = t^2 + \frac{3}{2}\cos(\sqrt{6}t) - \frac{3\sqrt{6}}{2}\sin(\sqrt{6}t) - \frac{3}{2}$

Conclusion

In this article, we have successfully solved for Y(s), the Laplace transform of the solution y(t), to the given initial value problem. We accomplished this by applying the Laplace transform to the differential equation, utilizing initial conditions, solving for Y(s), and employing partial fraction decomposition to facilitate the inverse Laplace transform. The final solution y(t) represents the time-domain response of the system described by the differential equation. The Laplace transform method provides a systematic approach to solving linear differential equations, making it a valuable tool in various engineering and mathematical applications.