Second order Linear Ordinary Differential Equations - PowerPoint Presentation | Calculus for MAT PDF Download

 Page 1


Advanced Engineering Mathematics             2. Second-order Linear ODEs 1
2.1 Homogeneous linear ODEs
2.2 Homogeneous linear ODEs with constant coefficients
2.3 Differential operators
2.5 Euler-Cauchy equation
2.6 Existence and uniqueness theory
2.7 Nonhomogeneous ODEs
2.10Solution by variation of parameters
2. Second-order Linear Ordinary 
Differential Equations
Advanced Engineering Mathematics             2. Second-order Linear ODEs 2
2.1  Homogeneous linear ODEs
? A linear second-order DE is formed of
y’’ + p(x)y’ + q(x)y = r (x)
? If r (x) ? 0  (i.e., r (x) = 0  for all x considered),
then the DE is called homogeneous.
If r (x) ? 0 , the DE is called nonhomogeneous.
? The functions p, q and r are called the coefficients of the 
equation.
? A solution of a second-order DE on some open interval
a < x < b is a function y(x) that satisfies the DE, for all x in 
that interval.
Page 2


Advanced Engineering Mathematics             2. Second-order Linear ODEs 1
2.1 Homogeneous linear ODEs
2.2 Homogeneous linear ODEs with constant coefficients
2.3 Differential operators
2.5 Euler-Cauchy equation
2.6 Existence and uniqueness theory
2.7 Nonhomogeneous ODEs
2.10Solution by variation of parameters
2. Second-order Linear Ordinary 
Differential Equations
Advanced Engineering Mathematics             2. Second-order Linear ODEs 2
2.1  Homogeneous linear ODEs
? A linear second-order DE is formed of
y’’ + p(x)y’ + q(x)y = r (x)
? If r (x) ? 0  (i.e., r (x) = 0  for all x considered),
then the DE is called homogeneous.
If r (x) ? 0 , the DE is called nonhomogeneous.
? The functions p, q and r are called the coefficients of the 
equation.
? A solution of a second-order DE on some open interval
a < x < b is a function y(x) that satisfies the DE, for all x in 
that interval.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 3
? At first we discuss the properties of solutions of second-
order linear DE, then we consider the solving strategies.
? Theorem 1 (Fundamental Theorem for homogeneous linear 
2nd-order DE)
For a homogeneous linear 2nd-order DE, any 
linear combination of two solutions on an open 
interval I is again a solution on I.  In particular, for 
such an equation, sums and constant multiples of 
solutions are again solutions.
Proof. 
Let y
1
and y
2
be solutions of y’’ + py’ + qy = 0 on I.
Substituting y = c
1
y
1
+ c
2
y
2
into the DE, we get 
y’’ + py’ + qy = (c
1
y
1
+ c
2
y
2
)’’ + p(c
1
y
1
+ c
2
y
2
)’ + q(c
1
y
1
+ c
2
y
2
)
= c
1
y
1
’’ + c
2
y
2
’’ + c
1
py
1
’ + c
2
py
2
’ + c
1
qy
1
+ c
2
qy
2
= c
1 
(y
1
’’ + py
1
’ + qy
1
) + c
2
(y
2
’’ + py
2
’ + qy
2
) = 0.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 4
? Ex.
y
1
= e 
x
and  y
2
= e 
-x
are two solutions of  y’’ – y = 0
y = -3y
1
+ 8y
2
= -3e 
x
+ 8e 
-x
is a solution of y’’ – y = 0
since (-3e 
x
+ 8e 
-x
) – ( -3e
x
+ 8e 
-x
) = 0.
? Caution! The linear combination solutions does not hold for 
nonhomogeneous or  nonlinear DE.
? Ex.2. y
1
= 1 + cosx and y
2
= 1 + sinx are solutions of 
nonhomogeneous DE  y’’ + y = 1.
but  2(1 + cosx)  and  (1 + cosx) + (1 + sinx)  are not
solutions of y’’ + y = 1.
? Ex.3. y
1
= x 
2
and  y
2
= 1  are solutions of nonlinear DE
y’’y – xy’ = 0, but  -x 
2
and  x 
2
+ 1  are not solution
of y’’y – xy’ = 0.
