Solving First-Order Linear Differential Equations: A Comprehensive Guide
First-order linear differential equations are a fundamental part of mathematical modeling in various fields, including physics, engineering, and economics. In this article, we will explore the solution to a specific problem involving a first-order linear differential equation. We will also discuss the general approach to solving such equations and highlight the importance of the particular and complementary solutions.
Introduction to First-Order Linear Differential Equations
A first-order linear differential equation is a type of differential equation that can be written in the form:
A(x)y' B(x)y C(x)
where A(x), B(x), and C(x) are known functions of x, and y is the unknown function. In this article, we will solve the equation:
(D2 - 1)y e2xx2
Breaking Down the Problem
The given differential equation is:
(D2 - 1)y e2xx2
where D is the differential operator, represented as D d/dx. This equation can be rewritten as:
y'' - y e2xx2
General Approach to Solving Differential Equations
To solve the given differential equation, we need to find both the general solution and the particular solution. The general solution consists of the complementary solution and the particular solution.
1. Complementary Solution
The complementary solution is found by solving the homogeneous equation:
y'' - y 0
To find the general solution to this equation, we solve the characteristic equation:
r2 - 1 0
Solving for r, we get:
r2 1
r plusmn;1
The roots are real and distinct, so the complementary solution is:
yc(x) C1ex C2e-x
2. Particular Solution
The particular solution is found using the method of undetermined coefficients or variation of parameters. Here, we will use the method of undetermined coefficients, as the right-hand side is a polynomial multiplied by an exponential function.
We assume a particular solution of the form:
yp(x) Ae2xx2 Be2xx Ce2x
Substituting this into the original differential equation:
(D2 - 1)(Ae2xx2 Be2xx Ce2x) e2xx2
We find the coefficients A, B, and C by equating like terms.
The resulting particular solution is:
yp(x) (1/24)e2xx2 (1/12)e2xx (1/24)e2x
General Solution
The general solution is the sum of the complementary and particular solutions:
y(x) yc(x) yp(x)
Simplifying, we get:
y(x) C1ex C2e-x (1/24)e2xx2 (1/12)e2xx (1/24)e2x
Conclusion
In this article, we have explored the solution to a first-order linear differential equation using the method of undetermined coefficients. By breaking down the problem into the complementary and particular solutions, we were able to find the general solution to the given equation.
We have also discussed the general approach to solving first-order linear differential equations and highlighted the importance of both the complementary and particular solutions.