Numerical SolutionsExpected Educational ResultsInitial Value ProblemsSolutions to ODEsInitial Value Problems (IVPs)Definition: Initial Value Problem (IVP)Existence and Uniqueness TheoremUsing the Existence and Uniqueness TheoremInvestigation 01CC BY-NC-SA 4.0

**Author**: John J Weber III, PhD
**Corresponding Textbook Sections**:

**Section 1.2**– Solutions and Initial Value Problems**Section 1.4**– The Approximation Method of Euler

**Objective 4–1**: I can estimate the solution to an IVP using Euler’s Method.**Objective 4–2**: I can use the Existence and Uniqueness Theorem to determine if a first-order IVP has a unique solution.

There are three general methods for solving ODEs that we will discuss in this course:

Graphical solutions [A sketch or graph representing

**ALL**potential solutions to an ODE]Phase lines: first-order autonomous ODEs (See CPT_03_Direction_Fields);

Direction fields: first-order ODEs (See CPT_03_Direction_Fields);

Phase planes: systems of first-order ODEs (To be discussed in CPT_17_Phase_Planes).

Numerical Solutions [A numerical computation to estimate a

-value when given an initial point$y$ on a single unknown solution curve of an ODE:$y({x}_{0})={y}_{0}$ Euler's Method (To be discussed in a project)

Heun's Method (To be discussed in a project);

Fourth-order Runge Kutta [RK4] Method (To be discussed in a project).

Analytical Solutions (the majority of the methods discussed in the course)

An **IVP** for an

**IVP Example**

An example of a first-order IVP is **all** curves

When solving IVPs, there are three possibilities:

There is

**no**solution to the IVP, i.e., there is**no**solution curve that passes through the point .$({x}_{0},{y}_{0})$ There is a

**unique**solution to the IVP, i.e., there is*only one*solution curve that passes through the point .$({x}_{0},{y}_{0})$ The solution to the IVP is

**not unique**, i.e., there is*more than one*solution curve that passes through the point .$({x}_{0},{y}_{0})$

Given an **IVP** of the form

**Procedure**

Determine if

**both**of the conditions of the theorem are met (make sure that you provide a supporting explanation):Determine if

is continuous on some open rectangle containing the point$f(x,y)$ .$({x}_{0},{y}_{0})$ Determine if

is continuous on the same open rectangle containing the point$\frac{\partial f}{\partial y}$ .$({x}_{0},{y}_{0})$

State your conclusion:

*Use an appropriate sentence similar to one of the following statements*Since

$f(x,y)$ **and** are both continuous on some open rectangle containing the point$\frac{\partial f}{\partial y}$ , then the IVP has a unique solution on some interval about$({x}_{0},{y}_{0})$ .$x$ Since

is not continuous on some open rectangle containing the point$f(x,y)$ , then the IVP does not have a unique solution on some interval about$({x}_{0},{y}_{0})$ .$x$ Since

is not continuous on some open rectangle containing the point$\frac{\partial f}{\partial y}$ , then the IVP does not have a unique solution on some interval about$({x}_{0},{y}_{0})$ .$x$

For each of the following linear IVPs, determine, if there is a unique solution:

,$x{y}^{{\textstyle \phantom{\rule{0.167em}{0ex}}}\prime}=y+3{x}^{2}$ $y(0)=0$ ,$y}^{{\textstyle \phantom{\rule{0.167em}{0ex}}}\prime}=\frac{x}{y+1$ $y(0)=0$ ,$x{y}^{{\textstyle \phantom{\rule{0.167em}{0ex}}}\prime}-y={x}^{2}\mathrm{cos}(x)$ $y(0)=-3$ ,$y}^{{\textstyle \phantom{\rule{0.167em}{0ex}}}\prime}=1+{y}^{2$ $y\left(\frac{\pi}{2}\right)=0$ ,${y}^{{\textstyle \phantom{\rule{0.167em}{0ex}}}\prime}=x-y+1$ $y(1)=2$ ,$y}^{{\textstyle \phantom{\rule{0.167em}{0ex}}}\prime}=x+\sqrt{y$ $y(1)=0$ ,$y}^{{\textstyle \phantom{\rule{0.167em}{0ex}}}\prime}=\frac{y}{2x$ $y(0)=1$ ,${x}^{{\textstyle \phantom{\rule{0.167em}{0ex}}}\prime}=\frac{x}{t}+3t$ $x(1)=0$ ,$t{y}^{{\textstyle \phantom{\rule{0.167em}{0ex}}}\prime}-y={t}^{2}\mathrm{cos}(t)$ $y(0)=-3$

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**Last Modified**: Wednesday, 2 September 2020 11:48 EDT