Page 3


Advanced Engineering Mathematics             2. Second-order Linear ODEs 1
2.1 Homogeneous linear ODEs
2.2 Homogeneous linear ODEs with constant coefficients
2.3 Differential operators
2.5 Euler-Cauchy equation
2.6 Existence and uniqueness theory
2.7 Nonhomogeneous ODEs
2.10Solution by variation of parameters
2. Second-order Linear Ordinary 
Differential Equations
Advanced Engineering Mathematics             2. Second-order Linear ODEs 2
2.1  Homogeneous linear ODEs
? A linear second-order DE is formed of
y’’ + p(x)y’ + q(x)y = r (x)
? If r (x) ? 0  (i.e., r (x) = 0  for all x considered),
then the DE is called homogeneous.
If r (x) ? 0 , the DE is called nonhomogeneous.
? The functions p, q and r are called the coefficients of the 
equation.
? A solution of a second-order DE on some open interval
a < x < b is a function y(x) that satisfies the DE, for all x in 
that interval.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 3
? At first we discuss the properties of solutions of second-
order linear DE, then we consider the solving strategies.
? Theorem 1 (Fundamental Theorem for homogeneous linear 
2nd-order DE)
For a homogeneous linear 2nd-order DE, any 
linear combination of two solutions on an open 
interval I is again a solution on I.  In particular, for 
such an equation, sums and constant multiples of 
solutions are again solutions.
Proof. 
Let y
1
and y
2
be solutions of y’’ + py’ + qy = 0 on I.
Substituting y = c
1
y
1
+ c
2
y
2
into the DE, we get 
y’’ + py’ + qy = (c
1
y
1
+ c
2
y
2
)’’ + p(c
1
y
1
+ c
2
y
2
)’ + q(c
1
y
1
+ c
2
y
2
)
= c
1
y
1
’’ + c
2
y
2
’’ + c
1
py
1
’ + c
2
py
2
’ + c
1
qy
1
+ c
2
qy
2
= c
1 
(y
1
’’ + py
1
’ + qy
1
) + c
2
(y
2
’’ + py
2
’ + qy
2
) = 0.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 4
? Ex.
y
1
= e 
x
and  y
2
= e 
-x
are two solutions of  y’’ – y = 0
y = -3y
1
+ 8y
2
= -3e 
x
+ 8e 
-x
is a solution of y’’ – y = 0
since (-3e 
x
+ 8e 
-x
) – ( -3e
x
+ 8e 
-x
) = 0.
? Caution! The linear combination solutions does not hold for 
nonhomogeneous or  nonlinear DE.
? Ex.2. y
1
= 1 + cosx and y
2
= 1 + sinx are solutions of 
nonhomogeneous DE  y’’ + y = 1.
but  2(1 + cosx)  and  (1 + cosx) + (1 + sinx)  are not
solutions of y’’ + y = 1.
? Ex.3. y
1
= x 
2
and  y
2
= 1  are solutions of nonlinear DE
y’’y – xy’ = 0, but  -x 
2
and  x 
2
+ 1  are not solution
of y’’y – xy’ = 0.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 5
? Definition
A general solution of a 1st-order DE involved one arbitrary
constant, a general solution of a 2nd-order DE will involve
two arbitrary constant, and is of the form
y=c
1
y
1
+c
2
y
2
,
where y
1
and y
2
are linear independent and called a basis
of the DE.
? Definition
A particular solution is obtained by specifying c
1
and c
2
.
? Two function y
1
(x) and y
2
(x) are linear independent, if
c
1
y
1
(x)+c
2
y
2
(x) = 0 implies c
1
=c
2
=0.
In other words, two functions are linear independent if and
only if they are not proportional.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 6
? Wronskian test for linear independent
Let y
1
(x) and y
2
(x) be solutions of y’’ + py’ + qy = 0 on an 
interval I.
Let  W [y
1
, y
2
] =                = y
1
y
2
’ – y
1
’y
2   
for x in I.
Then  1.  Either  W [ y
1
, y
2
] = 0   for all x in I
or    W [y
1
, y
2
] ? 0   for all x in I .
2.  y
1
(x) and y
2
(x) are linear independent on I
? W [y
1
, y
2
] ? 0   for some x in I.
The proof will be given later.
Page 4


Advanced Engineering Mathematics             2. Second-order Linear ODEs 1
2.1 Homogeneous linear ODEs
2.2 Homogeneous linear ODEs with constant coefficients
2.3 Differential operators
2.5 Euler-Cauchy equation
2.6 Existence and uniqueness theory
2.7 Nonhomogeneous ODEs
2.10Solution by variation of parameters
2. Second-order Linear Ordinary 
Differential Equations
Advanced Engineering Mathematics             2. Second-order Linear ODEs 2
2.1  Homogeneous linear ODEs
? A linear second-order DE is formed of
y’’ + p(x)y’ + q(x)y = r (x)
? If r (x) ? 0  (i.e., r (x) = 0  for all x considered),
then the DE is called homogeneous.
