Question

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

Short answer

Answer: The solution to the given initial value problem is \(y(t) = \frac{1}{\sqrt{2}}e^t\sin{\sqrt{2}t}\).

Step-by-step solution

Find the Laplace transform of each term

We find the Laplace transforms of all the terms in the given differential equation: $$ \mathcal{L}\{y^{\prime\prime}\}(s) = s^2Y(s) - sy(0) - y'(0) = s^2Y(s) - 1 $$ $$ \mathcal{L}\{-2y^{\prime}\}(s) = -2(sY(s) - y(0)) = -2sY(s) $$ $$ \mathcal{L}\{2y\}(s) = 2Y(s) $$

Replace the Laplace transform of the terms

Now we substitute the Laplace transforms obtained in Step 1 back into the given differential equation: $$ s^2Y(s) - 1 - 2sY(s) + 2Y(s) = 0 $$

Solve the transformed equation

Next, we solve the equation for the Laplace transform of the function \(Y(s)\): $$ (s^2 - 2s + 2) Y(s) = 1 $$ Divide by \((s^2 - 2s + 2)\) to get \(Y(s)\) as follows: $$ Y(s) = \frac{1}{s^2 - 2s + 2} $$

Find the inverse Laplace transform

Lastly, we find the inverse Laplace transform of \(Y(s)\) to obtain the solution, \(y(t)\): $$ y(t) = \mathcal{L}^{-1}\left\{\frac{1}{s^2 - 2s + 2}\right\} $$ To compute the inverse Laplace transform, we need to rewrite the denominator as a complete square: $$ y(t) = \mathcal{L}^{-1}\left\{\frac{1}{(s-1)^2 + 1}\right\} $$ Now, recognize that the inverse Laplace transform of the expression inside the curly brackets is \(\frac{1}{\sqrt{2}}e^t\sin{\sqrt{2}t}\). This is because the inverse Laplace transform of \(\frac{1}{(s-a)^2 + \omega^2}\) is \(e^{at}\sin{\omega t}\). Therefore, we have the final solution: $$ y(t) = \frac{1}{\sqrt{2}}e^t\sin{\sqrt{2}t} $$

Key concepts

Differential Equations

Differential equations are mathematical expressions that relate a function to its derivatives. Essentially, these equations involve rates at which things change—their derivatives. Such equations are extremely useful since they model real-world phenomena like heat transfer, vibrations, and even population growth. In the given problem, we are dealing with a second-order differential equation because it contains the second derivative, denoted as \( y'' \). The specific differential equation here is \( y^{\prime \prime}-2 y^{\prime}+2 y=0 \). It is a linear homogeneous differential equation because all terms are either derivatives of \( y \) or \( y \) itself, and they sum to zero. Homogeneous equations like this typically describe systems that change in predictable ways over time without external forces acting on them.

Initial Value Problem

An initial value problem is a type of differential equation that comes with initial conditions. These conditions specify the values of the function and its derivatives at the start, making it possible to find a unique solution. For instance, in our exercise, we have the initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \). This means when time \( t = 0 \), the function \( y \) is 0 and its first derivative, indicating the initial rate of change, is 1. These initial conditions are crucial because they narrow down the infinite number of possible solutions to one definite solution, giving us insight into how a system begins and subsequently evolves.

Inverse Laplace Transform

The Laplace transform is a powerful tool that converts differential equations into algebraic equations, which are easier to solve. However, to find the solution in terms of the original function \( y(t) \), we need to perform the inverse Laplace transform. In the given exercise, we started with \( Y(s) = \frac{1}{s^2 - 2s + 2} \). To find the inverse Laplace transform, we first complete the square in the denominator to rewrite it as \( (s-1)^2 + 1 \). The inverse operation then allows us to express \( Y(s) \) back in the time domain as a function of \( t \). The result of this transformation yields \( y(t) = \frac{1}{\sqrt{2}}e^t\sin{\sqrt{2}t} \), which matches standard forms of known inverse transforms for mathematical functions.

Solution Techniques for Differential Equations

There are several techniques for solving differential equations, and the choice of method depends largely on the nature of the equation itself. The Laplace transform is one of these methods, particularly useful when dealing with linear time-invariant systems. It works by transforming the differential equations into a simpler form, making complex initial value problems more manageable. Solving the transformed equations often involves algebraic manipulation instead of differential calculus, simplifying the process. Once you've worked out a solution in the transform space, the inverse Laplace transform converts the solution back into the original domain. This approach is particularly valuable since it bypasses some of the more challenging calculus-based approaches, making it an ideal solution technique for mechanical systems, electrical circuits, and more.