Problem 61 - Calculus 3

Question

Solve the following equations using the method of undetermined coefficients. $$y^{\prime \prime}+2 y^{\prime}+y=\sin x+\cos x$$

Step-by-Step Solution

Answer

The general solution is \( y = C_1 e^{-x} + C_2 x e^{-x} + 0.5 \sin x \).

Step 1: Solve the Homogeneous Equation

First, we solve the associated homogeneous equation \( y'' + 2y' + y = 0 \). The characteristic equation is \( r^2 + 2r + 1 = 0 \). Factor this equation as \((r + 1)^2 = 0\), giving a repeated root \( r = -1 \). Hence, the general solution to the homogeneous equation is \( y_h = C_1 e^{-x} + C_2 x e^{-x} \).

Step 2: Choose a Form for the Particular Solution

Since the non-homogeneous term is \( \sin x + \cos x \), we choose a particular solution of the form \( y_p = A \sin x + B \cos x \).

Step 3: Differentiate the Particular Solution

Compute the first and second derivatives of \( y_p \). This gives \( y_p' = A \cos x - B \sin x \) and \( y_p'' = -A \sin x - B \cos x \).

Step 4: Substitute into the Differential Equation

Substitute \( y_p, y_p' \) and \( y_p'' \) into the original differential equation: \(-A \sin x - B \cos x + 2(A \cos x - B \sin x) + A \sin x + B \cos x = \sin x + \cos x\).

Step 5: Simplify and Match Coefficients

Simplify the left side to match the terms \( \sin x \) and \( \cos x \) on the right side. This yields: \((2A - 2B)\cos x + (-2B + 2A)\sin x = \sin x + \cos x\). By matching coefficients, we get two equations: \( 2A - 2B = 1 \) and \( -2B + 2A = 1 \).

Step 6: Solve the System of Equations

Solve the system: \( 2A - 2B = 1 \) and \( -2B + 2A = 1 \). These equations are identical, meaning \( A - B = 0.5 \) and \( A = B + 0.5 \). Choose \( B = 0 \), so \( A = 0.5 \).

Step 7: Write the Particular Solution

Thus, the particular solution is \( y_p = 0.5 \sin x \).

Step 8: Write the General Solution

The general solution to the differential equation is the sum of the homogeneous solution and the particular solution: \( y = C_1 e^{-x} + C_2 x e^{-x} + 0.5 \sin x \).

Key Concepts

Homogeneous Equation

When solving a differential equation using undetermined coefficients, the first step is to solve the homogeneous equation. This equation is derived from the original differential equation by setting the non-homogeneous terms to zero. For example, if you have the equation \(y'' + 2y' + y = 0\), you take the right side of the equation, which equals zero here, and solve it as a standard homogeneous equation.
The purpose of solving the homogeneous equation is to find the complementary or general solution, denoted by \(y_h\). This solution captures the behavior of the system without external forces or non-homogeneous terms.

• Start by finding the characteristic equation, derived from the homogeneous equation.
• Solve the characteristic equation to find the roots.
• Use the roots to write the general solution based on whether they are real, complex, or repeated.

For our step-by-step example, the characteristic equation \(r^2 + 2r + 1 = 0\) led to a repeated root \(r = -1\), resulting in a solution of \(y_h = C_1 e^{-x} + C_2 x e^{-x}\).

Particular Solution

The particular solution, denoted \(y_p\), is crucial for solving non-homogeneous differential equations. It accounts for the specific form of the external forces or functions present in the equation. In our example problem, the non-homogeneous part is \(\sin x + \cos x\).
To find a particular solution, a common approach is to propose a specific form based on the type of non-homogeneous term. This is where the method of undetermined coefficients shines:

• Identify the standard form of the non-homogeneous part. Here, it involves trigonometric functions.
• Propose a possible form for \(y_p\), i.e., \(A \sin x + B \cos x\).
• Fill in the general form with undetermined coefficients, like \(A\) and \(B\), which you will determine in later steps by substituting into the differential equation.

In this case, after calculation, the particular solution came out to be \(y_p = 0.5 \sin x\).

Differential Equation

A differential equation is simply an equation that involves an unknown function and its derivatives. These equations are fundamental in describing systems and processes in fields such as physics, engineering, and economics. The type of differential equation determines the method of solution.
Let's break down the parts of the original differential equation given in our problem: \(y'' + 2y' + y = \sin x + \cos x\). This is a second-order linear non-homogeneous differential equation because:

• It includes the second derivative \(y''\).
• It has constant coefficients as seen in \(y''\), \(y'\), and \(y\).
• The equation on the right side is a non-zero function, making it non-homogeneous.

Thus, solving it requires finding both the homogeneous solution (using the characteristic equation) and the particular solution (through an educated guess for \(y_p\)). Both solutions combine to give the general solution.

Characteristic Equation

The characteristic equation is a tool for finding the general solution of a homogeneous linear differential equation. It stems from the idea that the solution to these equations can often be written in the form of exponential functions due to their properties.
For a differential equation like \(y'' + 2y' + y = 0\), creating the characteristic equation involves:

• Replacing the derivatives in the homogeneous equation with powers of \(r\) (e.g., \( y'' \rightarrow r^2 \), \( y' \rightarrow r \)).
• Solving the resulting polynomial equation for \(r\).

In our solved example, the characteristic equation was \(r^2 + 2r + 1 = 0\), which was factored to \((r + 1)^2 = 0\), giving a repeated root of \(r = -1\). This root led to the specific form of the general solution for the homogeneous equation: \(y_h = C_1 e^{-x} + C_2 x e^{-x}\). The characteristic equation thus plays a central role in determining the behavior of the system described by the differential equation.