1. Prof. Enrique Mateus Nieves
PhD in Mathematics Education.
1
HIGHER ORDER DIFFERENTIAL EQUATIONS
Homogeneous linear equations with constant coefficients of order two and
higher.
Apply reduction method to determine a solution of the nonhomogeneous equation given in the
following exercises. The indicated function y1(x), is a solution of the associated homogeneous
equation. Determine a second solution of the homogeneous equation and a particular solution
of the inhomogeneous ED
1.
x
1 ey;yy 2
24 
 2. 11  1y;yy
3.
x
1
x
ey;eyyy  3
523 4.
x
1 ey;xyyy 
 34
Problems for group discussion:
1. Make a convincing demonstration that the second order equation ;cyybyu 0
a, b, c, constant always has at least one solution of the form
xm
1 ey 1
 , where 1m is
a constant.
2. Two. Explain why E. D. 1st point must have, consequently, a second solution of the
form
xm
2 ey 2
 or form
xm
2 xey 2
 , where 1m y 2m are constants.
HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS
We have seen that the first order linear equation, 0 uy
dx
dy
, where a is a constant, has the
exponential solution
ax
1 ecy 1 ranging  ;.-  therefore as natural to try to determine if
there are exponential solutions  .- homogeneous linear equations of higher order type:
   
0012
1
1  
 yayayayaya n
n
n
n  (1)
Where the coefficients ,1,0,i,ai  n are real constants and 0na . To our surprise, all
solutions of the equation (1) are exponential functions or are formed from exponential
functions.

2. Prof. Enrique Mateus Nieves
PhD in Mathematics Education.
2
Method of solution: start with the special case of the second order equation ay” + by’ + cy =
0. (2) If we try a solution of the form
mx
ey  , thens
mx
mey  and
mx
emy 2
 , so that
the equation (2) becomes:, 0 mxmxmx2
cebmeeam , or
  02
 cbmamemx
. As
mx
e never zero when x has real value, the only way that the
exponential function satisfies the differential equation is choosing a m such that it is a root of
the quadratic equation   02
 cbmam (3).
This equation is called auxiliary equation or characteristic equation of the differential equation
(2). Examine three cases: the solutions of the auxiliary equation corresponding to distinct real
roots, real and equal roots and complex conjugate roots.
CASE I: distinct real roots:
If equation (3) has two distinct real roots, 1m y 2m , arrived at two solutions,
xm
1 ey 1
 and
xm
2 ey 2
 . These functions are linearly independent on  .- and, therefore, form a
fundamental set. Then, the general solution of equation (2) in this interval is
xmxm
H ececy 21
21  (4)
CASE II: Real Estate and equal
When 21 mm  we necessarily exponential only solution,
xm
1 ey 1
 . According to the
quadratic formula,
a
b
m1
2

 because the only way 21 mm  is 04  acb2
. Thus, a second
solution of the equation is:
  xmxm
xm
xm
xm
H xedxe
e
e
ey 11
1
1
1
2
2
(5)
In this equation we take that
a
b
2m1

 . The general solution is therefore
xmxm
H xececy 11
21  (6)

3. Prof. Enrique Mateus Nieves
PhD in Mathematics Education.
3
CASE III: complex conjugate roots.
If 21 m,,m  are complex, we can write im1   y im2   , where
0and  and p > 0 and they are real, and e .i2
1 There is no formal difference
between this case and case I, hence,
x)i(
ececy x)i(
H

 
 
21 However, in
practice it is preferred to work with real functions and not complex exponential. With this object
using Euler's formula: ,senicosei

 that  is a real number. The consequence of
this formula is that: ,xsenixcose xi

 and ,xsenixcose x-i

 (7) where we
have used x)(cosx)(-cos   and x)(senx)(-sen   . Note that if the first add and
then subtract the two equations (7), we obtain respectively:
,xcosee x-ixi

2 y .xsiniee x-ixi

2
As
x)i(
ececy x)i( 
 
 
21 is a solution to equation (2) for any choice of the
constants 1c and, 2c if 121  cc and ,c 11  12 c obtain the solutions:
x)i(
eey x)i(
1

 
 
y
x)i(
eey x)i(
2

 
 
But   xcoseeeey xxixix
1 
2 
and
  xsinieeeey xxixix
2 
2 
Accordingly, the results demonstrate that the last two real functions xcose x

and
xsene x

are solutions of the equation (2). Moreover, these solutions form a fundamental,
therefore .- , the general solution is:
 xsincxcosce
xsinecxcosecy
2
x
xx
H






1
21
Second order differential equations
Solve the following differential equations:
1. 0352  yyy

4. Prof. Enrique Mateus Nieves
PhD in Mathematics Education.
4
SOLUTION: I present the auxiliary equations, roots and corresponding general solutions.
   33120352 2
1
1
2
 2m;mmmmm Hence:
x
ececy
x
3
21
2


