462 DIFFERENTIAL EQUATIONS
For line 5,wherex 0 = 1 .8:
y 1 =y 0 +h(y′) 0= 5. 312 +(0.2)(2.488)=5.8096and (y′) 0 =3(1+ 1 .8)− 5. 8096 =2.5904
For line 6, wherex 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 ofyagainstx.
A graph of the solution ofdy
dx=3(1+x)−ywith initial conditionsx=1 andy=4 is shown in
Fig. 49.6.
1.0 1.2 1.4 1.6 1.8 2.0 x5.06.0y4.0Figure 49.6
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
dx+y= 2 x, given the initial conditions that at
x=0,y=1, for the rangex=0(0.2)1.0. Draw
the graph of the solution in this range.x=0(0.2)1.0 means thatxranges from 0 to 1.0 in
equal intervals of 0.2 (i.e.h= 0 .2 in Euler’s method).
dy
dx+y= 2 x,hencedy
dx= 2 x−y,i.e.y′= 2 x−yIf initiallyx 0 =0 andy 0 =1, then
(y′) 0 =2(0)− 1 =− 1.
Hence line 1 in Table 49.2 can be completed with
x=0,y=1 andy′(0)=−1.Table 49.2x 0 y 0 (y′) 0- 0 1 − 1
- 0.2 0.8 −0.4
- 0.4 0.72 0.08
- 0.6 0.736 0.464
- 0.8 0.8288 0.7712
- 1.0 0.98304
For line 2, wherex 0 = 0 .2 andh= 0 .2:y 1 =y 0 +h(y′), from equation (2)= 1 +(0.2)(−1)=0.8and (y′) 0 = 2 x 0 −y 0 =2(0.2)− 0. 8 =−0.4For line 3, wherex 0 = 0 .4:y 1 =y 0 +h(y′) 0= 0. 8 +(0.2)(− 0 .4)=0.72and (y′) 0 = 2 x 0 −y 0 =2(0.4)− 0. 72 =0.08For line 4, wherex 0 = 0 .6:y 1 =y 0 +h(y′) 0= 0. 72 +(0.2)(0.08)=0.736and (y′) 0 = 2 x 0 −y 0 =2(0.6)− 0. 736 =0.464For line 5, wherex 0 = 0 .8:y 1 =y 0 +h(y′) 0= 0. 736 +(0.2)(0.464)=0.8288and (y′) 0 = 2 x 0 −y 0 =2(0.8)− 0. 8288 =0.7712For line 6, wherex 0 = 1 .0:y 1 =y 0 +h(y′) 0= 0. 8288 +(0.2)(0.7712)=0.98304As the range is 0 to 1.0, (y′) 0 in line 6 is not needed.