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MTH 256 Supplement to Judson's “The Ordinary Differential Equations Project”

David Froemke, Heiko Spoddeck

Section 1.6 Existence and Uniqueness of Solutions

Exercises Exercises

Supplemental HW Problems.

ODE = Ordinary Differential Equation;
IVP = Initial Value Problem
1.
Consider the differential equation
\begin{equation*} y’ = 5y^{4/5} \end{equation*}
  1. Given the initial condition \(y(0) = 0\text{,}\) what can we say about the existence or uniqueness of the solutions of our differential equation?
  2. Given the initial condition \(y(0) = 1\text{,}\) what can we say about the existence or uniqueness of the solutions of our differential equation?
  3. If either of the above initial conditions give uniquely existing solutions, give a rectangle that we can use to contain those possible solutions.
2.
Why and how does the formal definition of Uniqueness and Existence work? We are not asking for a formal proof here – what are the ideas behind the proof? Why does the continuity of the differential function, and of the partial derivative with respect to \(y\text{,}\) mean the differential equation has one and only one solution for a given initial condition? (Hint: Talk about continuity here. If we have some function \(f(x, y, y’)\) that’s not continuous for some inputs, what does that mean? Why could that give something that’s not unique, or that doesn’t exist?)
3.
Find a rectangular interval in \(t\) and \(y\) in which the initial value ODE
\begin{equation*} ty’ + 2y = 4t^2, y(1) = 3 \end{equation*}
has a unique solution.
4.
Given the initial value ODE
\begin{equation*} y’ = \frac{t}{y-2}, y(-1) = 0\text{,} \end{equation*}
find a solution to this IVP and give a domain for where the solution exists uniquely. What happens as we approach the limit of its defined domain?
5.
Solve the initial value ODE
\begin{equation*} x’ = \frac{x}{t}, x(-1) = -2\text{.} \end{equation*}
Verify your solution – does it work? Give an interval for t for the solution you found.
6.
For the initial value ODE
\begin{equation*} x’ = x^{1/3}, x(0) = 0\text{,} \end{equation*}
Wolfram Alpha tells us that the solution is
\begin{equation*} x(t) = \frac{2}{3}\sqrt{\frac{2}{3}} t^{3/2}\text{.} \end{equation*}
Explain why that’s not correct using the techniques of this chapter.
7.
Using the same ODE
\begin{equation*} x’ = x^{1/3}\text{,} \end{equation*}
what solutions do we get for the initial conditions \(x(0) = 1\) and \(x(0) = -1\text{?}\) What kind of intervals of uniqueness do we get for these two different IVPs?
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