MAP 2302
Differential Equations
Part II
Summary 2- Higher Order Differential Equations
Sanchez
Linear combination of functions.
A function y=f(x) is said to be a linear combination of the functions y=g(x) and y=h(x), if there exist
constants c and d such that f(x)=cg(x) + dh(x)
Linearly independent set of functions: a set of functions are said to be linearly independent if none
of them can be expressed as a linear combination of the others.
That is, c1 f1 ( x)  c2 f 2 ( x)  c3 f 3 ( x)  ...  ck f k1 ( x)  0 then c1  c2  c3  c4  ...  ck  0
Wronskian of a set of functions.
The Wronskian of the set of functions f1(x), f2(x), f2(x),…, fn(x) is defined by the determinant
f1 ( x)
f 2 ( x)
...
f n ( x)
f1 ( x)
f 2 ( x)
...
f n ( x
.
.
.
.
W  (f1 , f 2 , . . ., f n ) 
provided the functions
.
.
.
.
.
(n 1)
f1
( x)
.
(n 1)
f2
( x)
.
.
...
(n 1)
f1
( x)
have n  1 derivative s.
Test for independence of a set of functions:
The set of functions f1 , f 2 , . . ., f n  is linearly independen t if the Wronskian W(f1 , f 2 , . . ., f n )  0
for every x in the int erval .
Problem 1: Use the Wronskian to show that the functions sinx and cosx are linearly independent.
W(sin x, cos x) 
sin x
cos x
cos x  sin x
  sin2 x  cos2 x  1  0 for all x valuew
Theorem : If a  b then the functions e ax and e bx are linearly independen t .
W( e ax , e bx ) 
e ax
ae
ax
e bx
be
bx
 e ax e bx
1 1
a b
 a ax b bx (b  a )  0 because e ax  0'
e bx  0 and b  a  0
Therefore , e ax and e bx are linearly independen t .
-1-
General Solution: a general solution of a DE is a family of solutions defined on some interval I that
contains all the solutions of the DE that are defined on the interval I.
Initial Value problem of an nth-order linear differential equation
Model : a n ( x)
dn y
 a n 1 ( x )
d n 1y
 . . .  a 2 ( x)
d2y
 a1 ( x)
dy
 a o ( x )y  g( x )
dx
dx n 1
dx 2
Initial conditions : y( x o )  y o , y ( x o )  y1 , y ( x o )  y 2 , . . . y (n 1) ( x o )  y n 1
dx n
(1)
Existence and uniqueness Theorem:
If a n ( x), a n 1 ( x), . . . , a1 ( x) , a o ( x) and gx  are all continuous on an int erval I, and a n ( x)  0
for every x in I, and x  x o belongs to I, then the solution of the initial value problem (1) exists
and it is unique .
Properties of the solutions of an nth order homogeneous linear differential equation:
1) Any linear combination of solutions is also a solution. That is,
if y1 and y 2 are solutions and c1 anf c 2 are any cons tan ts, then c1y1  c 2 y 2 is also a solution
Pr oof :
dn y
d n  1y
d2y
dy
1. a n ( x )
 a n 1 ( x )
 . . .  a 2 ( x)
 a1 ( x)
 a o ( x )y  0
n
n

1
2
dx
dx
dx
dx
d n y1
d n  1y 1
d 2 y1
dy
2. a n ( x )
 a n 1 ( x )
 . . .  a 2 ( x)
 a1 ( x ) 1  a o ( x )y1  0
n
n

1
2
dx
dx
dx
dx
dn y 2
d n  1y 2
d2y 2
dy
3. a n ( x )
 a n 1 ( x )
 . . .  a 2 ( x)
 a1 ( x) 2  a o ( x)y 2  0
n
n

