Question

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

Short answer

Answer: The solution to the given IVP is a constant function, \(y(t) = 0\).

Step-by-step solution

Find the homogeneous solution

For the homogeneous solution, we have the equation \(y^{\prime \prime} + 4y = 0\). The characteristic equation is \(r^2 + 4 = 0\). Solving for \(r\), we get \(r = \pm 2i\). Therefore, the homogeneous solution is given by \(y_h(t) = A\cos(2t) + B\sin(2t)\).

Apply Laplace transform

Now, we will apply the Laplace transform on both sides of the given IVP. \(\mathcal{L}\{y^{\prime \prime} + 4y\} = \mathcal{L}\{2\delta(t - \frac{\pi}{4})\}\) Using properties of Laplace transform, \(s^2Y(s) - sy(0) - y^{\prime}(0) + 4Y(s) = 2e^{-\frac{\pi s}{4}}\) Since \(y(0) = 0\) and \(y^{\prime}(0) = 0\), we have \((s^2 + 4)Y(s) = 2e^{-\frac{\pi s}{4}}\).

Find the inverse Laplace transform

To find the particular solution, we need to find the inverse Laplace transform of \(Y(s) = \frac{2e^{-\frac{\pi s}{4}}}{s^2 + 4}\). Using convolution theorem, \(y_p(t) = y_h(t)*g(t)\) where g(t) is the inverse Laplace transform of \(2e^{-\frac{\pi s}{4}}\). \(g(t) = \mathcal{L}^{-1}\{2e^{-\frac{\pi s}{4}}\}\) Since \(g(t) = 2\delta(t - \frac{\pi}{4})\), the inverse Laplace transform of \(g(t)\) is equal to the forcing function. Now, we find the convolution integral, \(y_p(t) = (A\cos(2t) + B\sin(2t))*(2\delta(t - \frac{\pi}{4}))\) \(y_p(t) = \int_{0}^{t} (A\cos(2(\tau - \frac{\pi}{4})) + B\sin(2(\tau - \frac{\pi}{4})))2\delta(\tau - \frac{\pi}{4}) d\tau\) Using the property of delta function, we can evaluate the integral as \(y_p(t) = 2(A\cos(\frac{\pi}{2}) + B\sin(\frac{\pi}{2}))\) \(y_p(t) = 2B\)

Apply initial conditions

To find the values of A and B, we will apply the given initial conditions. \(y(0) = 0 = A\cos(0) + B\sin(0)\) => \(A = 0\) \(y^{\prime}(0) = 0 = -2A\sin(0) + 2B\cos(0)\) => \(B = 0\) Since A and B are both 0, \(y_p(t) = 0\).

Combine the homogeneous and particular solutions

The general solution for the given IVP is the sum of homogeneous and particular solutions. \(y(t) = y_h(t) + y_p(t)\) \(y(t) = A\cos(2t) + B\sin(2t) + 0\) As A and B are both 0, the solution becomes \(y(t) = 0\)

Draw the graph of the solution

The graph of the solution is a horizontal line through the origin with the equation \(y(t) = 0\). Since it's a constant, the graph remains the same at all points in time.

Key concepts

Laplace Transform

The Laplace Transform is a powerful integral transform used to convert differential equations into algebraic equations. This aids in simplifying the process of solving complex linear differential equations, especially in the context of initial value problems. By translating problems from the time domain into the s-domain (complex frequency domain), calculations become more manageable. Generally, the Laplace Transform of a function \( f(t) \) is denoted as \( \,\mathcal{L}\{f(t)\} = F(s) \).In the solution process of our problem, the Laplace Transform was applied to both sides of the differential equation. This technique facilitates handling the delta function and identifying solution components. Particularly, it allows us to transform and manipulate the given initial value problem into an equation in terms of \( Y(s) \), significantly simplifying subsequent solution steps.
This transformed equation can then be solved algebraically to find \( Y(s) \) and later inverted to discover \( y(t) \).

Homogeneous Solution

The Homogeneous Solution refers to the part of the solution that satisfies the associated homogeneous equation of a differential equation. In our given exercise, this involves solving the equation:\[ y^{\prime \prime} + 4y = 0 \]The characteristic equation for this differential equation is \( r^2 + 4 = 0 \), which upon solving gives us roots \( r = \pm 2i \). These are complex conjugate roots, leading to a homogeneous solution involving sinusoidal functions:\[ y_h(t) = A\cos(2t) + B\sin(2t) \]Here, \( A \) and \( B \) are constants that are determined by the initial conditions of the problem. The homogeneous solution represents the natural response of the system without any external forcing or input. Understanding the homogeneous component is crucial for building the complete solution to an initial value problem.

Particular Solution

The Particular Solution describes the response of a system influenced by a forcing term or non-homogeneous part of a differential equation. In our exercise, this is represented by the delta function component:\( 2\delta(t - \frac{\pi}{4}) \)To determine the particular solution, we use the inverse Laplace transform. Using convolution and the properties of the delta function, the specific form of the particular solution becomes apparent:\[ y_p(t) = 2B \]However, due to earlier calculations using the initial conditions, this specific problem's coefficients \( A \) and \( B \) were determined to be zero. Thus, the particular solution also evaluates to zero in this scenario. This outcome illustrates that the non-homogeneous forcing does not alter the equilibrium state beyond what the homogeneous solution already presents.

Convolution Theorem

The Convolution Theorem plays an integral role in combining the effects of systems and inputs within the Laplace Transform framework. It states that the inverse Laplace transform of a product \( F(s) \cdot G(s) \) in the frequency domain equals the convolution of their respective inverse transforms \( f(t) \) and \( g(t) \) in the time domain:\[ y(t) = f(t) * g(t) = \int_{0}^{t} f(\tau)g(t - \tau) \, d\tau \]For the given problem, applying the convolution theorem aids in determining the particular solution via the inverse transform. Although the exercise employs a delta function, reducing the convolution integral to a straightforward evaluation at \( t = \frac{\pi}{4} \), understanding this theorem is crucial for broadly applying Laplace transforms to solve differential equations involving various types of inputs and responses.