Lecture Notes in Differential Equations - Bruce E. Shapiro

362 LESSON 33. NUMERICAL METHODS
We then observe that since y n ≈ y(t n ) and y n+1 ≈ y(t n+1 ), then (33.12)
follows immediately from (33.16).
If the scalar initial value problem of equation (33.4) is replaced by a systems
of equations
y ′ = f(t, y), y(t 0 ) = y 0 (33.17)
then the Forward Euler’s Method has the obvious generalization
y n+1 = yn + hf(t n , y n ) (33.18)
Example 33.1. Solve y ′ = y, y(0) = 1 on the interval [0, 1] using h = 0.25.
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:
⎫
(t 0 , y 0 ) = (0, 1) ⎪⎬
f(t, y) = y
⎪⎭
h = 0.25
We first compute the solution at t = t 1 .
(33.19)
y 1 = y 0 + hf(t 0 , y 0 ) = 1 + (0.25)(1) = 1.25 (33.20)
t 1 = t 0 + h = 0 + 0.25 = 0.25 (33.21)
(t 1 , y 1 ) = (0.25, 1.25) (33.22)
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 ) (33.23)
= 1.25 + (0.25)(1.25) = 1.5625 (33.24)
t 2 = t 1 + h = 0.25 + 0.25 = 0.5 (33.25)
(t 2 , y 2 ) = (0.5, 1.5625) (33.26)
y 3 = y 2 + hf(t 2 , y 2 ) (33.27)
= 1.5625 + (0.25)(1.5625) = 1.953125 (33.28)
t 3 = t 2 + h = 0.5 + 0.25 = 0.75 (33.29)
(t 3 , y 3 ) = (0.75, 1.953125) (33.30)