Solving Non-Homogeneous Differential Equations using Undetermined Coefficients: A Step-by-Step Guide

Differential equations are a fundamental tool in mathematics, used to model a wide range of phenomena in science and engineering. In this guide, we will explore a method to solve non-homogeneous differential equations using the method of undetermined coefficients. We'll apply this method to the following differential equation:

The Differential Equation to Solve

x^2 frac{d^2y}{dx^2} - 4x frac{dy}{dx} - 2y e^x

Step 1: Solving the Homogeneous Equation

First, we consider the associated homogeneous equation:

x^2 frac{d^2y}{dx^2} - 4x frac{dy}{dx} - 2y 0

This is a Cauchy-Euler equation, which can be solved by assuming a solution of the form y x^m. We will compute the derivatives and substitute them into the homogeneous equation.

Derivatives

frac{dy}{dx} mx^{m-1}

frac{d^2y}{dx^2} m(m-1)x^{m-2}

Substituting these into the homogeneous equation:

x^2 m(m-1)x^{m-2} - 4xmx^{m-1} - 2x^m 0

This simplifies to:

m(m-1)x^m - 4mx^m - 2x^m 0

Factoring out x^m:

(m(m-1) - 4m - 2)x^m 0

Setting the characteristic polynomial to zero:

m^2 - 3m - 2 0

Factoring or using the quadratic formula:

(m - 4)(m 1) 0 implies m 4, -1

General Solution to the Homogeneous Equation

The general solution to the homogeneous equation is:

y_h C_1 x^{-1} C_2 x^{-2}

where C_1 and C_2 are constants.

Step 2: Solving the Non-Homogeneous Equation

We need to find a particular solution y_p to the non-homogeneous equation. We will use the method of undetermined coefficients, assuming a particular solution of the form:

y_p Ae^x

where A is a constant to be determined.

Derivatives of the Particular Solution

frac{dy_p}{dx} Ae^x

frac{d^2y_p}{dx^2} Ae^x

Substituting y_p, frac{dy_p}{dx}, and frac{d^2y_p}{dx^2} into the left-hand side of the original equation:

x^2Ae^x - 4xAe^x - 2Ae^x e^x

This simplifies to:

x^2A - 4xA - 2A 1

Factoring out e^x, which is non-zero:

Ax^2 - 4Ax - 2A 1

Setting up a system of equations by equating coefficients:

For x^2: A 0

For x^1: 4A 0

For the constant term: 2A 1 implies A frac{1}{2}

The Particular Solution

Thus, we find:

A frac{1}{2}

Therefore, the particular solution is:

y_p frac{1}{2}e^x

Step 3: Combining Solutions

The general solution to the original non-homogeneous equation is the sum of the homogeneous and particular solutions:

y y_h y_p C_1x^{-1} C_2x^{-2} frac{1}{2}e^x

Final Solution

So, the final solution to the differential equation is:

y C_1x^{-1} C_2x^{-2} frac{1}{2}e^x

where C_1 and C_2 are constants determined by initial or boundary conditions.