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Solve, over the interval 0 ≤ t ≤ 20, the following first-order differential equa-
tion for a = 2 and a = 0.5:

dy
a + y = 1
dt
where y(0) = 0. (Physically, this would correspond to the charging of a capac-
itor from a dc source connected suddenly to the battery at time zero. Here, y
is the voltage across the capacitor, and a = RC.)
NOTE The analytic solution to this problem is y = 1 – exp(–t/a).

4.7.2 Higher-Order Iterators: The Runge-Kutta Method*
In this subsection, we want to explore the possibility that if we sampled the
function n-times per step, we will obtain a more accurate solution to the ODE
than that obtained from the first-order iterator for the same value of ∆t.
To focus the discussion, consider the ODE:

y′(t) = f(t, y(t)) (4.41)

Higher-order ODEs can be reduced, as will be shown at the end of the sub-
section, to a system of equations having the same functional form as Eq.
(4.41). The derivation of a technique using higher-order iterators will be
shown below in detail for two evaluations per step. Higher-order recipes can
be found in most books on numerical methods for ODE.
The key to the Runge-Kutta method is to properly arrange each of the eval-
uations in a particular step to depend on the previous evaluations in the
same step.
In the second-order model:

if: k = f t n y t n( ( ), ( ( )))(∆ t) (4.42)
1

then: k = f tn +( ( ) α∆ t y tn +, ( ( )) β k )(∆ t) (4.43)
2 1
and y t n(( + 1 )) = y t n(( )) + ak + bk (4.44)
1 2
where a, b, α, and β are unknown parameters to be determined. They should
3
be chosen such that Eq. (4.44) is correct to order (∆t) .
To find a, b, α, and β, let us compute y(t(n + 1)) in two different ways. First,
Taylor expanding the function y(t(n + 1)) to order (∆t) , we obtain:
2

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