Table of Contents
Differential Equations
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Topic PlaylistThese theorems are applicable to initial value problems
\begin{equation} y' = f(x,y) \qquad y(x_{0}) = y_{0} \end{equation}where the point $(x_{0},y_{0})$ is contained in an open rectangle $a < x < b$, $c < y < d$ called $R$.
Explanation VideoThe meaning of $A \Rightarrow B$, which is read as "A implies B" or "if A then B", has a specific meaning in mathematics that doesn't quite match up with normal conversation.
The short definition is when $A$ is a true statement, $B$ is a true statement. When $A$ is a false statement, $B$ could be either true or false.
A more concrete example is statement $A$ is "I ran over your laptop with my car." and statement $B$ is "Your laptop won't turn on." where your laptop is a normal laptop.
• If statement $A$ is true, then statement $B$ is guaranteed to be true. Normal laptops do not turn on after being crushed under the wheel of a car.
• If statement $A$ is false, then statement $B$ could be true or false. Just because I didn't run over your laptop with my car doesn't mean your laptop won't turn on. There are many reasons your laptop won't turn on, like the battery is dead, you spilled water on it, and I ran over your laptop with my car.
For $A \Rightarrow B$, when $A$ is a false statement, you know absolutely nothing about statement $B$.
Explanation Video$f(x,y)$ is continuous on the open rectangle $R \qquad \Rightarrow \qquad$ the initial value problem has at least one solution on some open subinterval of $(a,b)$ that contains $x_{0}$.
Explanation Video$f(x,y)$ and $\dfrac{\partial f}{\partial y}$ are continuous on the open rectangle $R \qquad \Rightarrow \qquad$ the initial value problem has exactly one solution on some open subinterval of $(a,b)$ that contains $x_{0}$.
Solution VideoDetermine when an initial value problem would exist or is unique based on the theorems. Use the Direction Field Plotter by Ariel Barton to plot some integral curves. Solve the ODE. Investigate what happens when either $f(x,y)$ or $\dfrac{\partial f}{\partial y}$ are not continuous, if ever.
\begin{equation} y' = 2xy \end{equation} Solution VideoDetermine when an initial value problem would exist or is unique based on the theorems. Use the Direction Field Plotter by Ariel Barton to plot some integral curves. Solve the ODE and determine why the left half of the parabola doesn't solve the ODE. Investigate what happens when either $f(x,y)$ or $\dfrac{\partial f}{\partial y}$ are not continuous, if ever.
\begin{equation} y' = 2\sqrt{y} \end{equation} Solution VideoDetermine when an initial value problem would exist or is unique based on the theorems. Use the Direction Field Plotter by Ariel Barton to plot some integral curves. Solve the ODE. Investigate what happens when either $f(x,y)$ or $\dfrac{\partial f}{\partial y}$ are not continuous, if ever.
\begin{equation} xy' + y = 3x^{2} \end{equation} Solution VideoDetermine when an initial value problem would exist or is unique based on the theorems, if ever. Use the Direction Field Plotter by Ariel Barton to plot some integral curves. Solve the ODE. Investigate what happens when either $f(x,y)$ or $\dfrac{\partial f}{\partial y}$ are not continuous, if ever.
\begin{equation} y' = \frac{y+1}{x+1} \end{equation} Solution VideoDetermine when an initial value problem would exist or is unique based on the theorems. Use the Direction Field Plotter by Ariel Barton to plot some integral curves. Do NOT solve the ODE. Investigate what happens when either $f(x,y)$ or $\dfrac{\partial f}{\partial y}$ are not continuous, if ever.
\begin{equation} y' = \frac{1}{1-x^{2}-y^{2}} \end{equation} Solution VideoDetermine when an initial value problem would exist or is unique based on the theorems. Use the Direction Field Plotter by Ariel Barton to plot some integral curves. Investigate what happens when either $f(x,y)$ or $\dfrac{\partial f}{\partial y}$ are not continuous, if ever. Do NOT attempt to solve the ODE in general, but set $y = mx$ and solve for the approximate value of $m$ that satisfies the ODE.
\begin{equation} y' = \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \end{equation}