The Computable Differential Equation Lecture ... - Bruce E. Shapiro

CHAPTER 3. APPROXIMATE SOLUTIONS 45
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Algorithm 3.1. Forward Euler Method To solve the initial value
problem
y ′ = f(t, y), y(t 0 ) = y 0
on an interval [t 0 , t max ] with a fixed step size h.
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1. input: f(t, y), t 0 , y 0 , h, t max
2. output: (t 0 , y 0 )
3. let t = t 0 , y = y 0
4. while t < t max
(a) let y = y + hf(t, y)
(b) let t = t + h
(c) let t n = t, y n = y
(d) output: (t n , y n )
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Example 3.1. Solve y ′ = y, y(0) = 1 on the interval [0, 1] using h = 0.25.
Solution. The exact solution is y = e x . We compute the values using Euler’s method.
For any given time point t k , the value y k depends purely on the values of t k1 and
y k1 . This is often a source of confusion for students: although the formula y k+1 =
y k + hf(t k , y k ) only depends on t k and not on t k+1 it gives the value of y k+1 .
We are given the following information:
We first compute the solution at t = t 1 .
(t 0 , y 0 ) = (0, 1) (3.16)
f(t, y) = y (3.17)
h = 0.25 (3.18)
y 1 = y 0 + hf(t 0 , y 0 ) = 1 + (0.25)(1) = 1.25 (3.19)
t 1 = t 0 + h = 0 + 0.25 = 0.25 (3.20)
(t 1 , y 1 ) = (0.25, 1.25) (3.21)
Then we compute the solutions at t = t 1 , t 2 , . . . until t k+1 = 1.
y 2 = y 1 + hf(t 1 , y 1 ) (3.22)
= 1.25 + (0.25)(1.25) = 1.5625 (3.23)
t 2 = t 1 + h = 0.25 + 0.25 = 0.5 (3.24)
(t 2 , y 2 ) = (0.5, 1.5625) (3.25)
c○2007, B.E.Shapiro
Last revised: May 23, 2007
Math 582B, Spring 2007
California State University Northridge