Question

In Exercises \(1-20\) use the Laplace transform to solve the initial value problem. Where indicated by \(\mathrm{C} / \mathrm{G}\), graph the solution. $$ y^{\prime \prime}+4 y=\left\\{\begin{array}{ll} t, & 0 \leq t<\frac{\pi}{2} \\ \pi, & t \geq \frac{\pi}{2} \end{array} \quad y(0)=0, \quad y^{\prime}(0)=0\right. $$

Short answer

In conclusion, the solution to the given initial value problem for the linear second-order differential equation with a piecewise forcing function is: $$ y(t) = \frac{1}{4}t\sin(2t) - \frac{1}{4}(t-\frac{\pi}{2})\sin(2(t-\frac{\pi}{2}))u(t-\frac{\pi}{2}) + \frac{\pi}{2}\sin(2t)u(t-\frac{\pi}{2}) $$ where \(u(t)\) is the unit step function.

Step-by-step solution

Apply Laplace Transform to the given differential equation

To solve the initial value problem, first take the Laplace transform of the given differential equation: $$ \mathcal{L}\{y''(t) + 4y(t)\} = \mathcal{L}\{f(t)\} $$ where \(f(t)\) is the piecewise forcing function: $$ f(t) = \begin{cases} t, & 0 \leq t < \frac{\pi}{2}\\ \pi, & t \geq \frac{\pi}{2} \end{cases} $$ Now apply the Laplace Transform to each term individually which gives us: \(\mathcal{L}\{y''(t)\} + 4\mathcal{L}\{y(t)\} = \mathcal{L}\{f(t)\}\) Use the properties of Laplace Transforms to transform the second-order derivative in the left-hand side: $$ s^2Y(s) - sy(0) - y'(0) + 4Y(s) = F(s) $$ Plugging in the initial conditions \(y(0) = 0\) and \(y'(0) = 0\), we get: $$ s^2Y(s) + 4Y(s) = F(s) $$

Find Laplace Transform of the forcing function and solve for Y(s)

Since the forcing function is piecewise, we can find its Laplace Transform by splitting it into two separate cases (before and after \(\frac{\pi}{2}\)): $$ F(s) = \mathcal{L} \{ f(t) \} = \mathcal{L} \left\\{ \begin{cases} t, & 0 \leq t < \frac{\pi}{2}\\ \pi, & t \geq \frac{\pi}{2} \end{cases} \right. $$ $$ F(s) = \frac{1}{s^2} - \frac{e^{-\frac{\pi}{2}s}}{s^2} + \frac{\pi}{s}e^{-\frac{\pi}{2}s} $$ Now plug \(F(s)\) into the transformed equation and solve for \(Y(s)\): $$ (s^2 + 4)Y(s) = \frac{1}{s^2} - \frac{e^{-\frac{\pi}{2}s}}{s^2} + \frac{\pi}{s}e^{-\frac{\pi}{2}s} $$ $$ Y(s) = \frac{1}{(s^2 + 4)s^2} - \frac{e^{-\frac{\pi}{2}s}}{(s^2 + 4)s^2} + \frac{\pi}{s(s^2 + 4)}e^{-\frac{\pi}{2}s} $$

Apply Inverse Laplace Transform to find y(t)

Now that we have Y(s), we need to find the inverse Laplace Transform to obtain y(t): $$ y(t) = \mathcal{L}^{-1} \{ Y(s) \} $$ $$ y(t) = \mathcal{L}^{-1} \left\\{ \frac{1}{(s^2 + 4)s^2} \right\\} - \mathcal{L}^{-1} \left\\{ \frac{e^{-\frac{\pi}{2}s}}{(s^2 + 4)s^2} \right\\} + \mathcal{L}^{-1} \left\\{ \frac{\pi}{s(s^2 + 4)}e^{-\frac{\pi}{2}s} \right\\} $$ Applying inverse Laplace transforms, partial fractions and convolution theorem, we can get: $$ y(t) = \frac{1}{4}t\sin(2t) - \frac{1}{4}(t-\frac{\pi}{2})\sin(2(t-\frac{\pi}{2}))u(t-\frac{\pi}{2}) + \frac{\pi}{2}\sin(2t)u(t-\frac{\pi}{2}) $$ Where \(u(t)\) is the unit step function. This is the final solution of the given initial value problem.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In the context of physics, engineering, and other sciences, these equations are used to describe the behavior of complex systems where change is involved. An ordinary differential equation (ODE) is an equation containing a function of one independent variable and its derivatives. The function's derivatives are taken with respect to that variable.

The importance of solving a differential equation in a particular context lies in the fact that it enables us to predict the future behavior of a system or determine the characteristics of a system from its rate of change. In the exercise provided, we have been given a second-order linear homogenous differential equation with constant coefficients. Such equations are commonly solved using the Laplace transform, a powerful technique that can greatly simplify the solution process by converting the differential equation into an algebraic equation.

Inverse Laplace Transform

The inverse Laplace transform is a method to revert the process of the Laplace transform and is denoted by \(\mathcal{L}^{-1}\{f(s)\}\), where \(f(s)\) is a function in the Laplace domain. This technique is vital for solving differential equations because it allows us to transform a complex differential equation in the time domain into a more manageable form in the frequency or s-domain.

After applying Laplace transforms to the differential equation and solving for \(Y(s)\), we need the inverse transform to get the solution, \(y(t)\), back in the time domain. It involves breaking down a given Laplace transform expression into simpler components that can be recognized as transforms of known functions. In the exercise, partial fraction decomposition, together with properties of Laplace transforms, such as linearity, and the use of convolution theorem for the piecewise function are required to express \(Y(s)\) in a form amenable to inverse transformation. The result, \(y(t)\), reflects the behavior of the original dynamic system for any time \(t\).

Piecewise Functions

A piecewise function is a function defined by multiple sub-functions, each applying to a certain interval of the main function's domain. These functions are particularly important for modeling situations where a parameter behaves differently under different conditions or at different times. Piecewise functions allow us to describe these variations within a single expression.

In the given initial value problem, the forcing function \(f(t)\) is piecewise. It takes a linear form \(t\) within the interval \(0 \leq t < \frac{\pi}{2}\) and is constant \(\pi\) from \(\frac{\pi}{2}\) onwards. To apply the Laplace transform to \(f(t)\), we consider each piece separately and use linearity to combine the transforms. This is pivotal for finding the Laplace Transform of \(F(s)\), which eventually leads to determining the solution \(Y(s)\) and, through inverse Laplace transform, to the solution in the time domain, \(y(t)\). In practice, especially in engineering and physics, piecewise functions are used to represent inputs like forces that change at certain times or locations.