2. 02510  yyy
SOLUTION:   50502510 1
22
 2mmmmm Hence:
xx
xececy 5
2
5
1 
3. 0 yyy
SOLUTION: ,i
2
1
-m,i
2
1
-mmm 2
2
3
2
3
01 1
2
 Hence:










x
2
3
sincx
2
3
coscey
x
H 21
2
4. Initial value problem. Solve the initial value problem
  20134  0y-1,(0)y;yyy
SOLUTION: The roots of the auxiliary equation
i32m,i32mmm 2  1
2
0134 so that
 x3sincx3coscey x
H 21
2

By applying the condition 10 )(y , we see that  0senc0cosce 21
0
1  and
11 c . We differentiate the above equation and then applying y’(O) = 2 we get
232 2  c or
3
4
2 c ; therefore, the solution is:






 x3senx3cosey x
H
3
42

5. Prof. Enrique Mateus Nieves
PhD in Mathematics Education.
5
The two differential equations, 0 pyy y 02  kyy , k real, are important in
applied mathematics. For the first, the auxiliary equation 022
 km has imaginary roots
ikm1  y ikm1  . According to equation (8), with 0 y k, the general solution is
kxsenckxcoscy 2 1 (9)
Auxiliary equation the second equation, 022
 km , has distinct real roots km1  y
km1  ; therefore its general solution is
-kx
2
kx
ececy  1 (10)
Note that if we choose 2
1
2  cc1 and then 2
1
22
1
 cyc1 in  ,10 particular solutions we
kxcosh
ee
y
-kxkx



2
and kx.senh
ee
y
-kxkx



2
inasmuch as kxcosh and
kxsenh are linearly independent in any range of the x axis, an alternative form of the
general solution of 0 pyy is kxsenhckxcoshcy 2
Higher-Order Equations
In general, to solve a differential equation of order n as
   
0012
1
1  
 yayayayaya n
n
n
n  (11)
Where n,20,1,i,ai  are real constants, we must solve a polynomial equation of degree
n:
   
001
2
2
1
1  
 amamamama n
n
n
n  (12)
If all the roots of equation (12) are real and distinct, the general solution of equation (11) is
.ecececy xm
n
xm
2
xm n21
 1
It's difficult to summarize the analogous cases II and III because the roots of an auxiliary
equation of degree greater than two can occur in many combinations. For example, a quintic
could have five distinct real roots, or three distinct real roots and two complex, or four real and
complex, five reals but equal, but two equals five reals, and so on. When 1m is a root of an
equation k multiplicity auxiliary degree n (ie roots equals k), one can show that the solutions
are linearly independent

6. Prof. Enrique Mateus Nieves
PhD in Mathematics Education.
6
.exex,xe,e xmkxmxmxm 1111 12 

Finally, remember that when the coefficients are real, complex roots of auxiliary equation
always appear in conjugate pairs. Thus, for example, a cubic polynomial equation may have
two complex zeros at most.
Third-order differential equation
Resolve 043  yyy y”’ + 3~” - 4y = 0.
SOLUTION: In reviewing 043 23
 mm we should note that one of its roots is 11m .
If we divide 43 23
 mm eight ,m 1 we see that
      ,mmmmmmm 22144143 223
 and then the other roots are
2 32 mm . Thus, the general solution is
.xecececy -2x-2x
2
x
31 
Fourth-order differential equation
Resolve 02 2
2
4
4
 y
dx
yd
dx
yd
SOLUTION: The auxiliary equation is   01013
2224
 mmm and has the
roots imm 31  y imm 42  . Thus, according to the case II, the solution is:
.ecxecececy -ix
4
ix-ix
2
ix
 31
According to Euler's formula, we can write the grouping
-ix
2
ix
ecec 1 in the form
xsencxcosc 21 With a change in the definition of the constants. equally,
 -ix
4
ix
ececx 3 can be expressed in the form  xsencxcoscx 43 Accordingly, the
general solution is.
x.sinxcxcosxcxsincxcoscy 42  31

7. Prof. Enrique Mateus Nieves
PhD in Mathematics Education.
7
General Exercises Nonhomogeneous linear equations with constant coefficients
of order two and higher.
For each of the following E. D. finds the general solution:
1. 04  yy 2. 052  yy
3. 036  yy 4. 08  yy
5. 09  yy 6. 03  yy
7. 06  yyy 8. 023  yyy
9. 01682
2
 y
dx
dy
dx
yd
10. 025102
2
 y
dx
dy
dx
yd
11. 053  yyy 12. 048  yyy
13. 02512  yyy 14. 028  yyy
15. 054  yyy 16 0432  yyy
17. 023  yyy 18. 022  yyy
19. 054  yyy 20. 044  yyy
21. 0 yy 22. 05  yy
23. 0935  yyyy 24. 01243  yyyy
25. 02  yyy 26. 04  yyy
27. 092
2
3
3
4
4
 y
dx
yd
dx
yd
dx
yd
28. 051025 2
2
3
3
4
4
5
5
 y
dx
dy
dx
yd
dx
yd
dx
yd
dx
yd