differential equation

462 DIFFERENTIAL EQUATIONS
For line 5, where x 0 = 1.8:
y 1 = y 0 + h(y ′ ) 0
= 5.312 + (0.2)(2.488) = 5.8096
and (y ′ ) 0 = 3(1 + 1.8) − 5.8096 = 2.5904
For line 6, where x 0 = 2.0:
y 1 = y 0 + h(y ′ ) 0
= 5.8096 + (0.2)(2.5904)
= 6.32768
(As the range is 1.0 to 2.0 there is no need to calculate
(y ′ ) 0 in line 6). The particular solution is given by the
value of y against x.
A graph of the solution of dy = 3(1 + x) − y
dx
with initial conditions x = 1 and y = 4 is shown in
Fig. 49.6.
y
x = 0(0.2)1.0 means that x ranges from 0 to 1.0 in
equal intervals of 0.2 (i.e. h = 0.2 in Euler’s method).
dy
dx + y = 2x,
dy
hence = 2x − y,
dx i.e. y′ = 2x − y
If initially x 0 = 0 and y 0 = 1, then
(y ′ ) 0 = 2(0) − 1 = −1.
Hence line 1 in Table 49.2 can be completed with
x = 0, y = 1 and y ′ (0) =−1.
Table 49.2
x 0 y 0 (y ′ ) 0
1. 0 1 −1
2. 0.2 0.8 −0.4
3. 0.4 0.72 0.08
4. 0.6 0.736 0.464
5. 0.8 0.8288 0.7712
6. 1.0 0.98304
6.0
5.0
4.0
Figure 49.6
1.0 1.2 1.4 1.6 1.8 2.0 x
In practice it is probably best to plot the graph as
each calculation is made, which checks that there is
a smooth progression and that no calculation errors
have occurred.
Problem 2. Use Euler’s method to obtain a
numerical solution of the differential equation
dy
+ y = 2x, given the initial conditions that at
dx
x = 0, y = 1, for the range x = 0(0.2)1.0. Draw
the graph of the solution in this range.
For line 2, where x 0 = 0.2 and h = 0.2:
y 1 = y 0 + h(y ′ ), from equation (2)
= 1 + (0.2)(−1) = 0.8
and (y ′ ) 0 = 2x 0 − y 0 = 2(0.2) − 0.8 = −0.4
For line 3, where x 0 = 0.4:
y 1 = y 0 + h(y ′ ) 0
= 0.8 + (0.2)(−0.4) = 0.72
and (y ′ ) 0 = 2x 0 − y 0 = 2(0.4) − 0.72 = 0.08
For line 4, where x 0 = 0.6:
y 1 = y 0 + h(y ′ ) 0
= 0.72 + (0.2)(0.08) = 0.736
and (y ′ ) 0 = 2x 0 − y 0 = 2(0.6) − 0.736 = 0.464
For line 5, where x 0 = 0.8:
y 1 = y 0 + h(y ′ ) 0
= 0.736 + (0.2)(0.464) = 0.8288
and (y ′ ) 0 = 2x 0 − y 0 = 2(0.8) − 0.8288 = 0.7712
For line 6, where x 0 = 1.0:
y 1 = y 0 + h(y ′ ) 0
= 0.8288 + (0.2)(0.7712) = 0.98304
As the range is 0 to 1.0, (y ′ ) 0 in line 6 is not needed.