Question

Use the Laplace transform to solve the initial value problem. $$y^{\prime \prime}+2 y^{\prime}+y=e^{-t}, y(0)=0, y^{\prime}(0)=1$$

Short answer

In this problem, we solved the given initial value problem involving a second-order linear differential equation with constant coefficients using the Laplace transform method. The initial value problem was: $$y^{\prime \prime}+2 y^{\prime}+y=e^{-t}, y(0)=0, y^{\prime}(0)=1$$ After performing the necessary steps and calculations, we found the solution to be: $$y(t) = te^{-t}$$

Step-by-step solution

Find the Laplace transform of the equation

Take the Laplace transform for the entire differential equation: $$\mathcal{L}\{y^{\prime\prime} + 2y^{\prime} + y = e^{-t}\}$$ Then, apply linearity and the Laplace transform properties to rewrite it as follows: $$\mathcal{L}\{y^{\prime\prime}\} + 2\mathcal{L}\{y^{\prime}\} + \mathcal{L}\{y\} = \mathcal{L}\{e^{-t}\}$$

Use the properties of Laplace transform to simplify the equation

Using the properties of Laplace transform (for the derivatives): $$s^2Y(s) - sy(0) - y^{\prime}(0) + 2[sY(s) - y(0)] + Y(s) = \frac{1}{s + 1}$$ Now substitute the initial values \(y(0) = 0\) and \(y^{\prime}(0) = 1\): $$s^2Y(s) - 0 - 1 + 2[sY(s) - 0] + Y(s) = \frac{1}{s + 1}$$

Solve for Y(s)

Combine the terms and isolate Y(s): $$(s^2Y(s) + 2sY(s) + Y(s)) = 1 + \frac{1}{s + 1}$$ Now factor out the Y(s): $$Y(s)(s^2 + 2s + 1) = 1 + \frac{1}{s + 1}$$ Then, divide both sides by \((s^2 + 2s + 1)\) to find Y(s): $$Y(s) = \frac{1 + \frac{1}{s+1}}{s^2 + 2s + 1}$$

Simplify the expression for Y(s)

Combine the fractions in the numerator: $$Y(s) = \frac{(s+1) + 1}{(s+1)(s^2 + 2s + 1)}$$ Now, simplify the expression: $$Y(s) = \frac{s + 2}{(s+1)^3}$$

Find the inverse Laplace transform of Y(s)

To find the solution to the original initial value problem, we need to find the inverse Laplace transform of Y(s): $$y(t) = \mathcal{L}^{-1}\{Y(s)\} = \mathcal{L}^{-1}\left\{\frac{s + 2}{(s+1)^3}\right\}$$ Using the inverse Laplace transform properties and table, we identify that the given Laplace transform resembles the derivative of a shifted unit step function. Applying the formula for the inverse Laplace transform of a derivative of a shifted unit step function, we get: $$y(t) = te^{-t}$$ The solution to the initial value problem is: $$y(t) = te^{-t}$$

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) involves a differential equation accompanied by specific values at the beginning of the interval. In an IVP, you are given a differential equation and initial conditions that help you find a unique solution. For example, in the problem above, the equation is:

• \(y'' + 2y' + y = e^{-t}\)

The initial conditions provided are:

• \(y(0) = 0\)
• \(y'(0) = 1\)

These conditions specify the value of the function and its first derivative at \(t=0\). By knowing the initial values, we can solve the differential equation uniquely using techniques like the Laplace transform. It is a crucial step because without these initial values, we could end up with multiple possible solutions. Thus, the initial value problem gives us a precise starting point from which to trace the solution curve through time.

Differential Equations

Differential equations are mathematical equations involving functions and their derivatives. They are used to describe how things change over time, like population growth, chemical reactions, or physical movements. The differential equation in our exercise is:

• \(y'' + 2y' + y = e^{-t}\)

It is a second-order linear differential equation because the highest derivative is the second derivative, \(y''\). Sometimes these equations look complex, but they represent real-world phenomena in terms of changes. The order of the differential equation is determined by the highest derivative present, while its linearity is dictated by the absence of products or higher powers of the function and its derivatives. Solving these equations often involves techniques like the Laplace transform, which transforms the differential equation into an algebraic equation, making it easier to solve.

Inverse Laplace Transform

The Inverse Laplace Transform is a mathematical process used to convert a function from the Laplace domain back to the time domain. After solving a differential equation using the Laplace Transform, we often end up with a function \(Y(s)\) in the Laplace domain. To interpret it in terms of the original variable, we apply the inverse process. In our exercise, the equation in the Laplace domain becomes:

• \(Y(s) = \frac{s + 2}{(s+1)^3}\)

We then use the inverse Laplace Transform to return to the time domain, represented by \(y(t)\). Tools such as tables and known transformations often assist in this step. For our problem, \(y(t)\) turns out to be:

• \(y(t) = te^{-t}\)

In simple terms, the inverse transform is like hitting the 'undo' button, bringing the function back to its original form, which provides the solution to our initial value problem. This step is essential for verifying the behavior of the solution in its native domain of time.