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
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.
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].