1
2
dx
dx
dx
dx
d n c1y1  c 2 y 2 
d n 1 c1y1  c 2 y 2 
d 2 c1y1  c 2 y 2 
3. a n ( x )
 a n 1 ( x )
 . . .  a 2 ( x)
dx n
dx n 1
dx 2
dc1y1  c 2 y 2 
 a1 ( x )
 a o ( x)c1y1  c 2 y 2 
dx
 c d n y  c d n y  
 c d n 1 y  c d n 1 y  
1  2
2 a
1  2
2   . . .  a ( x )c y   a ( x )c y 
 a n ( x) 1
( x) 1
n

1
o
1 1
o
2 2
n
n

1
n

1
 dx n



dx
dx


 dx

n
 c d n y  
 c d n 1 y  


1 a
1   . . .  a ( x )c y   a ( x) c1d y 2  
 a n ( x) 1
( x) 1
n

1
o
1
1
n
 dx n 
 dx n 1 
 dx n 






 c d n 1 y  
2   . . .  a ( x)c y  
 a n 1 ( x ) 1
o
1 2
 dx n 1 


 0  0  0  c1y1  c 2 y 2 is a solution .
-2-
2) An nth order homogeneous linear differential equation has n linearly independent solutions:
y1 , y 2 , y 3 , . . , y n .
3). The general solution of the homogeneous linear differential equation is given by
y h  c1y1  c 2 y 2  . . .  c n y n .
4) If Yh is the homogeneous solution of a non-homogeneous differential equation and Yp is a
particular solution of the non-homogeneous differential equation, then
Y = Yh + Yp is the general solution of the non-homogeneous differential equation.
Pr oof :
Step 1. Let a n ( x )
dn y
d n  1y
d2y
dy
 a n 1 ( x )
 . . .  a 2 ( x)
 a1 ( x )
 a o ( x )y  g ( x )
n
n

1
2
dx
dx
dx
dx
Step 2. Yh is the hom ogeneous solution
 a n ( x)
dn y h
d n  1y h
d2yh
dy
 a n 1 ( x )
 . . .  a 2 (x)
 a1 ( x ) h  a o ( x )y h  0
dx
dx n
dx n 1
dx 2
Step 3. Yp is the hom ogeneous solution
 a n ( x)
Step 4.
dn y p
dx n
 a n 1 ( x )




d n  1y p
dx n 1
 . . .  a 2 (x)

d2yp
dx 2
 a1 ( x )

dy p

dx
 a o ( x )y p  g ( x )

dn y h  y p
d n 1 y h  y p
d2 yh  yp
a n (x)
 a n 1 ( x )
 . . .  a 2 (x)
dx n
dx n 1
dx 2
d yh  yp
 a1 ( x )
 a o (x) y h  y p 
dx
n
2
 n

 n 1
 2
y h  d n 1 y p 
d y h  d y p 
d
d y h  d y p



a n (x)

 a n 1 ( x )

 . . .  a 2 (x)




n
n 
n 1
n 1 
2
dx
dx
dx
dx
dx
dx 2







 
 
 
 


 dy h  d y p 
 a1 ( x )

  a o ( x ) y h   ( y p ) 
dx 
 dx
 d n y  
 d n 1 y  
 2

h a
h   . . .  a ( x ) d y h    a ( x ) dy h    a ( x )y ) 
a n ( x)
(x)
n

1
2
1 
o
h

 dx n 
 dx n 1 
 dx 2 
 dx 






 dn y 
 d n 1 y 
d2 y 
d yp 
p 
p 
p 
a n ( x)
 a n 1 ( x  )
 . . .  a 2 ( x)
 a1 ( x )
  a o (x) y p ) 



n 
n 1 
2 
dx 

dx
dx
dx







 


 
 
 
0  g( x )
 g( x )
Therefore , y h  y p is a solution of the non  hom ogeneous differenti al equation
-3-
  