If r (x) ? 0 , the DE is called nonhomogeneous.
? The functions p, q and r are called the coefficients of the 
equation.
? A solution of a second-order DE on some open interval
a < x < b is a function y(x) that satisfies the DE, for all x in 
that interval.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 3
? At first we discuss the properties of solutions of second-
order linear DE, then we consider the solving strategies.
? Theorem 1 (Fundamental Theorem for homogeneous linear 
2nd-order DE)
For a homogeneous linear 2nd-order DE, any 
linear combination of two solutions on an open 
interval I is again a solution on I.  In particular, for 
such an equation, sums and constant multiples of 
solutions are again solutions.
Proof. 
Let y
1
and y
2
be solutions of y’’ + py’ + qy = 0 on I.
Substituting y = c
1
y
1
+ c
2
y
2
into the DE, we get 
y’’ + py’ + qy = (c
1
y
1
+ c
2
y
2
)’’ + p(c
1
y
1
+ c
2
y
2
)’ + q(c
1
y
1
+ c
2
y
2
)
= c
1
y
1
’’ + c
2
y
2
’’ + c
1
py
1
’ + c
2
py
2
’ + c
1
qy
1
+ c
2
qy
2
= c
1 
(y
1
’’ + py
1
’ + qy
1
) + c
2
(y
2
’’ + py
2
’ + qy
2
) = 0.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 4
? Ex.
y
1
= e 
x
and  y
2
= e 
-x
are two solutions of  y’’ – y = 0
y = -3y
1
+ 8y
2
= -3e 
x
+ 8e 
-x
is a solution of y’’ – y = 0
since (-3e 
x
+ 8e 
-x
) – ( -3e
x
+ 8e 
-x
) = 0.
? Caution! The linear combination solutions does not hold for 
nonhomogeneous or  nonlinear DE.
? Ex.2. y
1
= 1 + cosx and y
2
= 1 + sinx are solutions of 
nonhomogeneous DE  y’’ + y = 1.
but  2(1 + cosx)  and  (1 + cosx) + (1 + sinx)  are not
solutions of y’’ + y = 1.
? Ex.3. y
1
= x 
2
and  y
2
= 1  are solutions of nonlinear DE
y’’y – xy’ = 0, but  -x 
2
and  x 
2
+ 1  are not solution
of y’’y – xy’ = 0.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 5
? Definition
A general solution of a 1st-order DE involved one arbitrary
constant, a general solution of a 2nd-order DE will involve
two arbitrary constant, and is of the form
y=c
1
y
1
+c
2
y
2
,
where y
1
and y
2
are linear independent and called a basis
of the DE.
? Definition
A particular solution is obtained by specifying c
1
and c
2
.
? Two function y
1
(x) and y
2
(x) are linear independent, if
c
1
y
1
(x)+c
2
y
2
(x) = 0 implies c
1
=c
2
=0.
In other words, two functions are linear independent if and
only if they are not proportional.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 6
? Wronskian test for linear independent
Let y
1
(x) and y
2
(x) be solutions of y’’ + py’ + qy = 0 on an 
interval I.
Let  W [y
1
, y
2
] =                = y
1
y
2
’ – y
1
’y
2   
for x in I.
Then  1.  Either  W [ y
1
, y
2
] = 0   for all x in I
or    W [y
1
, y
2
] ? 0   for all x in I .
2.  y
1
(x) and y
2
(x) are linear independent on I
? W [y
1
, y
2
] ? 0   for some x in I.
The proof will be given later.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 7
? Find a basis if one solution is known (reduction of order)
Let y
1
be a solution of y” + py’ + qy = 0 on some interval I. 
Substitute y
2 
= uy
1
into the DE to get  y
2
’ = u’y
1
+ uy
1
’  and
y
2
” = u”y
1
+ 2u’y
1
’ + uy
1
”.
Then u”y
1
+ 2u’y
1
’ + uy
1
” + p(u’y
1 
+ uy
1
’) + quy
1 
= 0
? u”y
1
+ u’(2y
1
’ + py
1
) + u(y
1
” + py
1
’ + qy
1
) = 0
? u”y
1 
+ u’(2y
1
’ + py
1
) = 0
Let v = u’
Advanced Engineering Mathematics             2. Second-order Linear ODEs 8
? Ex. Find a basis of solution for the DE
x
2
y” – xy’ + y = 0
Solution.
