DETA - Numerical Solution Using Euler's Method
Numerical Solution Using Slope Fields
Euler's Method
Recall when the concept of a derivative was first introduced, it represented the slope of a tangent line to a curve at a given point. The equation of the corresponding tangent line now serves as the basis for finding the numerical solution of a differential equation using Euler's method. For the differential equation y'= f(x, y) with initial condition y(x0) = y0, the solution curve can be approximated by its linearization L(x) = y(x0) + y'(x0)(x - x0) or y1 = y0 + f(x0, y0)(x - x0). The function L(x) provides a good approximation to the solution y(x) over a solution y(x) over a short interval about x0. Moving a small amount (step size) from x0 to x1 = x0 + dx, where dx = Δx in differential notation, produces the next linearization L(x) = y1 + f(x1, y1)(x - x1) or y2 = y1 + f(x1, y1)dx. Continuing this process, the next step from x2, y2 with a slope f(x2, y2), produces the third approximation y3 = y2 + f(x2, y2)dx. In general, yn = yn-1 + f(xn-1, yn-1)dx.
Approximating a Particular Solution
Euler's method is based on starting at the point given by the initial value and proceeding in the direction indicated by the slope field, stopping at a nearby location and proceeding in that direction (midcourse correction), and continuing the process of stringing linearizations together to approximate the curve over a longer interval. Euler's method does not produce an exact solution to an initial-value problem, but rather approximations whose accuracy may be increased by decreasing the step size, which increases the number of midcourse corrections.
When an exact solution for an initial-value problem y'= f(x, y), y(x0) = y0 cannot be immediately found or is not required, a table of approximate numerical values of y for values of x in an appropriate interval can be easily generated using technological tools. The table is considered a numerical solution of the problem, and the procedure used to generate such a table is referred to as a numerical method. The next presentation includes a derivation of Euler's Method.
Using Euler's Method to find the solution of y' = 1 + y with initial conditions y(0) = 1 and step size 0.05 produces the following solution data points.
|
x |
y |
|---|---|
|
0.000000 |
1.000000 |
|
0.050000 |
1.100000 |
|
0.100000 |
1.205000 |
|
0.150000 |
1.315250 |
|
0.200000 |
1.431013 |
|
0.250000 |
1.552563 |
|
0.300000 |
1.680191 |
|
0.350000 |
1.814201 |
|
0.400000 |
1.954911 |
|
0.450000 |
2.102656 |
|
0.500000 |
2.257789 |
|
0.550000 |
2.420679 |
|
0.600000 |
2.591713 |
|
0.650000 |
2.771298 |
|
0.700000 |
2.959863 |
|
0.750000 |
3.157856 |
|
0.800000 |
3.365749 |
|
0.850000 |
3.584037 |
|
0.900000 |
3.813238 |
|
0.950000 |
4.053900 |
|
1.000000 |
4.306595 |
Based on this calculation, the particular solution of the differential equation is approximately 4.3066. It is important to remember that Euler's Method for finding a numerical solution is only an approximation of the exact solution. The error in Euler's method is the difference between the exact value and the approximate value.
|
x |
y |
y(exact) |
Error |
|---|---|---|---|
|
0.000000 |
1.000000 |
1.000000 |
0.000000 |
|
0.050000 |
1.100000 |
1.102542 |
0.002542 |
|
0.100000 |
1.205000 |
1.210342 |
0.005342 |
|
0.150000 |
1.315250 |
1.323668 |
0.008418 |
|
0.200000 |
1.431013 |
1.442806 |
0.011793 |
|
0.250000 |
1.552563 |
1.568051 |
0.015488 |
|
0.300000 |
1.680191 |
1.699718 |
0.019527 |
|
0.350000 |
1.814201 |
1.838135 |
0.023934 |
|
0.400000 |
1.954911 |
1.983649 |
0.028738 |
|
0.450000 |
2.102656 |
2.136624 |
0.033968 |
|
0.500000 |
2.257789 |
2.297443 |
0.039654 |
|
0.550000 |
2.420679 |
2.466506 |
0.045827 |
|
0.600000 |
2.591713 |
2.644238 |
0.052525 |
|
0.650000 |
2.771298 |
2.831082 |
0.059784 |
|
0.700000 |
3.027505 |
3.087505 |
0.067642 |
|
0.750000 |
3.234000 |
2.644238 |
0.076144 |
|
0.800000 |
3.365749 |
3.451082 |
0.085333 |
|
0.850000 |
3.584037 |
3.679294 |
0.095257 |
|
0.900000 |
3.813238 |
3.919206 |
0.105968 |
|
0.950000 |
4.053900 |
4.171419 |
0.117519 |
|
1.000000 |
4.306595 |
4.436564 |
0.129969 |
Based on an exact solution y = 2ex - 1, the error continues to increase and the percent error can be computed using the formula:
%error=|exact−approximate|exact⋅100
Thus the percent error is approximately 0.029 (0.129969/4.436564) or 2.9%. Additional calculations necessary to reduce the error must be carefully weighed in terms of accuracy required and the resulting accumulation of round-off errors.
Another example of using Euler's Method to solve a differential equation is presented below.
Numerical Solution Using Euler's Method Practice
Numerical Solution Using Euler's Method: Even More Problems!
Complete problems from your textbook and/or online resources as needed to ensure your complete understanding of the numerical solution of a differential equation using Euler's Method.
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