Slope fields and Euler’s method

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We describe numerical and graphical methods for understanding differential equations.

Slope fields

We cannot (yet!) solve the differential equation $y' = x+y$ However, from the equation alone, we can describe some facts about the solution.

Consider the solution $y$ to the differential equation $y'=x+y$ which passes through the point $(1,2)$ . Is $y$ increasing or decreasing at $x=1$ ?

increasing decreasing

By definition, a solution to this differential equation which passes through $(1,2)$ must have $y(1)=2$ , and $\begin{align*} y'(1) &= 1+y(1)\\ &=1+2\\ &=3. \end{align*}$ This is positive, so the function is increasing at $x=1$ .

In fact, we can say more. The differential equation $y' = x+y$ tells us the slope of any solution passing through a given point.

Consider the solution $y$ to the differential equation $y'=x+y$ which passes through the point $(1,-2)$ . What is the slope of the solution $y$ at $x=1?$

By definition, a solution to this differential equation which passes through $(1,-2)$ must have $y(1)=-2$ , and $\begin{align*} y'(1) &= 1+y(-1)\\ &= 1-2=-1. \end{align*}$ Hence the slope is $-1$ at $x=1$ .

The slope of $y$ at $x=1$ is $\answer {-1}$ .

Given a differential equation, say $y'=x+y$ , we can pick points in the plane and compute what the slope of a solution at those points will be. Repeating this process, we can generate a slope field. The slope field for the differential equation $y' = x+y$ looks like this:

Let’s be explicit:

A slope field, also called a direction field, is a graphical aid for understanding a differential equation, formed by:

Choosing a grid of points.
At each point, computing the slope given by the differential equation, using the $x$ and $y$ -values of the point.
At each point, drawing a short line segment with that slope.

Here is the slope field for the differential equation $y'=x+y$ , with a few solutions of the differential equation also graphed.

Notice that the slope field suggests one solution to this differential equation, which is a straight line.

Of the solutions shown in the slope field above, what is the formula for the linear solution? $y = \answer {-1-x}$

Autonomous differential equations

Consider the following differential equations $y' = x^3,\qquad y'=(y+1)(y-2), \qquad y'= x+y.$

The first differential equation, $y' = x^3$ , is rather easy to solve, we simply integrate both sides. This type of differential equation is called a pure-time differential equation. Pure-time differential equations express the derivative of the solution explicitly as a function of an independent variable. We can symbolically describe a pure-time differential equation as $\dd [y]{x} = F(x).$

On the other hand, the second differential equation, $y'=(y+1)(y-2)$ does not involve the independent variable, $x$ , at all! Such differential equations are called autonomous differential equations.

A differential equation that does not involve the independent variable is called an autonomous differential equation. We can symbolically write $\dd [y]{x} = F(y)$ to express an autonomous differential equation.

Finally the third differential equation, $y'= x+y$ , expresses $y'$ as a function of both $y$ and the independent variable $x$ . Such a differential equation is called a nonautonomous differential equation. We can symbolically describe a nonautonomous differential equation as $\dd [y]{x} = F(x,y).$

Which of the following are autonomous differential equations?

$y' = y^2+y$ $y'=x-y$ $y'=x$ $y'=y$ $y'=-x/y$

Since autonomous differential equations only depend on the function’s value their behavior does not depend on the independent variable,

Which of the five slope fields shown are for autonomous differential equations?

If a differential equation is autonomous, then all of the slopes will be the same on each horizontal line.

Which of the five slope fields shown are for pure-time differential equations?

If a differential equation is pure-time, then all of the slopes will be the same on each vertical line.

Which of the five slope fields shown are for differential equations that are neither autonomous nor pure-time?

If a differential equation is neither autonomous nor pure-time, then the slope will change along horizontal lines and vertical lines.

Consider the differential equation $y' = (y+1)(2-y)$ , whose slope field is given below. Which of the following statements appear to be true?

The solution passing through the origin has $\lim _{x \to \infty } f(x) = 2$ . The solution passing through $(0,3)$ has $\lim _{x \to \infty } f(x) = \infty$ . The solution passing through $(0,-3)$ is always decreasing. There are two solutions which are constant functions. Every solution has a vertical tangent line at $x=0$ .

