Initial Value Problems - Numerical Methods in Engineering with MATLAB - page 290

= = .

Using n 4 wehave h 0 125 and the midpointformulas become

y 1 = y 0 + hf 0 = 1 + 0 . 125 sin 1 . 0 = 1 . 105184

y 2 = y 0 + 2 hf 1 = 1 + 2(0 . 125)sin 1 . 105184 = 1 . 223 387

y 3 = y 1 + 2 hf 2 =

. + . . = .

1 105184 2(0 125)sin 1 223 387 1 340 248

y 4 = y 2 + 2 hf 3 = 1 . 223 387 + 2(0 . 125)sin 1 . 340 248 = 1 . 466 772

1
2 ( y 4 +

y h / 2 (0 . 5) = y 3 + hf 4 )

1
2 (1

= . 466 772 + 1 . 340 248 + 0 . 125 sin 1 . 466 772)

= .

1 465 672

Richardson extrapolationresults in

4(1 . 465 672) − 1 . 463 459

y (0 . 5) = = 1 . 466 410

3

which compares favorablywith the “true” solution y (0 . 5) = 1 . 466 404.

EXAMPLE 7.11

L

i

E ( t )

R

i

C

The differentialequations governing the loop current i and the charge q on the ca-
pacitor of the electriccircuit shown are
L di

q
C =

dq
dt dt +

Ri + E ( t ) = i

If the appliedvoltage E issuddenlyincreased from zeroto9 V, plot the resulting loop
current during the first ten seconds. Use R = 1 . 0 , L = 2 H and C = 0 . 45F.

Solution Letting

y 1
y 2

q
i

y = =

and substituting the givendata, the differentialequations become

y 1
y 2

y 2

y = =

− Ry 2 − y 1 / C + E ) / L

(