Section 1.2. The Classification of Differential Equations

Objective

1. The classification of differential equations: First Order, Second Order, Homogenous, Non-Homogeneous, Separable, Autonomous, Exact.

2. The Unique Solution Theorem

There are four important questions in the study of differential equations:
(a) Is there a solution? (Existence)
(b) If there is a solution, is it unique? (Uniqueness)
(c) If there is a solution, how do we find it?
(d) If there is a solution, what does this solution means for the model it presents?
In order to answer all questions above, we need to classify the differential equations first.

 

Definition:(a) An ordinary differential equations (ODE) is a differential equation such that the unknown function depends on a single variable, only ordinary derivatives appear in the equation.
(b) A partial differential equation (PDE) is a differential equation such that the unknown function depends on several variables, partial derivatives appear in the equation.
(c) An initial value problem (IVP) is an ODE or PDE with initial conditions given.

 

Example 1: The population model: [latex]y'=ay[/latex] is an ODE if [latex]y=f(t)[/latex] is the population function depending on the variable [latex]t[/latex], the time. If [latex]y(t_{0})=y_{0}[/latex] is given, the together with [latex]y'=ay[/latex], this is an initial value question. The heat conduction model: [latex]au_{tt}=u_{t}[/latex] is a PDE if [latex]u(x,t)[/latex] is the temperature function depending on the position, [latex]x[/latex] and the time, [latex]t[/latex]. If [latex]u(x_{0},t)=u_{1}[/latex] and [latex]u(x,t_{0})=u_{2}[/latex] are given, then it is an initial value problem.

 

 

 

 

Definition: The order of an ODE is the order of the highest derivative that appears in the equation.

 

Example 2: [latex]y^{(4)}+ty''+e^{t}yy'=t[/latex] is a 4th order differential equation. 

 

 

Definition: An ODE is linear if the differential equation is linear in the variables [latex]y,y',y'',...,y^{(n)}[/latex].

 

Example 3: (a) [latex]y^{(4)}+ty''+e^{t}yy'=t[/latex] is nonlinear as [latex]yy'[/latex] is not linear.

(b) [latex]y^{(4)}+ty''+e^{t}y=t[/latex] is linear even [latex]e^{t}[/latex] is nonlinear.

(c) [latex]y^{(4)}+t\text{sin}(y)+e^{t}y=t[/latex] is nonlinear as [latex]\text{sin}(y)[/latex] is not linear.

 

 

Exercise 3: Decide which ODE is linear and which is nonlinear. State which order the ODE is. 

(a) [latex]\text{cos}(t)y^{(5)}+ty''+2y'=\text{ln}(t)[/latex] 

(b) [latex]y^{(6)}+(2+t)y''-y=\text{ln}(y)[/latex] 

(c) [latex]y'''+ty+e^{y}=t[/latex] 

 

 

Theorem (Existence and Uniqueness): Let

$$y^{(n)}+p_{1}(t)y^{(n-1)}+p_{2}(t)y^{(n-1)}+\cdots+p_{n-1}(t)y’+p_{n}(t)y  =g(t),$$

$$y^{(n-1)}(t_{0})=y_{1},y^{(n-2)}(t_{0})=y_{2},…,y'(t_{0})=y_{n-1},y(t_{0})=y_{n}$$

be a linear [latex]n[/latex]-th order initial value problem. If the function [latex]p_{i}(t)[/latex]‘s and [latex]g(t)[/latex] are continuous on an open interval [latex]I[/latex]: [latex](a ,b)[/latex] containing the point [latex]t=t_{0}[/latex], then there exists a unique function [latex]y=f(t)[/latex] that satisfies the IVP for each [latex]t[/latex] in [latex]I[/latex].

 

 

 

Example 4: [latex](t+5)y''+\text{cos}(t)y'+\sqrt{t^{2}-9}=2\text{ln}(2-t)[/latex], [latex]y(-6)=4,y'(-6)=0[/latex] is an IVP. Determine an interval in which the solution of the given IVP is certain to exist.

 

 

 

Exercise 4: [latex](t-2)y''+\text{ln}(t)y'+e^{t}=\sqrt{9-t^{2}}[/latex], [latex]y(1)=4,y'(1)=2[/latex] is an IVP. Determine an interval in which the solution of the given IVP is certain to exist.

 

 

Group work

1. Decide which ODE is linear and which is nonlinear. State which

order the ODE is. 

(a) [latex]e^{y}y''+ty'+2y=\text{sin}(t)[/latex] 

(b) [latex]yy'=4t[/latex] 

(c) [latex]y'=(y-2)(y+1)[/latex] 

(d) [latex]y''-y'-2=0[/latex]

2. Determine an interval in which the solution of the given IVP is

certain to exist.

(a) [latex](t^{2}+1)y-4y'=-(t+4)(t-5)-(t+1)y''[/latex], [latex]y(-4)=4,y'(-4)=2[/latex].

(b) [latex]-(t+5)(t-1)y''-(t-5)y=t^{2}+4y'+16[/latex], [latex]y(-1)=2,y'(-1)=5[/latex].

(c) [latex](t^{2}-9)y''+\text{tan}(t)y'+e^{t}y=\sqrt{t^{2}+1}[/latex], [latex]y(2)=4,y'(2)=2[/latex].

License

Differential Equations Copyright © by Kuei-Nuan Lin. All Rights Reserved.

Share This Book