# First Order Ordinary Differential Equation

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```					Chapter 08.05
On Solving Higher Order Equations
for Ordinary Differential Equations

After reading this chapter, you should be able to:
1. solve higher order and coupled differential equations,

We have learned Euler’s and Runge-Kutta methods to solve first order ordinary differential
equations of the form
 f  x, y , y 0  y0
dy
(1)
dx
What do we do to solve simultaneous (coupled) differential equations, or differential
equations that are higher than first order? For example an n th order differential equation of
the form
dny            d n 1 y
 ao y  f x 
dy
a n n  a n 1 n 1    a1                                                       (2)
dx            dx            dx
with n  1 initial conditions can be solved by assuming
y  z1                                                                          (3.1)
dy dz1
        z2                                                               (3.2)
dx dx
d 2 y dz 2
        z3                                                             (3.3)
dx 2      dx

d n 1 y dz n 1
          zn                                                         (3.n)
dx n 1      dx
d n y dzn

dx n      dx
1            d n 1 y                    
 a0 y  f  x 
dy
   an 1 n 1   a1
                                      
an            dx       dx                 
=
1
 a n1 z n   a1 z 2  a0 z1  f x                            (3.n+1)
an
The above Equations from (3.1) to (3.n+1) represent n first order differential equations as
follows

08.05.1
08.05.2                                                                            Chapter 08.05

dz1
 z 2  f1 z1 , z 2 ,, x                                                (4.1)
dx
dz 2
 z 3  f 2 z1 , z 2 , , x                                              (4.2)
dx

dz n

1
 a n1 z n   a1 z 2  a0 z1  f x                             (4.n)
dx     an
Each of the n first order ordinary differential equations are accompanied by one initial
condition. These first order ordinary differential equations are simultaneous in nature but can
be solved by the methods used for solving first order ordinary differential equations that we

Example 1
Rewrite the following differential equation as a set of first order differential equations.
d2y
3 2  2  5 y  e  x , y 0  5, y 0  7
dy
dx      dx
Solution
The ordinary differential equation would be rewritten as follows. Assume
dy
 z,
dx
Then
d 2 y dz

dx 2 dx
Substituting this in the given second order ordinary differential equation gives
dz
3  2z  5 y  e x
dx

dx 3

dz 1  x
 e  2z  5 y  
The set of two simultaneous first order ordinary differential equations complete with the
initial conditions then is
 z, y 0  5
dy
dx
 e  2 z  5 y , z 0  7 .
dz 1  x
dx 3
Now one can apply any of the numerical methods used for solving first order ordinary
differential equations.

Example 2
Given
d2y
 2  y  e t , y 0  1,    0  2 , find by Euler’s method
dy                      dy
2
dt        dt                      dt
a) y 0.75 
Higher Order Equations                                                              08.05.3

b) the absolute relative true error for part(a), if y0.75 exact  1.668

c)
dy
0.75
dt
Use a step size of h  0.25 .
Solution
First, the second order differential equation is written as two simultaneous first-order
differential equations as follows. Assume
dy
z
dt
then
dz
 2 z  y  e t
dt
dz
 e t  2 z  y
dt
So the two simultaneous first order differential equations are
 z  f1 t,y,z, y (0)  1
dy
(E2.1)
dt
 et  2 z  y  f 2 t , y, z , z (0)  2
dz
(E2.2)
dt
Using Euler’s method on Equations (E2.1) and (E2.2), we get
y i 1  y i  f1 t i , y i , z i h                                       (E2.3)
z i 1  z i  f 2 t i , yi , z i h                                       (E2.4)
a) To find the value of y 0.75  and since we are using a step size of 0.25 and starting at
t  0 , we need to take three steps to find the value of y 0.75  .
For i  0, t0  0, y0  1, z0  2 ,
From Equation (E2.3)
y1  y 0  f1 t 0 , y 0 , z 0 h
 1  f1 0,1,20.25 
 1  20.25
 1.5
y1 is the approximate value of y at
t  t1  t 0  h  0  0.25  0.25
y1  y0.25   1.5
From Equation (E2.4)
z1  z 0  f 2 t 0 , y 0 , z 0 h
 2  f 2 0,1,20.25 
                
