Problem 27 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The solution is \( y = \sqrt{2}\sin(2x) - \frac{1}{2} \).

Step 1: Solve the Homogeneous Equation

First, solve the homogeneous equation, \[ y'' + 4y = 0. \] The characteristic equation is \( r^2 + 4 = 0 \), which gives: \[ r = rac{-4}{2} ightarrow r = ext{n }2i, \] since it involves imaginary numbers, the solutions are: \[ y_h = C_1 ext{cos}(2x) + C_2 ext{sin}(2x). \]

Step 2: Solve the Particular Solution

Since the non-homogeneous part is a constant (-2), assume a particular solution of the form: \[ y_p = A. \] Substitute into the differential equation: \[ 0 + 4A = -2 \] Solving gives \( A = -\frac{1}{2} \).So the particular solution is: \[ y_p = -\frac{1}{2}. \]

Step 3: Construct the General Solution

Combine the homogeneous and particular solutions: \[ y = C_1 ext{cos}(2x) + C_2 ext{sin}(2x) - \frac{1}{2}. \]

Step 4: Apply Initial Condition for y(π/8)

Plug \( y\left(\frac{\pi}{8}\right) = \frac{1}{2} \) into the general solution. \[ \frac{1}{2} = C_1 ext{cos}\left(\frac{\pi}{4}\right) + C_2 ext{sin}\left(\frac{\pi}{4}\right) - \frac{1}{2}. \] Simplify using \( \text{cos}(\frac{\pi}{4}) = \text{sin}(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \): \[ \frac{1}{2} = C_1 \frac{\sqrt{2}}{2} + C_2 \frac{\sqrt{2}}{2} - \frac{1}{2}. \] Simplifies to: \[ C_1 \frac{\sqrt{2}}{2} + C_2 \frac{\sqrt{2}}{2} = 1. \] Multiply through by \( \sqrt{2} \): \[ C_1 + C_2 = \sqrt{2}. \]

Step 5: Apply Initial Condition for y'(π/8)

Find the derivative of the general solution: \[ y' = -2C_1 ext{sin}(2x) + 2C_2 ext{cos}(2x). \] Plug \( y'(\frac{\pi}{8}) = 2 \): \[ 2 = -2C_1 \text{sin}\left(\frac{\pi}{4}\right) + 2C_2 \text{cos}\left(\frac{\pi}{4}\right). \] Again, use \( \text{cos}(\frac{\pi}{4}) = \text{sin}(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \): \[ 2 = -2C_1 \frac{\sqrt{2}}{2} + 2C_2 \frac{\sqrt{2}}{2}. \] Simplifies to: \[ 2 = -C_1 \sqrt{2} + C_2 \sqrt{2}. \] Divide through by \( \sqrt{2} \): \[ C_2 - C_1 = \frac{2}{\sqrt{2}} = \sqrt{2}. \]

Step 6: Solve the System of Equations

Solve the equation system:1. \( C_1 + C_2 = \sqrt{2} \)2. \( C_2 - C_1 = \sqrt{2} \)Add these two equations:\[ 2C_2 = 2\sqrt{2} \implies C_2 = \sqrt{2}. \]Substitute back into the first equation:\[ C_1 + \sqrt{2} = \sqrt{2} \implies C_1 = 0. \]

Step 7: Write the Final Solution

With \( C_1 = 0 \) and \( C_2 = \sqrt{2} \), substitute back into the general solution:\[ y = \sqrt{2}\text{sin}(2x) - \frac{1}{2}. \]

Key Concepts

Homogeneous Equations

Homogeneous equations in differential calculus are equations where every term is a function of the dependent variable and its derivatives. They do not have constant or external forces acting on them. A typical form looks like this: \[ y'' + a y' + b y = 0. \] In our exercise, the homogeneous equation is \( y'' + 4y = 0 \). To solve it, we find the characteristic equation, derived by replacing each derivative with \( r \), so it becomes \( r^2 + 4 = 0 \). The roots are found to be imaginary (\( r = \-2i \)), indicating oscillatory solutions involving sine and cosine functions. Thus, the solution is expressed as:

• \( y_h = C_1 \cos(2x) + C_2 \sin(2x) \)

This solution represents all possible combinations of sinusoidal waves that satisfy the equation without any external forces.

Particular Solutions

Particular solutions are solutions to non-homogeneous differential equations that account for external forces or inputs.These solutions do not include arbitrary constants (like \(C_1\) and \(C_2\)) because they are meant to fit specific conditions given in the problem.In the given differential equation, a constant forcing term, \-2, means our particular solution might also be a constant.By substituting \(y_p = A\) into the non-homogeneous equation \(y'' + 4y = -2\), we solve:

• \( 0 + 4A = -2 \)
• \( A = -\frac{1}{2} \)

Hence, the particular solution is \( y_p = -\frac{1}{2} \). It represents a constant shift applied to the homogeneous solution to satisfy the equation with the external input.

Initial-Value Problems

An initial-value problem (IVP) involves solving a differential equation subject to specific initial conditions. These conditions provide specific values for the solution and sometimes its derivatives at a particular point, ensuring a unique solution.In our problem, the conditions were provided as \( y(\pi/8) = 1/2 \) and \( y'(\pi/8) = 2 \).Initial conditions allow us to determine the constants in the general solution. This helps to form a specific solution that fits the behavior at the initial point as described.IVPs are essential because they transform the indefinite form into a concrete one, capturing any external dependencies particular to the problem.

General Solution

The general solution to a differential equation combines the homogeneous solution and the particular solution. This solution covers all possible behaviors of the equation, incorporating both the natural response to any internal dynamics and an external force. For the given problem, the general solution is:

• \( y = C_1 \cos(2x) + C_2 \sin(2x) - \frac{1}{2} \)

Here, \(C_1 \cos(2x) + C_2 \sin(2x)\) covers the homogenous part, and \(-\frac{1}{2}\) is the particular solution fitting the constant force.To derive a precise function that satisfies the initial conditions, we solve for \(C_1\) and \(C_2\) using those given conditions.

System of Equations

A system of equations helps solve for unknowns in problems involving multiple conditions or constraints. In particular, linear systems are useful for identifying constants in a differential equation's general solution. From our initial conditions and the equations derived from them, we had:

• \( C_1 + C_2 = \sqrt{2} \)
• \( C_2 - C_1 = \sqrt{2} \)

Solving these simultaneously means combining equations to eliminate one variable. By adding them, we found \(2C_2 = 2\sqrt{2}\), leading to \(C_2 = \sqrt{2}\). Substitute back to get \(C_1 = 0\). This step is crucial because it ensures the general solution fits the specific scenario set by initial conditions, yielding \( y = \sqrt{2} \sin(2x) - \frac{1}{2}\) as the precise answer.