Section 2.4. Distinct Solutions of Differential Equations

Objective:

1. Definition of the characteristic equation of a differential equation

2. Classify roots of a characteristic equation and their corresponding solutions of a differential equation

3. Find a fundamental set of a 2nd order differential equation when roots are distinct.

In Section 2.1, we introduce the characteristic equation of a differential equation. We use it to find the solutions of the ODE. Here, we are finalize that what we found using the characteristic equation are indeed all possible solutions when the roots of the characteristic equation are distinct.

Definition: The characteristic equation of the ODE [latex]ay''+by'+cy=0[/latex] is [latex]a\lambda^{2}+b\lambda+c=0[/latex]. 

Recall from algebra class, we know a quadric equation, [latex]a\lambda^{2}+b\lambda+c=0[/latex], it has roots

$$\lambda=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$

When [latex]b^{2}-4ac\gt 0[/latex], we have two distinct real number roots, [latex]\lambda_{1}[/latex] and [latex]\lambda_{2}[/latex]. When [latex]b^{2}-4ac\lt 0[/latex], we have two distinct complex number roots, [latex]\lambda_{1}=\frac{-b+(\sqrt{4ac-b^{2}})i}{2a}=\alpha+\beta i[/latex] and [latex]\lambda_{2}=\frac{-b-(\sqrt{4ac-b^{2}})i}{2a}=\alpha-\beta i[/latex]. When [latex]b^{2}-4ac=0[/latex], then we have repeated root, [latex]\frac{-b}{2a}[/latex]. In this section, we deal with distinct roots. We will deal with repeated root in Section 2.5.

Case 1: When [latex]b^{2}-4ac>0[/latex], and [latex]\lambda_{1}\neq\lambda_{2}[/latex] are two real distinct numbers, then we know from Section 2.1 that [latex]e^{\lambda_{1}t}[/latex] and [latex]e^{\lambda_{2}t}[/latex] are solutions of [latex]ay''+by'+cy=0[/latex]. We compute the Wronskian of [latex]e^{\lambda_{1}t}[/latex] and [latex]e^{\lambda_{2}t}[/latex]:

[latex]W=\left[\begin{array}{cc}e^{\lambda_{1}t} & e^{\lambda_{2}t}\\ \lambda_{1}e^{\lambda_{1}t} & \lambda_{2}e^{\lambda_{2}t} \end{array}\right]=\lambda_{2}e^{(\lambda_{1}+\lambda_{2})t}-\lambda_{1}e^{(\lambda_{1}+\lambda_{2})t}\neq0.[/latex]

From Section 2.3, we know that [latex]e^{\lambda_{1}t}[/latex] and [latex]e^{\lambda_{2}t}[/latex] form a fundament set of solutions. Hence [latex]y=c_{1}e^{\lambda_{1}t}+c_{2}e^{\lambda_{2}t}[/latex] is a general solution as [latex]W[/latex] is never zero as long as [latex]\lambda_{1}\neq\lambda_{2}[/latex]. 

Example 1: Find the general solution of the ODE, [latex]y''+y'-2y=0[/latex].

Exercise 1: Find the general solution of the ODE, [latex]y''+4y'+3y=0[/latex].

Example 2: Find the solution of the IVP, [latex]y''-5y'+4y=0[/latex], [latex]y(0)=2[/latex], [latex]y'(0)=-1[/latex].

Exercise 2: Find the solution of the IVP, [latex]y''+y'-6y=0[/latex], [latex]y(0)=1[/latex], [latex]y'(0)=3[/latex].

Case 2: When [latex]b^{2}-4ac\lt0[/latex], and [latex]\lambda_{1}=\alpha+\beta i\neq\lambda_{2}=\alpha-\beta i[/latex] are two complex distinct numbers where [latex]\beta\neq0[/latex]. We can show [latex]e^{\alpha t}\text{cos}(\beta t)[/latex] and [latex]e^{\alpha t}\text{sin}(\beta t)[/latex] are both solutions of [latex]ay''+by'+cy=0[/latex].  We compute the Wronskian of [latex]e^{\alpha t}\text{cos}(\beta t)[/latex] and [latex]e^{\alpha t}\text{sin}(\beta t)[/latex]:

[latex]W=\left[\begin{array}{cc} e^{\alpha t}\text{cos}(\beta t) & e^{\alpha t}\text{sin}(\beta t)\\ \alpha e^{\alpha t}\text{cos}(\beta t)-\beta e^{\alpha t}\text{sin}(\beta t) & \alpha e^{\alpha t}\text{sin}(\beta t)+\beta e^{\alpha t}\text{cos}(\beta t) \end{array}\right]=\beta e^{2\alpha t}\neq0.[/latex]

From Section 2.3, we know that [latex]e^{\alpha t}\text{cos}(\beta t)[/latex] and [latex]e^{\alpha t}\text{sin}(\beta t)[/latex] form a fundament set of solutions. Hence [latex]y=c_{1}e^{\alpha t}\text{cos}(\beta t)+c_{2}e^{\alpha t}\text{sin}(\beta t)[/latex] is a general solution as [latex]W[/latex] is never zero as long as [latex]\beta\neq0[/latex].

Example 3: Find the general solution of the ODE, [latex]y''+y'+2y=0[/latex].

Exercise 3: Find the general solution of the ODE, [latex]y''+2y'+3y=0[/latex].

Example 4: Find the solution of the IVP, [latex]y''+4y=0[/latex], [latex]y(0)=2[/latex], [latex]y'(0)=-1[/latex].

Exercise 4: Find the solution of the IVP, [latex]y''+9y=0[/latex], [latex]y(0)=1[/latex], [latex]y'(0)=3[/latex].

Group Work:

1. Find the solution of the IVP, [latex]y''+2y'-8y=0[/latex], [latex]y(0)=0[/latex], [latex]y'(0)=3[/latex].

2. Find the solution of the IVP, [latex]y''+2y'+6y=0[/latex], [latex]y(0)=0[/latex], [latex]y'(0)=1[/latex].

3. Find the solution of the IVP, [latex]y''+25y=0[/latex], [latex]y(\frac{\pi}{2})=2[/latex], [latex]y'(\frac{\pi}{2})=-1[/latex].

4. Find the solution of the IVP, [latex]y''+4y'=0[/latex], [latex]y(0)=-1[/latex], [latex]y'(0)=1[/latex].