 2  e 0  22  1 0.25 
1
dy
z1 is the approximate value of z (same as        ) at t  0.25
dt
08.05.4                                                           Chapter 08.05

z1  z 0.25   1
For i  1, t1  0.25, y1  1.5, z1  1 ,
From Equation (E2.3)
y 2  y1  f1 t1 , y1 , z1 h
 1.5  f 1 0.25,1.5,10.25 
 1.5  10.25 
 1.75
y 2 is the approximate value of y at
t  t 2  t1  h  0.25  0.25  0.50
y 2  y 0.5  1.75
From Equation (E2.4)
z 2  z1  f 2 t1 , y1 , z1 h
 1  f 2 0.25,1.5,10.25 
 1  e 0.25  21  1.50.25 
 1   2.7211 0.25 
= 0.31970
z 2 is the approximate value of z at
t  t 2  0.5
z2  z 0.5  0.3197 0
For i  2, t2  0.5, y2  1.75, z2  0.31970 ,
From Equation (E2.3)
y 3  y 2  f1 t 2 , y 2 , z 2 h
 1.75  f1 0.50,1.75,0.31970 0.25 
 1.75  0.31970 0.25 
 1.8299
y 3 is the approximate value of y at
t  t 3  t 2  h  0.5  0.25  0.75
y 3  y 0.75   1.8299
From Equation (E2.4)
z 3  z 2  f 2 t 2 , y 2 , z 2 h
 0.31972  f 2 0.50,1.75,0.31970 0.25 
 0.31972  e 0.50  20.31970   1.75 0.25 
 0.31972   1.7829 0.25 
 0.1260
z 3 is the approximate value of z at
t  t 3  0.75
z 3  z 0.75   0.12601
y 0.75   y3  1.8299
Higher Order Equations                                                            08.05.5

b) The exact value of y 0.75  is
y0.75 exact  1.668
The absolute relative true error in the result from part (a) is
1.668  1.8299
t                    100
1.668
= 9.7062%
c)
dy
0.75  z 3  0.12601
dx

Example 3
Given
d2y       dy                     dy
2
 2  y  e t ,y (0)  1, (0)  2 ,
dt       dt                     dt
find by Heun’s method
a) y 0.75 

b)
dy
0.75 .
dx
Use a step size of h  0.25 .
Solution
First, the second order differential equation is rewritten as two simultaneous first-order
differential equations as follows. Assume
dy
z
dt
then
dz
 2 z  y  e t
dt
dz
 e t  2 z  y
dt
So the two simultaneous first order differential equations are
 z  f1 t,y,z,y(0)  1
dy
(E3.1)
dt
 et  2 z  y  f 2 t, y, z , z (0)  2
dz
(E3.2)
dt
Using Heun’s method on Equations (1) and (2), we get
yi 1  yi  k1y  k2y h
1
(E3.3)
2
k 1y  f 1 t i , y i , z i                                             (E3.4a)

k 2y  f 1 t i  h, y i  hk 1y , z i  hk 1z                           (E 3.4b)

z i 1  z i 
2

1 z

k1  k 2z h                                             (E3.5)
08.05.6                                                        Chapter 08.05

k 1z  f 2 t i , y i , z i                               (E3.6a)
2          
k  f 2 t i  h, y i  hk , z i  hk
z
1
y   z
1               (E3.6b)
For i  0, to  0, yo  1, zo  2
From Equation (E3.4a)
k1y  f1 t o , y o , z o 
 f1 0,1,2
2
From Equation (E3.6a)
k1z  f 2 t 0 , y 0 , z 0 
 f 2 0,1,2
 e 0  22   1
= -4
From Equation (E3.4b)
k2y  f1 t0  h, y0  hk1y , z0  hk1z 
 f 1 0  0.25,1  0.25 2,2  0.25  4
 f 1 0.25,1.5,1
=1
From Equation (E3.6b)
k 2z  f 2 t 0  h, y 0  hk1y , z 0  hk1z 
 f 2 0  0.25,1  0.25 2,2  0.25  4
 f 2 0.25,1.5,1
 e 0.25  21  1.5
 2.7212
From Equation (E3.3)
1
2

y1  y0  k1y  k 2y h            
 1  2  10.25
1
2
 1.375
y1 is the approximate value of y at
t  t1  t 0  h  0  0.25  0.25
y1  y0.25   1.375
From Equation (E3.5)
1
2

