Problem 27 - Advanced Engineering Mathematics

Question

$$ \text { In Problems 27-36, solve the given initial-value problem. } $$ $$ y^{\prime \prime}+4 y=-2, \quad y(\pi / 8)=\frac{1}{2}, y^{\prime}(\pi / 8)=2 $$

Step-by-Step Solution

Answer

The solution to the initial-value problem is \( y(t) = \left(\frac{2-\sqrt{2}}{2}\right) \cos(2t) + \left(\frac{2+\sqrt{2}}{2}\right) \sin(2t) - \frac{1}{2} \).

Step 1: Find the Homogeneous Solution

First, solve the homogeneous equation associated with the differential equation: \[ y'' + 4y = 0. \] The characteristic equation is \( r^2 + 4 = 0 \). Solving for \( r \), we get \( r = \pm 2i \). Therefore, the homogeneous solution is \[ y_h(t) = c_1 \cos(2t) + c_2 \sin(2t). \]

Step 2: Find the Particular Solution

To find the particular solution, we use the method of undetermined coefficients. Assume a particular solution of the form \( y_p(t) = A \) since the right side of the differential equation is constant (-2). Substitute \( y_p'' + 4y_p = -2 \) to check: \[ 0 + 4A = -2. \] Thus, \( 4A = -2 \Rightarrow A = -\frac{1}{2} \). The particular solution is \( y_p(t) = -\frac{1}{2}. \)

Step 3: Form the General Solution

Combine the homogeneous and particular solutions to form the general solution: \[ y(t) = c_1 \cos(2t) + c_2 \sin(2t) - \frac{1}{2}. \]

Step 4: Apply Initial Conditions

Use the initial conditions \( y(\pi / 8) = \frac{1}{2} \) and \( y'(\pi / 8) = 2 \) to find \( c_1 \) and \( c_2 \). From the condition \( y(\pi / 8) = \frac{1}{2} \): \[ c_1 \cos\left(\frac{\pi}{4}\right) + c_2 \sin\left(\frac{\pi}{4}\right) - \frac{1}{2} = \frac{1}{2}. \] Simplify this: \[ c_1 \left(\frac{\sqrt{2}}{2}\right) + c_2 \left(\frac{\sqrt{2}}{2}\right) = 1. \] From the condition \( y'(\pi / 8) = 2 \), take the first derivative: \[ y'(t) = -2c_1 \sin(2t) + 2c_2 \cos(2t), \] substitute \( t = \frac{\pi}{8} \): \[ -2c_1 \sin\left(\frac{\pi}{4}\right) + 2c_2 \cos\left(\frac{\pi}{4}\right) = 2. \] Simplify this: \[ -c_1 \sqrt{2} + c_2 \sqrt{2} = 2. \]

Step 5: Solve the System of Equations

We have two equations: \[ \frac{c_1 \sqrt{2}}{2} + \frac{c_2 \sqrt{2}}{2} = 1 \] and \[ -c_1 \sqrt{2} + c_2 \sqrt{2} = 2. \]Solve this system: Equation 1: \( c_1 + c_2 = \sqrt{2} \). Equation 2: \( -c_1 + c_2 = 2. \)Add equations to eliminate \( c_1 \): \( 2c_2 = 2+\sqrt{2} \Rightarrow c_2 = \frac{2+\sqrt{2}}{2}. \)Substitute \( c_2 \) into the first equation: \( c_1 + \frac{2+\sqrt{2}}{2} = \sqrt{2} \Rightarrow c_1 = \sqrt{2} - \frac{2+\sqrt{2}}{2} = \frac{2-\sqrt{2}}{2}. \)

Step 6: Write the Final Solution

Using the values of \( c_1 \) and \( c_2 \), substitute back into the general solution: \[ y(t) = \left(\frac{2-\sqrt{2}}{2}\right) \cos(2t) + \left(\frac{2+\sqrt{2}}{2}\right) \sin(2t) - \frac{1}{2}. \]

Key Concepts

Homogeneous Solution

In the context of solving differential equations, a homogeneous solution refers to solving the associated homogeneous equation. For the example provided, the homogeneous differential equation is \( y'' + 4y = 0 \). This equation does not include any external forcing terms, meaning the right-hand side is equal to zero. The characteristic equation, derived from this homogeneous differential equation, is \( r^2 + 4 = 0 \). From this, we find the roots \( r = \pm 2i \). This suggests the presence of imaginary roots, indicative of oscillatory solutions. As a result, the homogeneous solution takes the form:

• \( y_h(t) = c_1 \cos(2t) + c_2 \sin(2t) \)

The coefficients \( c_1 \) and \( c_2 \) are constants that will be determined by initial or boundary conditions.

Particular Solution

Finding a particular solution involves solving the non-homogeneous part of the differential equation. This is achieved using a technique specific to the equation and the form of the non-homogeneous term. In this example, we are tasked with solving \( y'' + 4y = -2 \). The right-hand side \(-2\) is a constant, which simplifies the process. A suitable guess here is a constant function:

• Assume \( y_p(t) = A \)

Upon substituting into the differential equation \( y_p'' + 4y_p = -2 \), we equate the result to \(-2\), solving for \( A \). This leads to \( 4A = -2 \) or \( A = -\frac{1}{2} \). Therefore, the particular solution is:

• \( y_p(t) = -\frac{1}{2} \)

This function encompasses the forced response of the system to the external input.

Method of Undetermined Coefficients

The method of undetermined coefficients is a powerful technique for finding particular solutions to linear non-homogeneous differential equations. It works particularly well with equations where the non-homogeneous part (right-hand side) is a polynomial, exponential, sine, cosine, or their combinations.
In this method, we assume a form for the particular solution that mirrors the form of the non-homogeneous term. In our example, since the term \(-2\) is constant, we assumed \( y_p(t) = A \). The goal is to substitute this assumed form back into the differential equation and solve for the unknown coefficients (in our example, just \( A \)).

• Substitute \( y_p(t) = A \) into \( y'' + 4y = -2 \)
• Equate and solve for \( A \)
• Obtain \( A = -\frac{1}{2} \)

Thus, the undetermined coefficients have been determined, providing the particular solution.

System of Equations

A system of equations arises when applying initial or boundary conditions to determine the constants in the general solution of a differential equation. In our problem, once the homogeneous and particular solutions were combined, forming \( y(t) = c_1 \cos(2t) + c_2 \sin(2t) - \frac{1}{2} \), initial conditions were applied:

• \( y(\pi / 8) = \frac{1}{2} \)
• \( y'(\pi / 8) = 2 \)

By substituting these conditions into the general solution and its derivative, we derive a system of linear equations involving \( c_1 \) and \( c_2 \). Solving this system involves:

• Equation 1: \( c_1 + c_2 = \sqrt{2} \)
• Equation 2: \( -c_1 + c_2 = 2 \)

By adding and subtracting these equations, we can independently solve for each constant. This process transforms abstract constants into precise values, bringing us closer to the specific solution that fits the initial conditions.