Problem 2. Consider the differential equation y   y   2 y  sin x  3 cos x . If y  e 2 x and y  e  x
are solutions of the related homogeneous equation and y=cosx is a particular solution, find
a) the general solution
y  e 2x and y  e  x are solutions of the related hom ogeneous equation
 y h  c1e 2x  c 2 e  x is the hom ogeneous solution
y p  cos x  y  c1e 2x  c 2 e  x  cos x is the general solution
Answer : y  c1e 2x  c 2e  x  cos x
b) The IVP solution where y(0)=1 and y (0)  2
y ( 0)  1, y ( 0)  2 and y  c1e 2 x  c 2 e  x  cos x
 y   2c1e 2 x  c 2 e  x  sin x
2

 c1  3
1  c1  c 2  1 c1   c 2



2
 2  2c1  c 2
 3c1  2
c 2  
3

2
2
 y  e 2 x  e  x  cos x is the IVP solution
3
3
5) Theorem.
If y  e ax is a solution of the hom ogeneous differenti al equation with constant coefficients
an
dn y
dx n
 a n 1
d n  1y
dx n 1
 ...  a2
d2y
dx 2
 a1
dy
 a o y  0, then " a" is a solution of
dx
the polynomial function a n x n  a n 1x n 1  . . .  a 2 x 2  a1x  a 0  0
Pr oof : y  e ax  y   ae ax , y   a 2 e ax , . . . , y (n )  a n e an
  

a  a a  . . .  a a
 a n a n e an  a n 1 a n 1e an  . . .  a 2 a 2 e ax  a1ae ax  a o e ax  0
 an
n
n 1
n 1
2
2
 a1a  a o  0
Definition : the polynomial function a n x n  a n 1x n 1  . . .  a 2 x 2  a1x  a 0  0
is called the Characteri stic Equation of the linear hom ogeneous differenti al
equation with constant coefficients. The solutions are called the Eigenvalues of the DE .
Definition : a n D n  a n 1Dn 1  . . .  a 2 D 2  a1D  a 0  0
is called the differenti al operator of the linear hom ogeneous differenti al equation
with constant coefficients.
-4-
Pr oblem 3. Find the general solution of y   3y   4y   12y  0
Solution : y  e ax is a solution  y   ae ax , y   a 2 e ax , y   a 3e ax
 a 3e ax  3a 2 e ax  4ae ax  12e ax  0  a 3  3a 2  4a  12  0
 (a  2)(a  2)(a  3)  0  a  2 or a  2 or a  3
 y1  e 2x , y 2  e  2x and y 3  e  3x are linearly independen t solutions .
Therefore , the general solution is y  c1e 2x  c 2 e  2x  c 3e  3x
Pr oblem 4. Solve the DE y   y   6y  0
Solution : y  e ax is a solution  a 2  a  6  0  (a  3)(a  2)  0
 a  3 or a  2  y  c1e  2x  c 2 e 3x is the genereal solution
Problem 5. Find the general solution of the differential equation y ( 4)  y   4y   4y   0
y ( 4)  y   4y   4 y  0  a 4  a 3  4a 2  4a  0 is the characteristic equation.
a 3 (a  1)  4a(a  1)  0  a(a  1)(a 2  4)  0  a(a  1)(a  2)(a  2)  0
 a  0, a  1, a  2, and a  2
 y  c1e 0 x  c 2 e x  c 3e 2x  c 4 e  2 x
Note : y   c 2 e x  2c 3e 2x  2c 4 e  2 x
y   c 2 e x  4c 3e 2 x  4c 4 e  2 x
y   c 2 e x  8c 3e 2 x  8c 4 e  2x
Answer : y  c1  c 2 e x  c 3e 2 x  c 4 e  2 x is the general solution
b ) y (0)  2, y (0)  1, y (0)  0, Y (0)  0
 C1  1

4
 2  C1  C 2  C3  C 4
C


2
 1  C  2C  2C
3


2
3
4


1
 0  C 2  4C3  4C 4
 C3   4
 0  C 2  8C3  8C 4

1
C 4  

12
4
1
1
 y  1  e x  e 2 x  e  2 x is the IVP solution
3
4
12
-5-