Slope Field Review

Standard slope field example: $y^{\prime}=(y-3)(y+1)$, the equilibrium solution $=$ constant solution.
$$\notag y=C \quad \text { if and only if } \quad y^{\prime}=0$$

Thus to find equilibrium solution(s) if there are any, set $y^{\prime}=0$ :
$0=(y-3)(y+1)$ implies $y=3$ and $y=-1$
Since these are constant functions, the equilibrium solutions are $y=3$ and $y=-1$.

If $y^{\prime}=f(x)$ is a piecewise continuous function, the slope can only change from positive to negative and vice versa by passing through
1) a slope of 0 (horizontal tangent line) or
2) a slope of $\infty$ (vertical tangent line) or undefined.

import matplotlib.pyplot as plt
import numpy as np
from sympy import var, plot_implicit

#Seting up the grid for us to put the arrows on
nx, ny = .5, .5
x = np.arange(-5, 5, nx)
y = np.arange(-5, 5, ny)
X, Y = np.meshgrid(x, y)

#this is the original equation
dy = Y**2-2*Y-3
dx = np.ones(dy.shape)

#normalized rates of change
dyn = dy/np.sqrt(dy**2 +dx**2)
dxn = dx/np.sqrt(dy**2 +dx**2)



plt.rcParams['figure.figsize'] = [10, 10]
plot1 = plt.plot()
plt.quiver(X, Y, dxn, dyn, 
           color='Blue', 
           headlength=5) #this is the direction field with normalized arrow size
plt.show()

 

Definition: Initial value

A chosen point $(t_0,y_0)$ through which a solution must pass. 

I.e. $(t_0,y_0)$ lies on the graph of the solution that satisfies this initial value. 

Definition: Initial value problem (IVP)

A differential equation where initial value is specified.

An initial value problem can have 0 , 1 , or multiple equilibrium solutions (finite or infinite).

Long-term behavior 

Suppose a solution $y=f(t)$ to the differential equation $y^{\prime}=(y-3)(y+1)$ passes through the point $\left(t_0, y_0\right)$.

If $y_0>3$, then $\lim\limits_{t \to \infty} f(t) =$

If $y_0=3$, then $\lim\limits_{t \to \infty} f(t) =$

If $y_0<3$, then $\lim\limits_{t \to \infty} f(t) =$

 

Standard slope field example: $y^{\prime}=(y-3)(y+1)$
2.5 Preview:
$y=3$ is an unstable equilibrium solution
$y=-1$ is a stable equilibrium solution

Note: You don't need the slope field graph to determine stability.

Note also that $y^{\prime}=(y-3)(y+1)$ is autonomous. That is $y^{\prime}$ depends only on $y$ : $\quad y^{\prime}=f(y)$