Section 2.7. Solutions of non-homogeneous differential equations

Objective:

1. Definition of the complementary solution and a particular solution of a differential equation

2. Find the solution of a non-homogeneous differential equation

Recall for Section 2.1, this whole chapter is trying to solve the 2nd order differential equations, [latex]y''+p(t)y'+q(t)y=g(t)[/latex]. When [latex]g(t)[/latex] is zero, this type of differential equation is called homogeneous. When [latex]g(t)[/latex] is not zero, this type of differential equation is called non-homogenous. 

Definition: Given a differential equation, [latex]y''+p(t)y'+q(t)y=g(t)[/latex], the general solution of [latex]y''+p(t)y'+q(t)y=0[/latex] is called the complementary solution of the ODE. A particular [latex]y_{p}[/latex] such that [latex]y_{p}''+p(t)y'_{p}+q(t)y_{p}=g(t)[/latex] is called a particular solution of the ODE.

Example 1: Find the complementary solution of the ODE and a particular solution of the ODE. [latex]y''+y'-2y=e^{2t}[/latex].

Exercise 1: Find the complementary solution of the ODE and a particular solution of the ODE. [latex]y''-y'-6y=e^{4t}[/latex].

Theorem: If [latex]Y_{1}[/latex] and [latex]Y_{2}[/latex] are two solutions of [latex]y''+p(t)y'+q(t)y=g(t)[/latex], then [latex]Y_{1}-Y_{2}[/latex] is a solution of [latex]y''+p(t)y'+q(t)y=0[/latex]. In particular, if [latex]y_{1}[/latex] and [latex]y_{2}[/latex] form a fundamental set of solutions of [latex]y''+p(t)y'+q(t)y=0[/latex], then there are [latex]c_{1}[/latex] and [latex]c_{2}[/latex] such that [latex]Y_{1}-Y_{2}=c_{1}y_{1}+c_{2}y_{2}[/latex].

Proof: 

Theorem: The general solution of [latex]y''+p(t)y'+q(t)y=g(t)[/latex] is [latex]y=y_{c}+y_{p}[/latex] where [latex]y_{c}[/latex] is the general solution of [latex]y''+p(t)y'+q(t)y=0[/latex] and [latex]y_{p}[/latex] is a particular solution of [latex]y''+p(t)y'+q(t)y=g(t)[/latex]. 

Proof:

Example 2: Find a general solution of [latex]y''+2y'-3y=e^{2t}[/latex]. 

Exercise 2: Find a general solution of [latex]y''+5y'+6y=e^{2t}[/latex].

Example 3: Find a general solution of [latex]y''+2y'+4y=2t^{2}+t[/latex]. 

Exercise 3: Find a general solution of [latex]y''+6y'+9y=4t^{2}-1[/latex].

Example 4: Find a general solution of [latex]y''+4y'=\text{cos}(2t)[/latex]. 

Exercise 4: Find a general solution of [latex]y''+9y=\text{sin}(5t)[/latex].

Group Work

1. Find the solution of the IVP: [latex]y''-2y'-3y=e^{2t}+\text{cos}(t)[/latex], [latex]y(0)=1,[/latex] and [latex]y'(0)=-1[/latex].

2. Find a general solution of [latex]y''+y'+4y=\text{cos}(3t)+t[/latex]. 

3. Find a general solution of [latex]y''-4y'-5y=te^{-3t}[/latex].

4. Find a general solution of [latex]y''-4y'+4y=t\text{sin}(t)[/latex].