Problem 130 - Calculus 3

Question

For the following problems, find the general solution. $$y^{\prime \prime}=3 y^{\prime}+x e^{-x}$$

Step-by-Step Solution

Answer

The general solution is \( y(x) = C_1 + C_2 e^{3x} + x e^{-x} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( y'' = 3y' + xe^{-x} \). This is a linear non-homogeneous second-order differential equation due to the presence of the non-homogeneous term \( xe^{-x} \).

Step 2: Solve the Homogeneous Equation

First, find the solution to the homogeneous equation: \( y'' - 3y' = 0 \). The characteristic equation for this is \( r^2 - 3r = 0 \), which factors to \( r(r - 3) = 0 \). Thus, \( r = 0 \) and \( r = 3 \). The general solution to the homogeneous equation is \( y_h = C_1 + C_2 e^{3x} \).

Step 3: Find a Particular Solution to the Non-homogeneous Equation

Use the method of undetermined coefficients or variation of parameters to find a particular solution. We try a particular solution of the form \( y_p = Ax^2 e^{-x} + Bxe^{-x} \). Substituting into the differential equation and matching coefficients, you adjust \( A \) and \( B \) to satisfy the equation.

Step 4: Match Coefficients for Particular Solution

Differentiate \( y_p \) to find \( y_p'' \) and \( y_p' \) and substitute them back into the differential equation. After simplification and matching coefficients for terms involving \( x \) and constants, solve the system of equations to find values for \( A \) and \( B \). In this case, assuming \( y_p=Dx e^{-x} \) simplifies the process and can find \( D \).

Step 5: Verify and Simplify Particular Solution

Upon verification, \( y_p = x e^{-x} \). Substitute and verify whether this satisfies the original differential equation. Since it does, this confirmation allows us to combine the solutions.

Step 6: Write the General Solution

The general solution \( y(x) \) is the sum of the homogeneous and particular solutions:\[ y(x) = C_1 + C_2 e^{3x} + x e^{-x} \] where \( C_1 \) and \( C_2 \) are arbitrary constants.

Key Concepts

Second-order Differential Equations

Second-order differential equations involve second derivatives. Notated as \( y'' \) for example, these equations map how rapidly a rate of change itself changes. They are imperative in understanding dynamic systems in physics, engineering, and many more fields.

These equations can either be homogeneous or non-homogeneous, causing their solution methods to differ slightly.

• Homogeneous equations have a right side of zero, making them solvable with characteristic equations.
• Non-homogeneous equations, like the one discussed, have an additional term (often involving functions of x) that requires particular solutions.

For our exercise, with the differential equation \( y'' = 3y' + xe^{-x} \), it's a non-homogeneous second-order linear equation because of the \( xe^{-x} \) term. The order comes from \( y'' \), which tells us to consider up to the second derivative.

Method of Undetermined Coefficients

The Method of Undetermined Coefficients is a strategic way to find a particular solution of non-homogeneous linear differential equations. This method assumes a solution form based on the non-homogeneous part of the equation and determines its coefficients. It's particularly effective when the non-homogeneous term is a polynomial, exponential, or sine/cosine.

Here's how it unfolds:

• Choose a trial function (or guess) for the particular solution that resembles the structure of the non-homogeneous term.
• Differentiate it as needed and substitute back into the differential equation.
• Solve for the unknown coefficients by equating terms from both sides of the equation.

For example, in our problem, the non-homogeneous term is \( xe^{-x} \), leading us to choose a trial solution \( y_p = Ax^2 e^{-x} + Bxe^{-x} \). You substitute, differentiate, and finally solve for these coefficients, confirming our choice with verification.

Homogeneous Solutions

Homogeneous solutions form the foundation of solving non-homogeneous differential equations. They are solutions to the associated homogeneous equation, obtained by setting the non-homogeneous part to zero. In our example, the homogeneous equation is \( y'' - 3y' = 0 \).

Here's how you solve it:

• Find the characteristic equation by substituting \( y = e^{rx} \), yielding \( r^2 - 3r = 0 \).
• Solve for \( r \), giving roots \( r = 0 \) and \( r = 3 \).
• Build the general solution from these roots: \( y_h = C_1 + C_2 e^{3x} \).

These solutions, combined with particular solutions, lead to the general solution for the original non-homogeneous equation. Understanding this step is crucial because it represents the underlying behavior of the system when external influences are absent.