Question

Use the Laplace transform to solve the given initial value problem. $$ y^{\prime \prime}-4 y^{\prime}+4 y=0 ; \quad y(0)=1, \quad y^{\prime}(0)=1 $$

Short answer

Based on the provided step-by-step solution, a short answer is as follows: The given initial value problem is a second-order linear homogeneous differential equation with initial conditions. To solve the problem using the Laplace transform, perform the following steps: 1. Apply the Laplace transform to the given differential equation and initial conditions. 2. Solve the transformed equation for the Laplace transform of the solution Y(s). 3. Compute the inverse Laplace transform of Y(s) to obtain the solution for y(t). Following these steps, the solution to the given initial value problem is: $$ y(t) = e^{2t} - 1 $$

Step-by-step solution

Apply the Laplace Transform

We begin by applying the Laplace transform to the given differential equation, keeping in mind the properties of the transform, such as linearity and derivatives: $$ \mathcal{L}\{y''-4y'+4y\}=\mathcal{L}\{0\} $$ Using the linearity property, we get $$ \mathcal{L}\{y''\}-4\mathcal{L}\{y'\}+4\mathcal{L}\{y\}=0 $$ Now, let us denote the Laplace transform of y(t) as Y(s): $$ Y(s)=\mathcal{L}\{y(t)\} $$ and apply the Laplace Transform properties for derivatives: $$ s^2Y(s)-sy(0)-y'(0)-4[ sY(s)-y(0)] + 4Y(s) = 0 $$ Substituting the initial values, $$y(0)=1$$ and $$y'(0)=1$$, we obtain: $$ s^2Y(s)-s-1-4[sY(s)-1] + 4Y(s) = 0 $$

Solve the transformed equation for Y(s)

Now we have a transformed equation that we can solve for Y(s): $$ s^2Y(s)-4sY(s)+4Y(s)-s+4-1=0 $$ Grouping the terms and simplifying: $$ Y(s)(s^2-4s+4)-s+3=0 $$ Now, we can solve for $$Y(s)$$: $$ Y(s)=\frac{s-3}{(s-2)^2} $$

Calculate the inverse Laplace transform

To obtain the solution to the original problem, we now compute the inverse Laplace transform of Y(s): $$ y(t)=\mathcal{L}^{-1}\left\{\frac{s-3}{(s-2)^2}\right\} $$ This inverse Laplace transform can be computed using partial fraction decomposition and the inverse Laplace transform table: $$ \frac{s-3}{(s-2)^2} = A + \frac{B}{s-2} $$ Solving for A and B, we find that $$A=-1$$ and $$B=1$$. Thus, we have: $$ \frac{s-3}{(s-2)^2} = -1 + \frac{1}{s-2} $$ Now, using the inverse Laplace transform table, we get: $$ y(t) = \mathcal{L}^{-1}\{-1\} + \mathcal{L}^{-1}\left\{ \frac{1}{s-2} \right\} $$ Finally, applying the table and summing up the results, the solution is: $$ y(t)= -(1) + e^{2t} $$ Thus, the solution of the given initial value problem is: $$ y(t) = e^{2t} - 1 $$

Key concepts

Initial Value Problem

Understanding the initial value problem (IVP) is crucial when dealing with differential equations. An IVP consists of a differential equation along with a specified value, called the initial condition, which the solution must satisfy at a given point—in most cases, when time is zero. For example, if we have a second-order differential equation like the one in our exercise, the initial conditions would typically tell us the value of the function and its first derivative at time zero. In the given exercise, the IVP is defined by the differential equation \(y'' - 4y' + 4y = 0\) with initial conditions \(y(0)=1\) and \(y'(0)=1\). These conditions are essential for finding a unique solution that fits the specific scenario described by the problem.

Differential Equation Solution

Solving a differential equation means finding a function (or set of functions) that satisfies the equation, often subject to initial conditions. The Laplace transform is an efficient tool for solving linear differential equations, particularly when initial conditions are given. Through this method, a differential equation is transformed into an algebraic equation in the 's' domain, which is usually simpler to solve. The step-by-step solution presented transforms the original time-domain equation into an equation in terms of the Laplace variable 's'. By finding the function \(Y(s)\) in the 's' domain and then computing its inverse Laplace transform, we can get the time-domain solution, \(y(t)\), which is what we are ultimately interested in.

Inverse Laplace Transform

After solving a differential equation in the Laplace domain, the next step is to convert our solution back to the time domain. This is where the inverse Laplace transform comes into play. It reverses the process of the Laplace transform and is often represented by \(\text{\(\mathcal{L}^{-1}\)}\{Y(s)\}\). Applying the inverse transform to \(Y(s)\) yields the original function \(y(t)\).

Partial Fraction Decomposition

Often during the inverse transform process, we need to break down complex fractions into simpler parts, which is where partial fraction decomposition is useful. This technique helps us to express the function \(Y(s)\) as a sum of simpler fractions, making it easier to apply known inverse transforms from the Laplace transform table.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to simplify complex rational expressions, especially when dealing with inverse Laplace transforms. The goal is to break down a fraction into simpler 'partial fractions' that are easier to work with. This step often precedes the application of the inverse Laplace transform in order to correlate the decomposed terms with entries in the Laplace transform table. In our step-by-step solution, we decomposed \(\frac{s-3}{(s-2)^2}\) into \(-1 + \frac{1}{s-2}\). These simpler terms correspond to transform pairs that are easily looked up in the table, allowing for the final step, which is taking the inverse transform to determine \(y(t)\).