z1  z 0  k1z  k 2z h          
1
 2  ( 4  ( 2.7212))(0.25)
2
 1.1598
z1 is the approximate value of z at
t  t1  0.25
Higher Order Equations                                                                   08.05.7

z1  z 0.25   1.1598
For i  1, t1  0.25, y1  1.375 , z1  1.1598
From Equation (E3.4a)
k1y  f1 t1 , y1 , z1 
 f1 0.25,1.375 ,1.1598 
 1.1598
From Equation (E3.6a)
k1z  f 2 t1 , y1 , z1 
 f 2 0.25,1.375 ,1.1598 
 e 0.25  21.1598   1.375
 2.9158
From Equation (E3.4b)
k2y  f1 t1  h, y1  hk1y , z1  hk1z 
 f1 0.25  0.25,1.375  0.25 (1.1598 ),1.1598  0.25  2.9158 
 f1 0.50 ,1.6649 ,0.43087 
 0.43087
From Equation (E3.6b)
k 2z  f 2 t1  h, y1  hk1y , z1  hk1z 
 f 2 0.25  0.25,1.375  0.25 (1.1598 ),1.1598  0.25  2.9158 
 f 2 0.50 ,1.6649 ,0.43087 
 e 0.50  20.43087   1.6649
 1.9201
From Equation (E3.3)
1

y 2  y1  k1y  k 2y h
2
 1.375  1.1598  0.430870.25
1
2
 1.5738
y 2 is the approximate value of y at
t  t 2  t1  h  0.25  0.25  0.50
y2  y 0.50   1.5738
From Equation (E3.5)
 1

z 2  z1  k1z  k 2z h
2
1
 1.1598  (2.9158  (1.9201))(0.25)
2
 0.55533
z 2 is the approximate value of z at
t  t 2  0.50
z2  z 0.50   0.55533
08.05.8                                                                                 Chapter 08.05

For i  2, t2  0.50, y2  1.57384 , z2  0.55533
From Equation (E3.4a)
k1y  f1 t 2 , y 2 , z 2 
 f1 0.50,1.5738 ,0.55533 
 0.55533
From Equation (E3.6a)
k1z  f 2 t 2 , y 2 , z 2 
 f 2 0.50 ,1.5738 ,0.55533 
 e 0.50  20.55533   1.5738
 2.0779
From Equation (E3.4b)
k2y  f 2 t2  h, y2  hk1y , z2  hk1z 
 f1 0.50  0.25,1.5738  0.25 (0.55533 ),0.55533  0.25  2.0779 
 f1 0.75,1.7126 ,0.035836 
= 0.035836
From Equation (E3.6b)
k 2z  f 2 t 2  h, y 2  hk1y , z 2  hk1z 
 f 2 0.50  0.25,1.5738  0.25 (0.55533 ),0.55533  0.25  2.0779 
 f 2 0.75,1.7126 ,0.035836 
 e 0.75  20.035836   1.7126
 1.3119
From Equation (E3.3)
  1

y3  y 2  k1y  k 2y h
2
 1.5738  0.55533  0.0358360.25
1
2
 1.6477
y 3 is the approximate value of y at
t  t 3  t 2  h  0.50  0.25  0.75
y3  y 0.75   1.6477
b) From Equation (E3.5)
  1

z 3  z 2  k1z  k 2z h
2
1
 0.55533  (2.0779  (1.3119))(0.25)
2
 0.13158
z 3 is the approximate value of z at
t  t 3  0.75
z 3  z 0.75   0.13158
The intermediate and the final results are shown in Table 1.
Higher Order Equations                                                     08.05.9

Table 1 Intermediate results of Heun’s method.
i     0            1                2
ti    0            0.25             0.50
yi   1            1.3750           1.5738
zi       2          1.1598        0.55533
y
k   1    2          1.1598        0.55533
k   1
z
4          2.9158        2.0779
y
k   2    1          0.43087       0.035836
k   2
z
 2.7211    1.9201        1.3119
y i 1   1.3750     1.5738        1.6477
zi 1    1.1598     0.55533       0.13158

ORDINARY DIFFERENTIAL EQUATIONS
Topic        Higher Order Equations
Summary      Textbook notes on higher order differential equations
Major        General Engineering
Authors      Autar Kaw
Last Revised April 13, 2011
Web Site     http://numericalmethods.eng.usf.edu

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