Differential Equations 3 – Answers and some Solutions
1.
2
t
x
dt
dx

=
,
(
29
1
0
=
x
The general form of the d.e. for Euler’s method is
(
29
x
t
f
dt
dx
,
=
In this example, the function,
(
29
2
,
t
x
x
t
f

=
The initial conditions are written as
0
0
=
t
,
1
0
=
w
where
0
t
is the initial time and
0
w
is the solution at this time.
Euler’s method is defined by:
(
29
w
w
hf t
w
i
i
i
i
+
=
+
1
,
and
i
takes the values 0, 1, 2, … in turn
So the solution to the equation is built up in steps.
Start with
i
= 0.
This is the initial time at which the solution is known.
The new time is always
h
t
t
i
i
+
=
+
1
and the solution at this time is given by
(
29
w
w
hf t
w
i
i
i
i
+
=
+
1
,
So with
i
= 0, the new time is
1
.
0
0
1
=
+
=
h
t
t
and
(
29
0
0
0
1
,
w
t
hf
w
w
+
=
(
29
(
29
1
.
1
0
1
1
.
0
1
1
,
0
1
.
0
1
2
=

+
=
+
=
f
(
29
1
,
0
f
means the function
(
29
2
,
t
x
x
t
f

=
evaluated using 0 for the time,
t
,
and 1 for the solution,
x
.
Now move on to the next step.
i
= 1, the new time is
2
.
0
1
2
=
+
=
h
t
t
, and the solution at this time is
(
29
(
29
(
29
209
.
1
1
.
0
1
.
1
1
.
0
1
.
1
1
.
1
,
1
.
0
1
.
0
1
.
1
,
2
1
1
1
2
=

+
=
+
=
+
=
f
w
t
hf
w
w
One more step is required to obtain the solution at time 0.3.