Second Order Differential Equations
3.3 Nonhomogeneous Linear Second-order Differential Equations
A. General Solution of Nonhomogeneous Equations
In this section, we explore the nonhomogeneous linear second-order differential equation of the form:
[asciimath]y''+p(x)y'+q(x)y=f(x)[/asciimath] (3.3.1)
Uniqueness Theorem. If [asciimath]p(x)[/asciimath] and [asciimath]q(x)[/asciimath] are continuous on an open interval [asciimath](a,b)[/asciimath] and [asciimath]x_0[/asciimath] is in the interval, then the initial value problem has a unique solution within [asciimath]( a, b )[/asciimath].
To solve Equation 3.3.1, we first need the solutions to the associated homogeneous equation
[asciimath]y''+p(x)y'+q(x)y=0[/asciimath] (3.3.2)
We refer to Equation 3.3.2 as the complementary equation for Equation 3.3.1.
General Solution Theorem. [asciimath]y_p[/asciimath] is a particular solution to the nonhomogeneous Equation 3.3.1, and [asciimath]{y_1, y_2}[/asciimath] is a fundamental set of solutions to the complementary Equation 3.3.2, then the general solution of the nonhomogenous equation is
[asciimath]y = y_p + c_1 y_1 + c_2 y_2[/asciimath]. (3.3.3)
Here [asciimath]c_1 y_1 + c_2 y_2[/asciimath] represents the solution to the associated complementary equation, commonly referred to as [asciimath]y_c[/asciimath]. Therefore, Equation 3.3.3 often expressed as
[asciimath]y=y_p+y_c[/asciimath]
B. Superposition Principle
The superposition principle is a powerful tool that allows us to simplify solving nonhomogeneous equations. It works by dividing the forcing function into simpler components, finding a particular solution for each component, and then adding those solutions together to form a complete solution to the original equation.
Theorem. If [asciimath]y_(p1)[/asciimath] is a particular solution to the differential equation
[asciimath]yprimeprime+p(x)yprime+q(x)y=f_1(x)[/asciimath]
and [asciimath]y_(p2)[/asciimath] is a particular solution to the differential equation
[asciimath]yprimeprime+p(x)yprime+q(x)y=f_2(x)[/asciimath]
Then for any constants [asciimath]k_1[/asciimath] and [asciimath]k_2[/asciimath], [asciimath]y_p=k_1y_(p1)+k_2y_(p2)[/asciimath] is a particular solution to the differential equation
[asciimath]yprimeprime+p(x)yprime+q(x)y=k_1f_1(x)+k_2f_2(x)[/asciimath]
Given [asciimath]y_(p1)=3x/2-9/4[/asciimath] is a particular solution to [asciimath]y''+3y'+2y=3x[/asciimath] (i) and [asciimath]y_(p2)=e^(3x)/2[/asciimath] is a particular solution to [asciimath]y''+3y'+2y=10e^(3x)[/asciimath] (ii), find a particular solution to [asciimath]y''+3y'+2y=12x-20e^(3x)[/asciimath] (iii).
Show/Hide Solution
- The forcing function of equation (i): [asciimath]f_1(x)=3x[/asciimath]
- The forcing function of equation (ii): [asciimath]f_2(x)=10e^(3x)[/asciimath]
- The forcing function of equation (iii): [asciimath]f_3(x)=12x-20e^(3x)[/asciimath]
Looking at the right-hand side of the equations, we notice that [asciimath]f_3(x)=4f_1(x)-2f_2(x)[/asciimath]. Therefore, the same linear combination of [asciimath]y_(p1)[/asciimath] and [asciimath]y_(p2)[/asciimath] yields a particular solution for equation (iii):
[asciimath]y_(p3)=4y_(p1)-2y_(p2)[/asciimath]
[asciimath]=4((3x)/2-9/4)-2(e^(3x)/2)[/asciimath]
[asciimath]=6x-9-e^(3x)[/asciimath]
Try an Example
Section 3.3 Exercises
- Given [asciimath]y_(p1)=1/3 e^(-5 x)[/asciimath] is a particular solution to [asciimath]y''+6y'+8y = e^(-5x)[/asciimath], and [asciimath]y_(p2) = -5/8x+15/32[/asciimath] is a particular solution to [asciimath]y''+6y'+8y = -5x[/asciimath], use the method of superposition to find a particular solution to
[asciimath]y''+6y'+8y = -3 e^(-5x)+10x[/asciimath]
Show/Hide Answer
[asciimath]y_p=-e^(-5x)+5/4x-15/16[/asciimath]
- Given [asciimath]y_(p1)=-2/15 e^(3x)[/asciimath] is a particular solution to [asciimath]y''-4y'-12y = 2e^(3x)[/asciimath], and [asciimath]y_(p2) = 1/20sin(2x)-1/40cos(2x)[/asciimath] is a particular solution to [asciimath]y''-4y'-12y = -sin(2x)[/asciimath], use the principle of superposition to find a particular solution to
[asciimath]y''-4y'-12y = -5 e^(3x)+4sin(2x)[/asciimath]
Show/Hide Answer
[asciimath]y_p = 1/3e^(3x)-1/5sin(2x)+1/10cos(2x)[/asciimath]
