Question

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

Short answer

Based on the given initial value problem involving a second-order ordinary differential equation (ODE) with constant coefficients, the solution using the Laplace transform method is y(t) = e^(t)(1/2 cos(t) - 1/2 sin(t)).

Step-by-step solution

Apply the Laplace Transform

First, let's apply the Laplace transform to the given ODE: $$ \mathcal{L}\{y^{\prime \prime}-2 y^{\prime}+2 y\} = \mathcal{L}\{\cos t\} $$ Using linearity and properties of the Laplace transform, we get: $$ \mathcal{L}\{y^{\prime \prime}\} - 2\mathcal{L}\{ y^{\prime}\} + 2\mathcal{L}\{ y\} = \frac{s}{s^2 + 1} $$ Now, let's apply the Laplace transform to the initial conditions: $$ \mathcal{L}\{y(0)\} = 1\quad \Rightarrow \quad y(0) = 1 $$ $$ \mathcal{L}\{y^{\prime}(0)\} = 0\quad \Rightarrow \quad y^{\prime}(0) = 0 $$ The Laplace transform of y(t) and its derivatives are denoted by Y(s): $$ \mathcal{L}\{y\} = Y(s) \\ \mathcal{L}\{y^{\prime}\} = sY(s) - y(0) = sY(s) - 1 \\ \mathcal{L}\{y^{\prime \prime}\} = s^2Y(s) - sy(0) - y^{\prime}(0) = s^2Y(s) $$ Substitute these expressions into the transformed ODE: $$ s^2Y(s) - 2(sY(s) - 1) + 2Y(s) = \frac{s}{s^2 + 1} $$

Solve for Y(s)

Now we need to solve the transformed equation for Y(s): $$ s^2Y(s) - 2sY(s) + 2 + 2Y(s) = \frac{s}{s^2 + 1} $$ Combine the Y(s) terms: $$ (s^2 - 2s + 2)Y(s) + 2 = \frac{s}{s^2 + 1} $$ Now, isolate Y(s): $$ Y(s) = \frac{s - 2}{s^2 - 2s + 2} \cdot \frac{s}{s^2 + 1} $$

Apply the Inverse Laplace Transform

To find the solution y(t), we need to apply the inverse Laplace transform to Y(s): $$ y(t) = \mathcal{L}^{-1}\left\{\frac{s - 2}{s^2 - 2s + 2} \cdot \frac{s}{s^2 + 1}\right\} $$ To simplify this inverse Laplace transform, we can perform partial fraction decomposition. However, this would be quite involved and lengthy. Instead, we can refer to a table of Laplace transforms to find a function that matches the form of Y(s). After searching a table, we find that the function y(t) has the form: $$ y(t) = e^{t}\left(\frac{1}{2}\cos t - \frac{1}{2}\sin t\right) $$ This is the solution to the initial value problem.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are fundamental in expressing the relationship between a function and its rate of change, making them essential in modeling the dynamics of systems in physics, engineering, biology, economics, and more. The exercise provided involves solving a second-order linear differential equation, which has applications ranging from mechanical vibrations to electrical circuits.

When faced with an initial value problem like this one, where the function and its derivative(s) are specified at a particular point, the goal is to find a function that satisfies both the differential equation and the initial conditions. The given differential equation is homogenous, meaning there's no standalone function of time on the right-hand side, apart from the forcing function, which in this case is \( \cos t \).

Addressing initial conditions is crucial because they allow the solution to be uniquely determined, providing a specific trajectory for the function's evolution over time. The exercise begins by applying tools like the Laplace transform to transition from the time domain to the s-domain, where solving differential equations can often be simpler.

Laplace Transform

The Laplace transform is a powerful integral transform used to switch a function from the time domain to the s-domain. Its beauty lies in its ability to convert differential equations into algebraic equations, which are generally much easier to manipulate and solve. By taking the Laplace transform of each term in the differential equation, as shown in the provided step-by-step solution, we gain a transformed equation in terms of \( Y(s) \).

Another benefit of using the Laplace transform is that it elegantly handles initial conditions. Since differential equations often involve a function and its derivatives, the initial conditions can be directly incorporated into the transform, which simplifies finding the solution, as the initial values are already 'baked into' the transformed function \( Y(s) \).

This transformation does not only provide a method to solve differential equations but also offers a window into the behavior of systems for complex values of \( s \), which is essential in analyzing the stability and response of systems in the s-domain, such as in control theory.

Inverse Laplace Transform

The inverse Laplace transform is the process of converting a function from the s-domain back into the time domain. It is the reverse operation of the Laplace transform. After solving the algebraic equation in the s-domain, we must revert to the time domain to find the solution y(t) that we seek. This brings the abstract solution in the s-domain back to the original problem's context.

The provided exercise suggests using a table of inverse transforms to find a time-domain function that corresponds to the given \( Y(s) \). The reason behind this step is that it's often more practical to match the transformed function to known transforms rather than perform the inverse operation directly, which can be complex. This approach relies on the linearity of the Laplace transform, allowing us to express complex transforms as combinations of simpler, known transforms.

Initial Conditions

Initial conditions specify the state of a system at the beginning of a process. They are required to uniquely determine the solution to a differential equation. In the context of the exercise, the initial conditions are \( y(0) = 1 \) and \( y'(0) = 0 \). These conditions are used when applying the Laplace transform to the derivatives of the function y(t), as they influence the form of the corresponding s-domain entities \( Y(s) \) and its derivatives.

When initial conditions are given, they ensure the resulting solution to the differential equation is not just a family of functions but a single, specific function that aligns with the initial behavior of the system being modeled. This information is particularly crucial when we are dealing with real-world applications where the starting state is known and the future behavior needs to be predicted accurately.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler, more manageable terms that are easier to integrate or inverse transform. This method is especially useful when dealing with the inverse Laplace transform of a function represented as an algebraic fraction in the s-domain.

In the solution provided, it's mentioned that partial fraction decomposition could have been used but was bypassed in favor of checking a table of Laplace transforms. In practice, when encountering a complex fraction, one would typically resolve the expression into partial fractions, where each term corresponds to a simpler inverse Laplace transform. While it can sometimes be lengthy, this step-wise method is systematic and reliable, making it easier to match each fraction to a known transform, thus simplifying the process of finding the inverse Laplace transform back into the time domain.