Euler's method is a method for approximating the solution to an initial value problem like the following IVP that comes from Newton's Model of cooling.
dy
-- = 0.05 (65 - y), y(0) = 180
dt
on the interval [0, 30]
or, more generally, an IVP of the form
dy
-- = f(t, y), y(a) = w
dt
on the interval [a, b]
The basic idea is simple. This differential equation tells us how rapidly the variable y is changing and the initial condition tells us where y starts. By evaluating the right hand side of the differential equation at the initial point (a, w) we can determine the initial rate at which y is changing. For example, for the IVP
dy
-- = 0.05 (65 - y), y(0) = 90
dt
on the interval [0, 30]
the initial rate of change of y is
0.05(65 - 180) = -5.75
Click here to open a new window with a Java applet. Arrange the two windows so that they overlap and you can move easily back-and-forth between the two windows by clicking on the exposed portion of the inactive window to make it active. Eventually when you are done with the Java applet close its window.
When this applet begins it is set up to study the initial value problem above. The range for t-axis is zero to 30 and the range for the y-axis is zero to 200. The function f(t, y) = 0.05 (65 - y) is entered in the box at the top of the applet. Notice that you need to use * to denote multiplication. You can change all of these things later but for now leave them as is.
Move your cursor on the Java applet to the point (0, 180). This is the initial value ( t = 0 and y = 180) for this initial value pronlem. Click the mouse button with your cursor at this point. You should see a dot and a line through the dot appear when you click the mouse at this point.
The slope of the line is determined by the right hand side of the differential equation -- that is, by the calculation above that showed y' = -5.75. This line gives us a rough approximation for the temperature y(t). It is exactly right at the initial point but as we move away from the initial it is less accurate. The figure below illustrates what is happening.
The blue dot at the point (0, 180) marks the initial condition -- the point at which the "cork is figuartively placed in the water." The red curve is the exact solution. The blue line is the tangent line to the real solution at the initial point. We computed the slope of this line above. Notice this gives us a very rough approximation to the exact solution. Like all tangent lines, it is close to the curve for a while. This differential equation describes the temperature of a cup of coffee that starts out at 180 degrees Fahrneheit in a room whose temperature is 65 degrees. The green line indicates the ambient temperature, 65 degrees.
We can use this tangent line to get a rough estimate for the temperature of the coffee after 30 minutes. This estimate is indicated by an arrow on the graph above. Note that it isn't a very good estimate. In fact, it predicts the coffee will freeze!
We can obtain a better approximation by making a "midcourse correction." Instead of continuing with the same slope (the same rate of change for y ) for the entire interval [0, 30] we will just follow this tangent until time t = 15. Return to the applet and click the mouse button at the point where the tangent line crosses the line t = 15. The portion of the tangent line between the initial point and the point at which you clicked should turn red and another line should be drawn going through the point at which you clicked. The slope of this line is determined by evaluating the right hand side of the differential equation at this point. Moving along this new line to t = 30 takes us to the point shown in the figure below.
Notice this approximation is considerably better than our first approximation. By making one midcourse correction we improved our results quite a bit. We can do even better by making more midcourse corrections. The figures below show several examples.
Return to the Java applet. Click anyplace on the gray border to start over again. Construct a solution by making starting at the same initial point and making midcourse corrections every three minutes.
Use the Java applet in your other window to estimate solutions to the initial value problems below.
dy
-- = 0.05 (65 - y), y(0) = 10
dt
on the interval [0, 30]
dy
-- = (y - 2) (y - 8), y(0) = 3
dt
on the interval [0, 10]
dy
-- = (y - 2) (y - 8), y(0) = 7
dt
on the interval [0, 10]
dy
-- = (y - 2) (y - 8), y(0) = 1
dt
on the interval [0, 10]
dy
-- = (y - 2) (y - 8), y(0) = 8.1
dt
on the interval [0, 10]