Math 3400
Homework 4
Section 2.7
1. a) Find approximate values of the solution of the given initial value problem using
Euler’s method with h = 0.1
y 0 = 3 + t − y, y(0) = 1.
We define approximate values of the solution at t = 0, 0.1, 0.2, 0.3, 0.4 by using
yn+1 = yn + hf (tn , yn ).
In this case
f (t, y) = 3 + t − y.
We have y0 = y(0) = 1 and
y1 = 1 + 0.1(3 + 0 − 1) = 1.2,
y2 = 1.2 + 0.1(3 + 0.1 − 1.2) = 1.39,
y3 = 1.39 + 0.1(3 + 0.2 − 1.39) = 1.571,
y4 = 1.571 + 0.1(3 + 0.3 − 1.571) = 1.7439.
d) The equation is first order linear,
y 0 + y = 3 + t.
Using the integrating factor µ(t) = et we find
(et y)0 = (3 + t)et ,
et y = 3et + tet − et + C,
y(t) = 2 + t + Ce−t .
The initial condition gives C = −1. The solution we want is
y(t) = 2 + t − e−t .
A comparison of values gives
y(0) = 1,
y0 = 1
y(0.1) = 1.195,
y1 = 1.2
y(0.2) = 1.381, y2 = 1.39
y(0.3) = 1.559, y3 = 1.571
y(0.4) = 1.730, y4 = 1.7439
Section 2.8
It is probably easier to start by solving the equations exactly.
3.
y 0 = 2(y + 1), y(0) = 0.
Rewrite as
y 0 − 2y = 2.
1
The integrating factor is µ(t) = e−2t , giving
(e−2t y)0 = 2e−2t ,
e−2t y = −e−2t + C.
The initial condition gives C = 1. The solution we want is
y(t) = e2t − 1.
(3. a) The Picard iteration scheme gives
φ0 = 0,
and
Z
t
φn+1 = φ0 +
f (s, φn (s)) ds.
t0
In this case
Z
φ1 =
t
Z
2 ds = 2t,
φ2 =
0
t
2(2s + 1) ds = 2t + 2t2 ,
0
Z
φ3 =
0
t
4
2(2s + 2s2 + 1) = 2t + 2t2 + t3 .
3
This is just the truncated Taylor series for y(t) = e2t − 1, which gives
φn =
n
X
(2t)k
k=1
k!
,
and
lim φn = e2t − 1.
n→∞
2