Problem 3 - A First Course in Differential Equations with Modeling Applications

Question

Solve each differential equation by variation of parameters. $$y^{\prime \prime}+y=\sin x$$

Step-by-Step Solution

Answer

The general solution is \(y = C_1 \cos(x) + C_2 \sin(x) - \frac{x}{2} \cos(x) + \frac{1}{4}\sin(2x)\sin(x) - \frac{1}{4}\sin(x)\cos(2x)\).

Step 1: Identify the Homogeneous Equation

First, consider the homogeneous equation associated with the given differential equation: \[ y'' + y = 0 \]This equation will help us find the complementary solution (\(y_c\)).

Step 2: Find the Complementary Solution

The characteristic equation for the homogeneous differential equation is:\[ r^2 + 1 = 0 \]Solving for \(r\), we get \(r = \pm i\). Therefore, the complementary solution is:\[ y_c = C_1 \cos(x) + C_2 \sin(x) \]

Step 3: Apply Variation of Parameters Formula

For a differential equation of the form \( y'' + y = \,f(x) \), the particular solution \(y_p\) can be found using:\[ y_p = u_1 y_1 + u_2 y_2 \]Where \(y_1 = \cos(x)\) and \(y_2 = \sin(x)\) are solutions to the homogeneous equation, and \(u_1\) and \(u_2\) are functions to be determined.

Step 4: Determine Functions \(u_1\) and \(u_2\)

The functions \(u_1\) and \(u_2\) are given by:\[ u_1' = -\frac{y_2 f(x)}{W} = -\frac{\sin(x) \sin(x)}{1} = -\sin^2(x) \]\[ u_2' = \frac{y_1 f(x)}{W} = \frac{\cos(x) \sin(x)}{1} = \sin(x) \cos(x) \]where \(W\), the Wronskian of \(y_1\) and \(y_2\), equals 1.

Step 5: Integrate to Find \(u_1\) and \(u_2\)

Integrate \(u_1'\) and \(u_2'\) to find \(u_1\) and \(u_2\):\[ u_1 = \int -\sin^2(x) \, dx = \int -\frac{1-\cos(2x)}{2} \, dx = -\frac{x}{2} + \frac{1}{4}\sin(2x) + C_1 \]\[ u_2 = \int \sin(x)\cos(x) \, dx = \int \frac{1}{2}\sin(2x) \, dx = -\frac{1}{4}\cos(2x) + C_2 \]

Step 6: Construct the Particular Solution

Substitute \(u_1\) and \(u_2\) into the particular solution:\[ y_p = \left(-\frac{x}{2} + \frac{1}{4}\sin(2x)\right) \cos(x) + \left(-\frac{1}{4}\cos(2x)\right) \sin(x) \]Simplify to obtain:\[ y_p = -\frac{x}{2}\cos(x) + \frac{1}{4}\sin(x)\cos(x)\sin(2x) - \frac{1}{4}\cos(2x)\sin(x) \]

Step 7: Write the General Solution

The general solution to the differential equation is the sum of the complementary and particular solutions:\[ y = y_c + y_p = C_1 \cos(x) + C_2 \sin(x) - \frac{x}{2}\cos(x) + \frac{1}{4}\sin(x)\cos(x)\sin(2x) - \frac{1}{4}\cos(2x)\sin(x) \]

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. These types of equations are significant in many fields like physics, engineering, and economics because they model real-world phenomena such as motion, heat flow, or population dynamics.
A fundamental characteristic of differential equations is their ability to describe relationships involving a function and its rates of change with respect to one or more variables. In this exercise, the differential equation is given as: \[ y'' + y = \sin x \] This indicates a relationship involving a function \( y(x) \), its second derivative \( y'' \), and a non-homogeneous term \( \sin x \), which is the driving force or input of the system.

• "Homogeneous" equations have all terms dependent on the function and its derivatives.
• "Non-Homogeneous" includes an additional term like \( \sin x \) that doesn't solely depend on \( y \) or its derivatives.

This understanding sets the groundwork for solving the equation, as you will differentiate between the general behavior of the function described by a complementary solution and the particular effects imposed by the non-homogeneous part of the equation.

Complementary Solution

The complementary solution (\( y_c \)) addresses the homogeneous part of a differential equation, which is obtained by setting the non-homogeneous term to zero. For the equation \( y'' + y = 0 \), you start by finding its characteristic equation:\[ r^2 + 1 = 0 \]Solving gives the roots \( r = \pm i \), which suggests that the solutions are complex conjugates.
These complex roots translate to:

A trigonometric form using sine and cosine functions.

The complementary solution is:\( y_c = C_1 \cos(x) + C_2 \sin(x) \), where \( C_1 \) and \( C_2 \) are constants.

This solution captures the behavior of the "system" without any external force or input \( \sin x \). Think of it as the natural state of a swinging pendulum with no external push.

Wronskian

The Wronskian is an essential concept when using the method of variation of parameters to find particular solutions. It is a determinant calculated from two functions and their derivatives (in this exercise \( y_1 = \cos(x) \) and \( y_2 = \sin(x) \)). The purpose is to verify that these functions are linearly independent. For the Wronskian \( W \) of \( y_1 \) and \( y_2 \):\[ W = \begin{vmatrix} \cos(x) & \sin(x) \ -\sin(x) & \cos(x) \end{vmatrix} = \cos^2(x) + \sin^2(x) = 1 \]

• A non-zero Wronskian confirms linear independence, a crucial condition for applying variation of parameters.
• It's a handy tool to ensure two solutions can form a general solution, including new terms introduced by the particular solution.

The determinant here simplifies to 1, indicating our solutions are correctly characterized to seek a particular solution using variation of parameters.

Particular Solution

A particular solution (\( y_p \)) involves finding a specific solution to the non-homogeneous differential equation that accounts for external input or forcing functions—here, the \( \sin x \). Variation of parameters is a method to find this, unlike other methods (such as undetermined coefficients) because it does not assume the form of the solution.
To find the particular solution:

• Use the functions \( y_1 = \cos(x) \) and \( y_2 = \sin(x) \) as a basis derived from the complementary solution.
• Determine functions \( u_1 \) and \( u_2 \) such that:\[ y_p = u_1 y_1 + u_2 y_2 \]
• Calculate \( u_1' = -\frac{\sin(x) \sin(x)}{1} = -\sin^2(x) \) and \( u_2' = \frac{\cos(x) \sin(x)}{1} = \sin(x) \cos(x) \).

The next step is integrating these derivatives to find \( u_1 \) and \( u_2 \) and substituting back into the expression for \( y_p \). This constructive process resolves the final part of the general solution, representing the differential equation's total response, including its input effects.