Question

Use the Laplace transforms to solve each of the initial-value. \(y^{\prime \prime}+2 y^{\prime}+y=t e^{-2 t}\) \(y(0)=1, \quad y^{\prime}(0)=0\)

Short answer

The short answer to the initial-value problem is: \[y(t) = te^{2t} + e^{2t}\]

Step-by-step solution

Apply Laplace Transform

We first apply the Laplace transform on both sides of the ODE. Recall that the Laplace transform of the \(n\)-th derivative of a function is given by: \[\mathcal{L}\{y^{(n)}(t)\} = s^n Y(s) - s^{n-1} y(0) - ... - y^{(n-1)}(0)\] where \(Y(s)\) is the Laplace transform of \(y(t)\). Applying the Laplace transform on both sides, we have: \[\mathcal{L}\{y^{\prime \prime}+2 y^{\prime}+y\}(s) = \mathcal{L}\{t e^{-2 t}\}(s)\]

Use Laplace Transform Properties

Now, let's use the property of linearity and applying it to the left-hand side: \[\mathcal{L}\{y^{\prime \prime}\}(s) + 2\mathcal{L}\{y^{\prime}\}(s) + \mathcal{L}\{y\}(s) = \mathcal{L}\{t e^{-2 t}\}(s)\] We substitute the initial conditions and the Laplace transforms of each term: \[(s^2Y(s) - sy(0)- y'(0)) + 2(sY(s) - y(0)) + Y(s)=\frac{1}{(s+2)^2}\]

Solve for Y(s)

Now, we need to solve for \(Y(s)\), the Laplace transform of \(y(t)\). We plug in the initial conditions and simplify: \[(s^2Y(s) - s(1)-0) + 2(sY(s) - 1) + Y(s)=\frac{1}{(s+2)^2}\] Lifting the parentheses and rearranging terms, we get: \[(s^2+2s+1)Y(s) = \frac{1}{(s+2)^2} + s + 2\]

Determine Y(s)

Dividing both sides by \((s^2 + 2s + 1)\) and simplifying, we find the expression for \(Y(s)\): \[Y(s) = \frac{1+(s+2)^3}{(s+2)^2}\]

Find Inverse Laplace Transform

Finally, we will find the inverse Laplace transform of \(Y(s)\) to get the solution \(y(t)\). Taking the inverse Laplace transform and simplifying the expression, we get: \begin{align*} y(t) &= \mathcal{L}^{-1}\left\{ \frac{1+(s+2)^3}{(s+2)^2}\right\} \\ &= \mathcal{L}^{-1}\left\{\frac{1}{(s+2)^2}\right\} + \mathcal{L}^{-1}\left\{(s+2)^3\cdot\frac{1}{(s+2)^2}\right\} \\ &= te^{2t} + e^{2t} \end{align*} Therefore, the solution of the initial-value problem is: \[y(t) = te^{2t} + e^{2t}\]

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations that involve a function of one variable and its derivatives. They are called "ordinary" because they contain derivatives with respect to only one variable in contrast to partial differential equations which may involve multiple variables. ODEs are fundamental in understanding many phenomena because they mathematically represent physical laws, such as Newton's laws of motion. In basic terms, a typical ODE involves some unknown function, its derivatives, and possibly other functions or constants. For example, the equation given in this exercise, \(y'' + 2y' + y = te^{-2t}\), is an example of a second-order linear ODE, because the highest derivative is the second derivative (\(y''\)). This differential equation involves the function \(y(t)\) and its derivatives with respect to \(t\). ODEs can range from simple solvable forms to complex ones that may require advanced techniques, such as Laplace transforms, to solve.

Initial Value Problems

Initial Value Problems (IVPs) are a type of problem that involve finding a function that satisfies a differential equation and fulfills given conditions at a particular point in the domain. These conditions are called initial conditions, and they specify values of the function and possibly some of the derivatives at a certain point, often when the independent variable is zero, i.e., \(t = 0\).

In our problem, the initial conditions are \(y(0) = 1\) and \(y'(0) = 0\).

These conditions are crucial because they determine the specific solution to the differential equation and distinguish it from a family of potential solutions.

By setting these conditions, we define a unique path or curve—the solution—that the function \(y(t)\) must follow, starting from the initial point specified by \(t = 0\). IVPs are widely used in physics and engineering to model real-world systems that require the system's initial state to predict its future behavior.

Inverse Laplace Transform

The Inverse Laplace Transform is a method used to transform a function from the Laplace space back into the time domain. This technique is particularly useful when we have a Laplace-transformed function that represents an ordinary function in the time domain. To solve differential equations using Laplace transforms, one typically takes the Laplace transform to convert differential equations into algebraic equations, solves these easier algebraic equations for the transformed function, and then uses the inverse Laplace transform to revert the function back into its original form.

Understanding the Transform

After determining the Laplace transform \(Y(s)\) in our solved problem, the final task is to revert it to find \(y(t)\). This involves recognizing standard forms or occasionally using partial fraction decomposition and applying inverse transform tables. For example, \[y(t) = \mathcal{L}^{-1}\left\{\frac{1+(s+2)^3}{(s+2)^2}\right\}\], where we separated the terms to simplify the inverse transformation. This step is crucial as it provides the explicit solution to the original differential equation in the time domain.

Linear Differential Equations

Linear Differential Equations are a class of differential equations that involve linear combinations of the function and its derivatives. Specifically, these equations are 'linear' in the sense that each term is either a constant or the product of a constant and the function \(y\) or its derivatives. In such equations, there are no products or powers of the function and its derivatives. A hallmark of linear differential equations is their structure, which often enables the use of superposition, meaning the sum of the solutions of two linear equations is also a solution.

Characteristics of Linear Differential Equations

They provide a straightforward path to applying methods like Laplace Transforms.

They preserve linearity in operations, such as addition and scalar multiplication.

In our original exercise, the differential equation \(y'' + 2y' + y = te^{-2t}\) is linear since it only contains the first and second derivatives of \(y\) with no nonlinear terms like \(y^2\) or \(yy'\). This linearity is what permits the straightforward manipulation using Laplace transforms that helped us solve the equation by transforming the problem from the time domain to the frequency domain and back again.