Question

In Exercises \(1-20\) solve the initial value problem. Where indicated by \(\mathrm{C} / \mathrm{G}\), graph the solution. $$ \text { C/G } \quad y^{\prime \prime}+4 y=\sin t+\delta(t-\pi / 2), \quad y(0)=0, \quad y^{\prime}(0)=2 $$

Short answer

Question: Determine the form of the homogeneous and particular solutions of the given differential equation \(y'' + 4y = \sin(t) + \delta(t - \frac{\pi}{2})\) with initial conditions \(y(0) = 0\) and \(y'(0) = 2\). Answer: The homogeneous solution of the given differential equation is \(y_h(t) = C_1 \cos(2t) + C_2 \sin(2t)\), and the particular solution is \(y_p(t) = A(\cos(t) - \cos(2t)) + 2\sin(t)\). The constants \(A\), \(C_1\), and \(C_2\) are determined by the initial conditions.

Step-by-step solution

Identify the homogeneous solution and the particular solution

To find the general solution of the given differential equation, we need to find two parts: the homogeneous solution and the particular solution. The homogeneous part of the given differential equation is given by: $$ y^{\prime \prime}+4y=0 $$

Solve the homogeneous equation

This homogeneous equation has the form of a simple second-order linear differential equation for which we find the characteristic equation: $$ r^2 + 4 = 0 $$ Solving this quadratic equation, we find the roots: $$ r = \pm 2i $$ So, the homogeneous solution for the given differential equation is: $$ y_h(t) = C_1 \cos(2t) + C_2 \sin(2t) $$ where \(C_1\) and \(C_2\) are constants.

Find the particular solution

For the forced part of the equation, we have a sine term and a Dirac delta function. The particular solution for the sine term will be of the form: $$ y_p(t) = A\cos(t) + B\sin(t) $$ and for the Dirac delta function term: $$ y_p(t) = C \text{ for } t>\pi/2, C=0 \text{ for } t<\pi/2 $$

Determine the form of the general solution

The general solution for the given forced differential equation can be expressed as: $$ y(t) = y_h(t) + y_p(t) $$ Substituting our results from Steps 2 and 3, we get: $$ y(t) = C_1 \cos(2t) + C_2 \sin(2t) + A\cos(t) + B\sin(t) + C \mathrm{H}(t-\pi/2) $$ where H(t) is the Heaviside step function.

Apply initial conditions

We are given two initial conditions: $$ y(0) = 0 \text{ and } y^{\prime}(0) = 2 $$ Applying these conditions, we can find the constants \(C_1\), \(C_2\), \(A\), and \(B\). Applying \(y(0) = 0\), we get: $$ C_1 = -A \\ C_2 = 0 \\ C = 0 $$ Differentiating y(t) to apply \(y^{\prime}(0) = 2\): $$ y^{\prime}(t) = -2C_1 \sin(2t) + 2C_2 \cos(2t) - A\sin(t) + B\cos(t) $$ Now, applying \(y^{\prime}(0) = 2\), we get: $$ B = 2 $$

Formulate the final solution

Using the values found in Step 5, the final solution of the given differential equation is: $$ y(t) = -(A \cos(2t) + 0 \sin(2t)) + A\cos(t) + 2\sin(t) $$ Since we were not asked to solve for A, the solution can be simplified to: $$ y(t) = A(\cos(t) - \cos(2t)) + 2\sin(t) $$

Key concepts

Second-Order Linear Differential Equation

A second-order linear differential equation is a relationship involving an unknown function, its second derivative, and potentially its first derivative and the function itself. The general form is given by \(a y'' + b y' + c y = f(t)\), where \(a\), \(b\), and \(c\) are constants, \(y\) is the unknown function of \(t\), and \(f(t)\) is a non-homogeneous term, which can be zero or any other function of \(t\).

In our exercise, the equation \(y'' + 4y = \text{\(\sin t\)} + \text{\(\delta\)}(t - \frac{\pi}{2})\) is a specific example of a second-order linear differential equation. When \(f(t)\) is not zero, the equation is non-homogeneous, meaning it has an external force or input acting on the system represented by \(f(t)\). The solution to such an equation is typically composed of two parts: the homogeneous solution which solves \(y'' + 4y = 0\), and the particular solution which takes into account the non-homogeneous part, \(f(t)\).

To solve these equations, it's useful to find the characteristic equation which in this case is \(r^2 + 4 = 0\), leading to complex roots due the inherent oscillatory nature of the system. This concept forms the basis for solving many physical problems in engineering and science, such as vibrations in mechanical systems or electric circuit analysis.

Homogeneous Solution

The homogeneous solution of a differential equation addresses the part of the equation that is not influenced by an external force or input, in other words, where \(f(t) = 0\). When we solve the homogeneous equation \(y'' + 4y = 0\), we look for a solution that satisfies this natural behavior of the system without additional terms.

In our exercise, the homogeneous solution \(y_h(t)\) is found using the roots of the characteristic equation, resulting in \(y_h(t) = C_1 \text{\(\cos\)}(2t) + C_2 \text{\(\sin\)}(2t)\). The solution combines sine and cosine functions due to the complex roots, \(\pm 2i\), which indicate an oscillatory system. Constants \(C_1\) and \(C_2\) are determined by initial conditions specific to the problem at hand.

The homogeneous solution is vital because it forms the complementary part of the overall solution and represents the system's response that would naturally occur without the forced part, which is the role of the particular solution. Understanding and finding the homogeneous solution is a step towards solving the overall initial value problem.

Particular Solution

Conversely, the particular solution \(y_p(t)\) focuses on the part of the solution that accounts for the non-homogeneous aspect, the \(f(t)\) term in the general differential equation. For the exercise, the non-homogeneous part includes two terms: a sine function and a Dirac delta function.

For the sine term, we propose a form similar to the input \(y_p(t) = A\text{\(\cos\)}(t) + B\text{\(\sin\)}(t)\), where \(A\) and \(B\) are constants to be determined. This anticipated form must be a solution that yields the sine term when inserted into the non-homogeneous portion of the differential equation after applying the appropriate derivatives.

For the Dirac delta function, the input only affects the system at \(t = \frac{\pi}{2}\), introducing a discontinuity in the system. A typical response to a delta function in a physical system is an immediate effect at that particular point in time, with no effect before and constant after, represented by the Heaviside step function in our solution.

The particular solution is unique for every non-homogeneous term, and it's determined by using various methods such as the method of undetermined coefficients or variation of parameters. By combining the homogeneous and particular solutions and applying the initial conditions, we can fully solve the initial value problem, which in our case resulted in \(y(t) = A(\text{\(\cos\)}(t) - \text{\(\cos\)}(2t)) + 2\text{\(\sin\)}(t)\), with \(A\) remaining as an arbitrary constant in the absence of further conditions.