082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

616 Chapter 20 Ordinary differential equations II
with initial condition
y(x 0
) =y 0
isEuler’s method which weshall study inthe next section.
20.6 EULER’SMETHOD
You will recall, from Chapter 12, that given a functiony(x), the quantity dy
dx represents
thegradientofthatfunction.Soif dy
dx = f (x,y)andweseeky(x),weseethatthedifferential
equation tells us the gradient of the required function. Given the initial condition
y =y 0
whenx =x 0
wecanpicturethissinglepointasshowninFigure20.10.Moreover,
we know the gradient of the solution here. Because dy
dx = f(x,y)weseethat
dy
dx∣ =f(x 0
,y 0
)
x=x0
which equals the gradient of the solution at x = x 0
. Thus the exact solution passes
through (x 0
,y 0
)andhasgradient f (x 0
,y 0
)there.Wecandrawastraightlinethroughthis
point with the required gradient to approximate the solution as shown in Figure 20.10.
Thisstraightlineapproximatesthetruesolution,butonlynear (x 0
,y 0
)because,ingeneral,
the gradient is not constant but changes. So, in practice we only extend it a short
distance,h,alongthexaxistowherex =x 1
.Theycoordinateatthispointisthentaken
asy 1
.Wenowdevelopanexpressionfory 1
.Thestraightlinehasgradient f (x 0
,y 0
)and
passes through (x 0
,y 0
). Itcan beshown thatits equation istherefore
y=y 0
+(x−x 0
)f(x 0
,y 0
)
Whenx =x 1
theycoordinate isthen given by
y 1
=y 0
+(x 1
−x 0
)f(x 0
,y 0
)
and sincex 1
−x 0
=hwe find
y 1
=y 0
+hf(x 0
,y 0
)
Thisequationcanbeusedtofindy 1
.Wethenregard (x 1
,y 1
)asknown.Fromthisknown
point the whole process isthen repeated using the formula
y i+1
=y i
+hf(x i
,y i
)
y
y(x 1 )
y 1
True solution
Tangent line
approximation
y 0
x 0 x 1 x
0
Figure20.10
Approximation usedin Euler’smethod.