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We have three equations in four unknowns; the usual convention is to fix a
= 1/2, giving for the other quantities:

b = 12/; α = 1; β = 1 (4.52)

finally leading to the following expressions for the second-order iterator and
its parameters:

k = f t n y t n( ( ), ( ( )))(∆ t) (4.53a)
1

k = f tn +( ( ) ∆ t y tn +, ( ( )) k )(∆ t) (4.53b)
2 1
k + k
ytn(( + 1 )) = ytn(( )) + 1 2 (4.53c)
2

Next, we give, without proof, the famous fourth-order iterator Runge-
Kutta expression, one of the most widely used algorithms for solving ODEs
in the different fields of science and engineering:

k = f(t(n), y(n))(∆t) (4.54a)
1

k = f tn +( ( ) ∆ t 2/ , y tn +( ( )) k / 2)(∆ t) (4.54b)
2 1
k = f tn +( ( ) ∆ t 2/ , y tn +( ( )) k / 2)(∆ t) (4.54c)
3 2

k = f tn +( ( ) ∆ t y tn +, ( ( )) k )(∆ t) (4.54d)
4 3

k + 2 k + 2 k + k 4
1
3
2
ytn(( + 1 )) = ytn(( )) + (4.54e)
6
The last point that we need to address before leaving this subsection is
what to do in case we have an ODE with higher derivatives than the first. The
answer is that we reduce the n -order ODE to a system of n first-order ODEs.
th
Example 4.10
Reduce the following second-order differential equation into two first-order
differential equations:

ay″ + by′ + cy = sin(t) (4.55)

with the initial conditions: y(t = 0) = 0 and y′(t = 0) = 0

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