Week #16 - Differential Equations (Euler s Method) Section 11.3

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1 Week #16 - Differential Equations (Euler s Method) Section 11.3 From Calculus, Single Variable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 2005 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc. SUGGESTED PROBLEMS 1. Consider the differential equation y = x y whose slope field is in Figure Use Euler s method with x = 0.1 to estimate y when x = 0.4 for the solution curves satisfying (a) y(0) = 1 (b) y( 1) = 0 = x + y (a) So y(0.4) (b) Keep in mind that we don t usually ask questions that require this many steps. If we do, it s usually because there is a simple repeating pattern, as in this example. = x + y So y(0.4) = Consider the differential equation y = (sin x)(sin y). (a) Calculate approximate y-values using Euler s method with three steps and x = 0.1, starting at each of the following points: (i) (0, 2) (ii). (0, π). 1

2 (b) Use the slope field in Figure to explain your solution to part (a)(ii). (a) Remember to use radians in your calculator. = sin(x)sin(y) (i) (i) = sin(x)sin(y) 0 π π π π (b) If we consider the slope field at height y = π, the slopes will always be horizontal there because sin(π) = 0, so = sin(x)sin(y) = 0 for all y = π. We see this constant solution coming out of Euler s method in (ii). 10. Consider the differential equation = f(x) with initial value y(0) = 0. Explain why using Euler s method to approximate the solution curve gives the same results as using left Riemann sums to approximate x 0 f(t)dt. If we use Euler s method, with y = f(x), we will start at some x = x 0, and count up by intervals of x. This will produce estimates of y which will have the following form: y 1 = y 0 + f(x 0 ) x y 2 = y 1 + f(x 1 ) x... y n = y n 1 + f(x n 1 ) x where x n is where we want to stop, y n is our estimate of the function there, and each x i+1 = x i + x. Note that if we combine all of these terms together, we get y n = y n 1 + f(x n 1 ) x = (y n 2 + f(x n 2 x)) +f(x }{{} n 1 ) x y n 1... = y 0 + f(x 0 ) x + f(x 1 ) x f(x n 1 ) x = f(x 0 ) x + f(x 1 ) x f(x n 1 ) x since y(0) = 0 If instead we try to estimate the integral 2 xn 0 f(x) using rectangles, we will use

3 height is f(x 0 ) or f(x 1 ), or..., or f(x n 1 ) width of x Adding up the area of these rectangles gives us the Riemann sum Area = f(x 0 ) x + f(x 1 ) x f(x n 1 ) x which is exactly the same as the value calculated by Euler s method. QUIZ PREPARATION PROBLEMS 3. Consider the solution of the differential equation y = y passing through y(0) = 1. (a) Sketch the slope field for this differential equation, and sketch the solution passing through the point (0, 1). (b) Use Euler s method with step size x = 0.1 to estimate the solution at x = 0.1,0.2,...,1. (c) Plot the estimated solution on the slope field; compare the solution and the slope field. (d) Check that y = e x is the solution of y = y with y(0) = 1. (a) The slopes are steeper further away from the x axis, and zero along that axis. Slopes are positive above and negative below the x axis. (b) = y

4 (c) The points seem to go up in the same way that the graph does: small changes in y at first, followed by gradually larger and larger steps. (d) We can check that y = e x is the solution to = y,y(0) = 1 by seeing that y(0) = e 0 = 1 (initial condition) Left side = = d ex = e x = y = Right side 4. (a) Use Euler s method to approximate the value of y at x = 1 on the solution curve to the differential equation = x3 y 3 that passes through (0, 0). Use x = 1/5 (i.e., 5 steps). (b) Using Figure 11.28, sketch the solution that passes through (0, 0). Show the approximation you made in part (a). (c) Using the slope field, say whether your answer to part (a) is an overestimate or an underestimate. (a) Figure 11.28: Slope field for = x3 y 3. = x3 y So y(1) (b) It is too difficult to show the solution from part (a) in such a small diagram. Here is a sketch of the solution curve, though. 4

5 (c) Because the solution curve is convex up between x = 0 and x = 1, the solution from part (a) will be an underestimate of the real solution. 5. Consider the differential equation = 2x, with initial condition y(0) = 1. (a) Use Euler s method with two steps to estimate y when x = 1. Now use four steps. (b) What is the formula for the exact value of y? (c) Does the error in Euler s approximation behave as predicted in the box? (a) With two intervals between x = 0 and x = 1, we have x = 0.5. = 2x With four intervals between x = 0 and x = 1, we have x = = 2x (b) If = 2x, we can integrate directly, so y = x2 +C. Since we want a solution going through y = 1, C = 1 so our analytic (exact) solution becomes y = x (c) The text states that the error in Euler s method is roughly proportional to 1/(# of intervals). In that case, doubling the number of intervals should reduce the error by one half. At x = 1, Error = ((1) 2 + 1) }{{} }{{} 1.5 = 0.5 when x = 0.5. y(1) exact y(1) estimate At x = 1, Error = 0.25 when x = Thus doubling the number of intervals halved the error, as expected. 5