Question

Consider the initial value problem \(y^{\prime \prime}+\alpha y^{\prime}+\beta y=0, y(0)=y_{0}, y^{\prime}(0)=y_{0}^{\prime} .\) The Laplace transform of the solution, \(Y(s)=\mathcal{L}\\{y(t)\\}\), is given. Determine the constants \(\alpha, \beta, y_{0}\), and \(y_{0}^{\prime}\). $$Y(s)=\frac{s}{(s+1)^{2}}$$

Short answer

Question: Determine the values of constants α, β, y₀, and y₀′ in the given ODE and initial conditions, where the solution Y(s) has a Laplace transform of Y(s) = s/((s+1)^2) and y(t) = te^(-t). Answer: The values of the constants are α = 1, β = 1, y₀ = -y₀′, however, y₀ and y₀′ remain unspecified.

Step-by-step solution

Inverse Laplace Transform of Y(s)

To find the time-domain function y(t), we must find the inverse Laplace transform of Y(s). This can be found by using the partial fraction expansion method or by referring to a table of Laplace transforms. $$ y(t) = \mathcal{L}^{-1}\\{Y(s)\\} = \mathcal{L}^{-1}\\{\frac{s}{(s+1)^{2}}\\} $$ $$ y(t) = te^{-t} $$

Represent the ODE and initial conditions using Laplace transforms

Now let's represent the given ODE and initial conditions using Laplace transforms: ODE: \(y^{\prime \prime}+\alpha y^{\prime}+\beta y = 0\) Taking the Laplace transform of the ODE, we get: \((s^2Y(s) - sy(0) - y'(0)) + \alpha (sY(s) - y(0)) + \beta Y(s) = 0\) Now, substitute the given initial conditions y(0) = y₀ and y′(0) = y₀′: \((s^2Y(s) - sy_0 - y_0') + \alpha (sY(s) - y_0) + \beta Y(s) = 0\) Now we plug the given expression for Y(s): \((s^2\frac{s}{(s+1)^{2}} - sy_0 - y_0') + \alpha (s\frac{s}{(s+1)^{2}} - y_0) + \beta \frac{s}{(s+1)^{2}} = 0\)

Compare expressions to determine values of α, β, y₀, and y₀′

Now let's compare the expressions obtained in steps 1 and 2. Note that the inverse Laplace transform of \(te^{-t}\) is \(\frac{s}{(s+1)^2}\). Hence we can write the equations as: \( sy_0 + y_0' = 0 \quad \text{(constant part)}\) \( \beta = 1, \quad \alpha = 1 \quad \text{(coefficient of } \frac{s}{(s+1)^2}\text{)}\) Now we have determined the constants α, β, y₀, and y₀′: $$\alpha = 1, \quad \beta = 1, \quad y_{0} = -y_{0}', \quad \text{but } y_{0} \text{ and } y_{0}' \text{ remain unspecified.}$$

Key concepts

Inverse Laplace Transform

When solving an initial value problem using Laplace transforms, we often start by finding a solution in the 's-domain' or frequency domain. However, to interpret this solution in the 'time domain', which is usually the domain we experience in physical problems, we must perform an inverse Laplace transform.

The inverse Laplace transform is a process that converts a complex-valued function of a complex variable 's' back into a real-valued function of time 't'. For example, if you're presented with a function of 's', like the one in our exercise \(Y(s)=\frac{s}{(s+1)^{2}}\), the inverse Laplace transform helps to revert it to a time-based function, which, through the process, turns out to be \(y(t) = te^{-t}\). This function is now something we can use to model real-time dynamics or understand the behavior of physical systems. In homework problems, tables of standard Laplace transforms are often used to simplify this process.

Partial Fraction Expansion

When working with Laplace transformations, sometimes we encounter complex functions in 's' that are not easy to invert back into the time domain. Partial fraction expansion is a method that decomposes such complex rational expressions into simpler, more manageable terms that correspond to known inverse transforms.

For instance, the function \(\frac{s}{(s+1)^{2}}\) can be tricky to invert directly, but by breaking it down into partial fractions, it can be expressed as a sum of simpler fractions whose inverses we know or can easily find. Although this step is not explicitly shown in the solution, understanding how to perform partial fraction expansion is crucial, especially when dealing with more complicated expressions.

Ordinary Differential Equation (ODE)

An ordinary differential equation (ODE) is an equation that involves functions and their derivatives. It is 'ordinary' because it deals with functions of a single variable, unlike partial differential equations that involve multiple variables. ODEs form the backbone of classical mathematics and appear frequently in many scientific fields, such as physics, engineering, and economics.

Our original problem presents an ODE where the unknown function (usually depicting physical quantities like displacement or concentration over time) and its derivatives are related through coefficients. This relation reflects the underlying laws of motion or change. The ODE from the exercise, \(y'' + \alpha y' + \beta y = 0\), is linear and homogenous, meaning its solution entails finding a function \(y(t)\) that satisfies this relation given its initial conditions.

Initial Conditions

When solving an ODE, we typically have initial conditions that provide specific values of the unknown function and/or its derivatives at a certain time, typically taken as \(t=0\). These conditions are vital for finding a unique solution to an ODE since, without them, an infinite number of functions could satisfy the equation.

The given initial conditions, \(y(0)=y_{0}\) and \(y'(0)=y_{0}'\), fix our solution to the particular scenario described by these starting values. In our exercise, even though the explicit values are not given initially, we deduce information about the relationship between \(y_0\) and \(y_0'\) from the structure of the ODE and the form of its Laplace transform. Initial conditions are often measurements or assumed values based on the physical context of the problem.