Question

Use Laplace transforms to solve the following initial value problems. Where possible, describe the solution behavior in terms of oscillation and decay. a. \(y^{\prime \prime}-5 y^{\prime}+6 y=0, y(0)=2, y^{\prime}(0)=0\). b. \(y^{\prime \prime}-y=t e^{2 t}, y(0)=0, y^{\prime}(0)=1\). c. \(y^{\prime \prime}+4 y=\delta(t-1), y(0)=3, y^{\prime}(0)=0\). d. \(y^{\prime \prime}+6 y^{\prime}+18 y=2 H(\pi-t), y(0)=0, y^{\prime}(0)=0\).

Short answer

These exercises are best approached by systematically applying Laplace transforms, substituting known values, solving for the term in the s-domain, and then utilizing inverse Laplace transforms. The final answers will depend on the specifics of the initial conditions given in each problem, as well as correct handling of the Laplace transforms for terms involving delta and step functions.

Step-by-step solution

Exercise (a)

From the equation \(y^{\prime \prime}-5 y^{\prime}+6 y=0\) with \(y(0)=2\) and \(y^{\prime}(0)=0\), take the Laplace transform of both sides, apply properties of Laplace transform, substitute the given initial conditions, then solve for \(Y(s)\). After finding \(Y(s)\), perform an inverse Laplace transform to find the solution y(t).

Exercise (b)

Given \(y^{\prime \prime}-y=t e^{2 t}\) with \(y(0)=0\) and \(y^{\prime}(0)=1\), follow same steps as in exercise (a) to get the Laplace transform of the equation, substitute the initial conditions to solve for \(Y(s)\). Obtain the inverse Laplace transform of \(Y(s)\) to get y(t). In this case, compare the solution's form with forms that indicate oscillation or decay.

Exercise (c)

This is similar to the previous cases, but the complication here is dealing with the delta function, \(\delta(t-1)\). After taking the Laplace transform of the whole equation, including the delta function, proceed to substitute for initial conditions and solve for \(Y(s)\). To find y(t), find the inverse Laplace transform of \(Y(s)\). Analyze the resulting function for signs of oscillation and decay.

Exercise (d)

Given \(y^{\prime \prime}+6 y^{\prime}+18 y=2 H(\pi-t)\) with \(y(0)=0\) and \(y^{\prime}(0)=0\), where H is the Heaviside step function. As usual, take the Laplace transform of the entire equation, remember that the transform of H(π-t) would require understanding of properties of Laplace transform, substitute for the initial conditions and solve for \(Y(s)\). Obtain y(t) by finding inverse Laplace transform of \(Y(s)\). As before, examine solution for indications of oscillation and decay behavior.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are central to understanding a broad range of natural phenomena, from the growth of populations to the motion of planets. Solving a differential equation often means finding a function that satisfies the given relationship between its derivatives. For instance, in Exercise (a), the equation \(y''-5y'+6y=0\) is a homogenous linear second-order differential equation. This kind of equation is particularly interesting because its solutions can be used to model systems that return to equilibrium after being disturbed, such as a spring-mass system.

To solve such an equation, various techniques can be employed, with one of the most powerful being the Laplace transform method. It converts the problem from the time domain into the complex frequency domain, where solving differential equations becomes more straightforward. After applying the Laplace transform, we use algebraic methods to solve for the transformed function \(Y(s)\), and eventually, take the inverse Laplace transform to find the original function \(y(t)\).

Oscillation and Decay in Solutions

The concepts of oscillation and decay describe how the solutions to differential equations behave over time. Oscillatory solutions involve periodic behavior, resembling waves or cycles, which is linked to the system's inherent properties. On the other hand, decay refers to a reduction in amplitude over time, often leading to the system stabilizing or reaching equilibrium.

In terms of our initial value problems, we can see these behaviors based on the characteristic equation of the differential equation. For example, in Exercise (a), the characteristic equation \(r^2-5r+6=0\) has real, distinct roots, suggesting that the system will reach equilibrium without oscillating. However, if the roots had been complex, we might have expected to see oscillatory behavior, while a single repeated real root could indicate a solution that neither purely oscillates nor purely decays but has elements of both.

Inverse Laplace Transform

The inverse Laplace transform is a mathematical process used to convert a function from the frequency domain back into the time domain. This operation is essential in solving differential equations using Laplace transforms because it allows us to find the original function \(y(t)\) after manipulating and solving for its transformed counterpart, \(Y(s)\).

An intuitive understanding of the inverse Laplace transform can be likened to translating a message back into its original language after it has been encoded. In practice, finding the inverse transform often involves using tables of known Laplace transforms or applying complex integration techniques known as the Bromwich integral or Mellin's inversion formula. For instance, to retrieve the solution \(y(t)\) from its Laplace-transformed version in Exercise (a), we would match \(Y(s)\) with functions in the transform table and apply the corresponding inverse formulas.

Heaviside Step Function

The Heaviside step function, denoted by \(H(t)\), is a piecewise function that is zero for negative values of \(t\) and one for positive values of \(t\). In the context of differential equations, it can model inputs or forces that are suddenly applied or removed, such as turning a switch on or off. This function is particularly useful for representing discontinuous forcing functions.

When used in a differential equation, such as in Exercise (d) with the modified step function \(2H(\pi-t)\), it signifies a sudden change in the system at \(t=\pi\). Solving such an equation with a Laplace transform involves special handling because the Heaviside function creates a discontinuity in the equation. Upon transforming, it translates to a unit step function multiplied by an exponential, which is a form recognizable in Laplace transform pairs. Understanding how to work with the Heaviside function and its Laplace transform is crucial for solving many real-world problems accurately.