Lesson 3 - Direction Fields

Recall from Calculus: Given a function $f(x)$, the derivative of $f_y$ at a specific point $(x,y)$ yield a slop at that specific point, where $y(a)=$ slope of the tangent line at $x=a$.

Similarly, the slope of a solution curve can be graphed, and will give us different solution curves at various points. This is known as the direction field.

Use the link to analyze the direction field of $y^{\prime }=\frac{-y}{x}$.

Notice that:

For $y>0$:
$\underset{x\to 0^{+}}{lim}$, $-\frac{y}{x}=-\mathrm{\infty }$.
$\underset{x\to 0^{-}}{lim}$, $-\frac{y}{x}=\mathrm{\infty }$
$\underset{x\to \mathrm{\infty }}{lim}$, $-\frac{y}{x}=0$
$\underset{x\to -\mathrm{\infty }}{lim}$, $-\frac{y}{x}=0$

Note: The solution curves can never intersect as it would contradict existence and uniqueness theorem.

Example

The logistic equation for the population (in thousands) of a certain species is given by $\frac{dp}{dt}=3p-2p^2$.

a) Use the tool to plot the direction field. Because population can never be negative, we are only focused in the first quadrant.

b) If the initial population is 3000, what can you say about the limiting population $\underset{t\to \mathrm{\infty }}{lim}p(t)$?

Since we are in thousands, $p=3$. By analyzing the direction field, we can see that as $t$ goes to $\mathrm{\infty },$ $p(t)=3$ decreases and becomes steady at $p(t)=1.5$.

c) If $p(0)=0.8$, what is $\underset{t\to \mathrm{\infty }}{lim}p(t)$?

As t goes to infinity, $p(0)$ = 0.8 increases and becomes steady at $p(t)=1.5$.

d) Can a starting population of 2000 ever decline to 800?

The answer here is no because there is a horizontal asymptote at $p(t)=1.5$. This also known as convergence. A solution curve can never cross the line of convergence.

Try For Yourself:

Consider the DE: $\frac{dy}{dx}=x+\sin y$

a) A solution curve passes through the point $(1,\frac{\pi }{2})$. What is the slope at this point?
b) Show that every solution curve is increasing for x>1.
c) Show that the second derivative of every solution satisfied $\frac{d^{2}y}{d{x}^2}=1+x\cos y+\frac{1}{2}\sin 2y$.
d) A solution curve passes through $(0,0)$. Prove that this curve has a relative minimum at $(0,0)$.

Next Lesson: Lesson 4 - Euler's Method

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