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

620 Chapter 20 Ordinary differential equations II
20.7 IMPROVEDEULERMETHOD
YouwillrecallthatEuler’smethodisobtainedbyfirstfindingtheslopeofthesolutionat
(x 0
,y 0
) and imposing a straight line approximation (often called a tangent line approximation)
there as indicated in Figure 20.10. If we knew the gradient of the solution at
x =x 1
in addition to the gradient atx =x 0
, a better approximation to the gradient over
the whole interval might be the mean of the two. Unfortunately, the gradient atx = x 1
which is
dy
dx∣ =f(x 1
,y 1
)
x=x1
cannot be obtained until we knowy 1
. What we can do, however, is use the value ofy 1
obtained by Euler’s method in estimating the gradient atx = x 1
. This gives rise to the
improved Euler method:
y 1
=y 0
+h×(average of gradients atx 0
andx 1
)
{ } f(x0 ,y
=y 0
+h× 0
)+f(x 1
,y 1
)
2
=y 0
+ h 2 {f(x 0 ,y 0 )+f(x 1 ,y 0 +hf(x 0 ,y 0 ))}
Then, knowingy 1
the whole process isstartedagain tofindy 2
, etc. Generally,
y i+1
=y i
+ h 2 {f(x i ,y i )+f(x i+1 ,y i +hf(x i ,y i ))}
It can be shown that, like Euler’s method, the improved Euler method is equivalent to
truncating the Taylor series expansion, inthis case afterthe third term.
Example20.6 Use the improved Euler method to solve the differential equationy ′ = −xy 2 ,y(0) = 2,
inExample20.4.Asbeforetakeh = 0.25,butworkthroughouttofourdecimalplaces.
Solution Here f(x,y) = −xy 2 ,y 0
= 2,x 0
= 0.Wefind f(x 0
,y 0
) = f(0,2) = −0(2 2 ) = 0.The
improved Euler method
yields
y 1
=y 0
+ h 2 {f(x 0 ,y 0 )+f(x 1 ,y 0 +hf(x 0 ,y 0 ))}
y 1
=2+ 0.25 {0 + f(0.25,2)}
2
Now f (0.25,2) = −0.25(2 2 ) = −1,so that
y 1
= 2 +0.125(−1) = 1.875
WeshallsetoutthecalculationsrequiredinTable20.3.Youshouldworkthroughthenext
fewstagesyourself.Comparingthevaluesobtainedinthiswaywiththeexactsolution,
we see that, over the interval of interest, our solution is usually correct to two decimal
places (Table 20.4).