Question

Find the solution of the given initial value problem and draw its graph. \(y^{\prime \prime}+4 y=\delta(t-4 \pi) ; \quad y(0)=1 / 2, \quad y^{\prime}(0)=0\)

Short answer

#Question#: Consider the given initial value problem (IVP): \(y^{\prime\prime}(t) = 4y(t) + \delta(t-4\pi)\), with initial conditions \(y(0) = \frac{1}{2}\) and \(y'(0) = 0\), where \(\delta(t)\) represents the Dirac delta function. Find the solution \(y(t)\) for \(t>0\), and explain how to draw the graph of the solution. #Answer#: The solution for the given IVP is \(y(t) = \frac{1}{2}\cos(2t) + \frac{1}{4}e^{-4\pi t}(\sin(2t) - \cos(2t))\). To draw the graph of the solution, you can use a graphing calculator or software by plugging in the expression for \(y(t)\). Alternatively, you can sketch the graph manually by finding the period, peaks, and troughs of the cosine and sine terms, and noting that the second term approaches zero as \(t\) becomes larger to get an understanding of the shape of the graph.

Step-by-step solution

Apply Laplace Transform to the given equation

Start by applying the Laplace Transform to both sides of the given differential equation. Use the fact that the Laplace Transform of a derivative is given by: $$ \mathcal{L}\{y'(t)\} = sY(s) - y(0) $$ and $$ \mathcal{L}\{y^{\prime\prime}(t)\} = s^2Y(s) - sy(0) - y'(0), $$ where \(Y(s)\) is the Laplace Transform of the function \(y(t)\) and \(\mathcal{L}\{\cdot\}\) denotes the Laplace Transform operator. Also, remember that the Laplace Transform of a Dirac delta function at a point \(t = 4\pi\) is given by: $$ \mathcal{L}\{\delta(t-4\pi)\} = e^{-4\pi s}. $$

Solve for Y(s)

Applying the Laplace Transform to the given differential equation and using the properties from step 1, we get: $$ s^2Y(s) - s\cdot\frac{1}{2} = 4Y(s) + e^{-4\pi s}. $$ Now, solve for \(Y(s)\): $$ Y(s) = \frac{1}{2}\frac{s+e^{-4\pi s}}{s^2+4}. $$

Apply the inverse Laplace Transform

Apply the inverse Laplace Transform to find \(y(t)\): $$ y(t) = \mathcal{L^{-1}}\left\{\frac{1}{2}\frac{s+e^{-4\pi s}}{s^2+4}\right\}. $$ To find the inverse Laplace Transform, we recognize two parts of the fraction: $$ y(t) = \frac{1}{2}\left(\mathcal{L^{-1}}\left\{\frac{s}{s^2+4}\right\}+\mathcal{L^{-1}}\left\{\frac{e^{-4\pi s}}{s^2+4}\right\}\right). $$ For the first term, we have: $$ \mathcal{L^{-1}}\left\{\frac{s}{s^2+4}\right\} = \cos(2t), $$ and for the second term, use a Laplace Transform table or properties of convolution to find: $$ \mathcal{L^{-1}}\left\{\frac{e^{-4\pi s}}{s^2+4}\right\} = \frac{1}{2}e^{-4\pi t}(\sin(2t)-\cos(2t)). $$ Combining the terms, we find the solution: $$ y(t) = \frac{1}{2}\cos(2t) + \frac{1}{4}e^{-4\pi t}(\sin(2t) - \cos(2t)). $$

Graph the solution

To draw the graph of the solution, you can use a graphing calculator or software (such as Desmos, MATLAB, or WolframAlpha) by plugging in the expression for \(y(t)\). Or you can sketch the graph manually by finding the period, peaks, and troughs of the cosine and sine terms, noting that the second term approaches zero as \(t\) becomes larger. This will give you a rough understanding of the shape of the graph.

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with specified values at a starting point, known as initial conditions. In the context of our exercise, we have been given a second-order differential equation, specifically \(y'' + 4y = \delta(t-4\pi)\).
The initial conditions provided are \(y(0) = \frac{1}{2}\) and \(y'(0) = 0\).
This means that at \(t = 0\), the function \(y(t)\) starts at a value of \(\frac{1}{2}\) and its rate of change, the derivative, starts at 0. With these initial conditions, we can solve the differential equation and then determine the particular solution which fits these starting values.
It's like setting a starting line for your solution and mapping its journey from there. It's important as it ensures the solution is unique and applicable to the scenario at hand.

Dirac Delta Function

The Dirac delta function, denoted as \(\delta(t-a)\), is not a traditional function but rather a distribution or a "pseudo-function." It has an important property: it is zero everywhere except at \(t=a\), where it is infinitely high in such a way that its integral over the entire real line is 1.
In our problem \(\delta(t-4\pi)\), the delta function is centered at \(t = 4\pi\). This aspect plays a crucial role in engineering and physics for modeling instantaneous impulses or forces applied at specific points in time.
The presence of the delta function in the equation can be interpreted as an impulse applied at \(t = 4\pi\). When solving with Laplace Transforms, it simplifies because it is represented as \(e^{-4\pi s}\), allowing us to work algebraically to obtain the solution.

Inverse Laplace Transform

The inverse Laplace Transform is a technique used to revert a function from its Laplace transformed form back to the time domain. It is crucial in finding the actual solution \(y(t)\) once we have solved for \(Y(s)\) in the Laplace domain.
After applying the Laplace Transform to our differential equation and solving algebraically for \(Y(s)\), we get an expression that includes terms like \(\frac{s}{s^2+4}\) and \(\frac{e^{-4\pi s}}{s^2+4}\).

To transform back to the time domain, we leverage known inverse Laplace Transforms:

• The inverse of \(\frac{s}{s^2+4}\) is \(\cos(2t)\).
• The inverse of \(\frac{e^{-4\pi s}}{s^2+4}\) requires understanding convolution or using a transform table, resulting in \(\frac{1}{2}e^{-4\pi t}(\sin(2t)-\cos(2t))\).

Combining these, we find our time-domain solution \(y(t)\), which fully describes how the system evolves over time from its initial conditions and impulse input.