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MCA
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**Master of Computer Applications**Views:
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**2 months ago**

Runge-Kutta 2nd Order Method for Ordinary Differential Equations
Autar Kaw
After reading this chapter, you should be able to:
1. understand the Runge-Kutta 2nd order method for ordinary differential equations
and how to use it to solve problems.
What is the Runge-Kutta 2nd order method?
The Runge-Kutta 2nd order method is a numerical technique used to solve an ordinary
differential equation of the form
dy
= f (x, y ), y (0 ) = y 0
dx
Only first order ordinary differential equations can be solved by using the Runge-Kutta
2nd order method. In other sections, we will discuss how the Euler and Runge-Kutta
methods are used to solve higher order ordinary differential equations or coupled
(simultaneous) differential equations.
How does one write a first order differential equation in the above form?
Example 1
Rewrite
dy
+ 2 y = 1.3e − x , y (0 ) = 5
dx
in
dy
= f ( x, y ), y (0) = y 0 form.
dx
Source URL: http://numericalmethods.eng.usf.edu/
Saylor URL: http://www.saylor.org/courses/me205/
Attributed to: University of South Florida: Holistic Numerical Methods Institute
Saylor.org
Page 1 of 17

Solution
dy
+ 2 y = 1.3e − x , y (0 ) = 5
dx
dy
= 1.3e − x − 2 y, y (0 ) = 5
dx
In this case
f ( x, y ) = 1.3e − x − 2 y
Example 2
Rewrite
ey
dy
+ x 2 y 2 = 2 sin(3 x), y (0 ) = 5
dx
in
dy
= f ( x, y ), y (0) = y 0 form.
dx
Solution
ey
dy
+ x 2 y 2 = 2 sin(3 x), y (0 ) = 5
dx
dy 2 sin(3 x) − x 2 y 2
=
, y (0 ) = 5
dx
ey
In this case
2 sin(3 x) − x 2 y 2
f ( x, y ) =
ey
Runge-Kutta 2nd order method
Euler’s method is given by
Source URL: http://numericalmethods.eng.usf.edu/
Saylor URL: http://www.saylor.org/courses/me205/
Attributed to: University of South Florida: Holistic Numerical Methods Institute
Saylor.org
Page 2 of 17

yi +1 = yi + f ( xi , yi )h
(1)
where
x0 = 0
y 0 = y ( x0 )
h = xi +1 − xi
To understand the Runge-Kutta 2nd order method, we need to derive Euler’s method
from the Taylor series.
yi +1 = y i +
dy
dx
(xi +1 − xi ) + 1 d
2
2! dx
xi , y i
= y i + f ( xi , y i )( xi +1 − xi ) +
y
3
y
3! dx
3
(xi +1 − xi )2 + 1 d
2
xi , y i
(xi +1 − xi )3 + ...
xi , y i
1
1
2
3
f ' ( xi , y i )( xi +1 − xi ) + f ' ' ( xi , y i )( xi +1 − xi ) + ... (2)
2!
3!
As you can see the first two terms of the Taylor series
yi +1 = y i + f ( xi , yi )h
are Euler’s method and hence can be considered to be the Runge-Kutta 1st order
method.
The true error in the approximation is given by
Et =
f ′( xi , yi ) 2 f ′′( xi , yi ) 3
h +
h + ...
2!
3!
(3)
So what would a 2nd order method formula look like. It would include one more term of
the Taylor series as follows.
yi +1 = y i + f ( xi , y i )h +
1
f ′( xi , y i )h 2
2!
(4)
Let us take a generic example of a first order ordinary differential equation
Source URL: http://numericalmethods.eng.usf.edu/
Saylor URL: http://www.saylor.org/courses/me205/
Attributed to: University of South Florida: Holistic Numerical Methods Institute
Saylor.org
Page 3 of 17

dy
= e − 2 x − 3 y, y (0 ) = 5
dx
f ( x , y ) = e −2 x − 3 y
Now since y is a function of x,
f ′( x, y ) =
∂f ( x, y ) ∂f ( x, y ) dy
+
∂y dx
∂x
(5)
=
[(
)](
∂ −2 x
∂ −2 x
e − 3y +
e − 3 y e −2 x − 3 y
∂x
∂y
(
)
(
= −2e −2 x + (−3) e −2 x − 3 y
)
)
= −5e −2 x + 9 y
The 2nd order formula for the above example would be
yi +1 = y i + f ( xi , y i )h +
(
1
f ′( xi , y i )h 2
2!
)
= yi + e − 2 xi − 3 yi h +
(
)
1
− 5e −2 xi + 9 yi h 2
2!
However, we already see the difficulty of having to find f ′( x, y ) in the above method.
What Runge and Kutta did was write the 2nd order method as
yi +1 = y i + (a1 k1 + a 2 k 2 )h
(6)
where
k1 = f ( x i , y i )
k 2 = f ( xi + p1h, y i + q11 k1 h )
(7)
This form allows one to take advantage of the 2nd order method without having to
calculate f ′( x, y ) .
Source URL: http://numericalmethods.eng.usf.edu/
Saylor URL: http://www.saylor.org/courses/me205/
Attributed to: University of South Florida: Holistic Numerical Methods Institute
Saylor.org
Page 4 of 17

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