Consider the autonomous differential equation: $y'=(y+1)(2-y)$ The constant functions $f(x) = -1$ and $f(x) =2$ are solutions to this differential equation. In fact, for any autonomous differential equation $y' = F(y)$ , where $F$ is a function of $y$ , if $F(c) = 0$ for any constant $c$ , then $y = c$ will be a constant solution to the differential equation. These constant solutions are also known as equilibrium solutions. We can witness these solutions if we inspect the slope field:

Consider the autonomous differential equation $y'=\cos (y)$ . Which of the following are equilibrium solutions?

$\pi$ $\frac {\pi }{2}$ $0$ $\frac {-\pi }{2}$ $-\pi$

Consider the slope field of $y'=\cos (y)$ :

Finally, let’s work an example problem:

Consider the differential equation: $y' = y^2 - 3y+2$ Find all equilibrium solutions.

First note that this differential equation is pure-timeautonomousnonautonomous , and hence any solution where $y'=\answer [given]{0}$ , will be an equilibrium solution. Hence we must solve the equation $y^2 -3y + 2 = \answer [given]{0}.$ Write with me to factor $y^2 -3y + 2 = (y-1)(\answer [given]{y-2}),$ hence $y= 1$ and $y=\answer [given]{2}$ are equilibrium solutions to this differential equation.

Euler’s Method

In science and mathematics, finding exact solutions to differential equations is not always possible. We have already seen that slope fields give us a powerful way to understand the qualitative features of solutions. Sometimes we need a more precise quantitative understanding, meaning we would like numerical approximations of the solutions.

Again, suppose you have set up the following differential equation $y' = x+y$ If we know that $y$ solves this differential equation, and $y(0) = 1$ , how might we go about approximating $y(1)$ ? One idea is to repeatedly use linear approximation.

Let us approximate just using the two subintervals $\left [0,\frac {1}{2}\right ]\qquad \text {and}\qquad \left [\frac {1}{2},1\right ].$ Since $y'(0) = 1$ , we know that $y\left (\frac {1}{2}\right )\approx 1 \cdot \frac {1}{2} + 1 = \frac {3}{2}$ by linear approximation.

But now, we have $y'\left (\frac {1}{2}\right ) \approx \frac {1}{2}+\frac {3}{2} = 2$ so $y(1) \approx \frac {1}{2} \cdot 2 + y\left (\frac {1}{2}\right )\approx \frac {5}{2}$

Plotting our approximation with the actual solution we find:

This approximation could be improved by using more subintervals. We will now formalize the method of using repeated linear approximation to approximate solutions to differential equations, and call it Euler’s Method.

Euler’s Method Consider a differential equation on the interval $[a,b]$ $y' = F(x,y)$ where $F$ is a function of two variables and it is given that $y(a) = y_0$ . To approximate $y$ on the interval:

Set $x_0=a$ and $y_0 = y(a)$ .
Decide either a step-size $\d x$ or how many subintervals $n$ you want to divide the interval $[a,b]$ into. Either way: $n\cdot \d x = b-a \qquad \text {or}\qquad \frac {b-a}{\d x} = n$ and $x_k = k\cdot \d x + a$ .
Iteratively define $y_{k+1} = y_k + F(x_k,y_k)\d x$ for $k=0,1,\dots ,n-1$ .

Then $y(x_k) \approx y_k$ , and so $y(b) \approx y_{n}$ .

Let’s try an example.

Use Euler’s method to approximate $y$ on $[0,2]$ where $\d x = .5$ for the differential equation $y'=x+y$ where $y(0) = 1$ .

Using the notation from the definition, we have $F(x,y) = \answer [given]{x+y}$ Fill out the following table for Euler’s method using $4$ subintervals. $\begin {array}{l|l|l} k & x_k & y_k \\ \hline 0 & 0 & 1 \\ 1 & 0.5 & \answer [given]{1.5} \\ 2 & \answer [given]{1} & \answer [given]{2.5} \\ 3 & 1.5 & 4.25 \\ 4 & 2 & \answer [given]{7.125} \end {array}$ Thus $y(2) \approx y_4 =\answer [given]{7.125}$ .