One solution is y
1
= x
Since ?vdx = u can’t be a constant, y
1
and y
2
form a basis 
of solution.        
Page 5


Advanced Engineering Mathematics             2. Second-order Linear ODEs 1
2.1 Homogeneous linear ODEs
2.2 Homogeneous linear ODEs with constant coefficients
2.3 Differential operators
2.5 Euler-Cauchy equation
2.6 Existence and uniqueness theory
2.7 Nonhomogeneous ODEs
2.10Solution by variation of parameters
2. Second-order Linear Ordinary 
Differential Equations
Advanced Engineering Mathematics             2. Second-order Linear ODEs 2
2.1  Homogeneous linear ODEs
? A linear second-order DE is formed of
y’’ + p(x)y’ + q(x)y = r (x)
? If r (x) ? 0  (i.e., r (x) = 0  for all x considered),
then the DE is called homogeneous.
If r (x) ? 0 , the DE is called nonhomogeneous.
? The functions p, q and r are called the coefficients of the 
equation.
? A solution of a second-order DE on some open interval
a < x < b is a function y(x) that satisfies the DE, for all x in 
that interval.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 3
? At first we discuss the properties of solutions of second-
order linear DE, then we consider the solving strategies.
? Theorem 1 (Fundamental Theorem for homogeneous linear 
2nd-order DE)
For a homogeneous linear 2nd-order DE, any 
linear combination of two solutions on an open 
interval I is again a solution on I.  In particular, for 
such an equation, sums and constant multiples of 
solutions are again solutions.
Proof. 
Let y
1
and y
2
be solutions of y’’ + py’ + qy = 0 on I.
Substituting y = c
1
y
1
+ c
2
y
2
into the DE, we get 
y’’ + py’ + qy = (c
1
y
1
+ c
2
y
2
)’’ + p(c
1
y
1
+ c
2
y
2
)’ + q(c
1
y
1
+ c
2
y
2
)
= c
1
y
1
’’ + c
2
y
2
’’ + c
1
py
1
’ + c
2
py
2
’ + c
1
qy
1
+ c
2
qy
2
= c
1 
(y
1
’’ + py
1
’ + qy
1
) + c
2
(y
2
’’ + py
2
’ + qy
2
) = 0.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 4
? Ex.
y
1
= e 
x
and  y
2
= e 
-x
are two solutions of  y’’ – y = 0
y = -3y
1
+ 8y
2
= -3e 
x
+ 8e 
-x
is a solution of y’’ – y = 0
since (-3e 
x
+ 8e 
-x
) – ( -3e
x
+ 8e 
-x
) = 0.
? Caution! The linear combination solutions does not hold for 
nonhomogeneous or  nonlinear DE.
? Ex.2. y
1
= 1 + cosx and y
2
= 1 + sinx are solutions of 
nonhomogeneous DE  y’’ + y = 1.
but  2(1 + cosx)  and  (1 + cosx) + (1 + sinx)  are not
solutions of y’’ + y = 1.
? Ex.3. y
1
= x 
2
and  y
2
= 1  are solutions of nonlinear DE
y’’y – xy’ = 0, but  -x 
2
and  x 
2
+ 1  are not solution
of y’’y – xy’ = 0.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 5
? Definition
A general solution of a 1st-order DE involved one arbitrary
constant, a general solution of a 2nd-order DE will involve
two arbitrary constant, and is of the form
y=c
1
y
1
+c
2
y
2
,
where y
1
and y
2
are linear independent and called a basis
of the DE.
? Definition
A particular solution is obtained by specifying c
1
and c
2
.
? Two function y
1
(x) and y
2
(x) are linear independent, if
c
1
y
1
(x)+c
2
y
2
(x) = 0 implies c
1
=c
2
=0.
In other words, two functions are linear independent if and
only if they are not proportional.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 6
? Wronskian test for linear independent
Let y
1
(x) and y
2
(x) be solutions of y’’ + py’ + qy = 0 on an 
interval I.
Let  W [y
1
, y
2
] =                = y
1
y
2
’ – y
1
’y
2   
for x in I.
Then  1.  Either  W [ y
1
, y
2
] = 0   for all x in I
or    W [y
1
, y
2
] ? 0   for all x in I .
2.  y
1
(x) and y
2
(x) are linear independent on I
? W [y
1
, y
2
] ? 0   for some x in I.
The proof will be given later.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 7
? Find a basis if one solution is known (reduction of order)
Let y
1
be a solution of y” + py’ + qy = 0 on some interval I. 
Substitute y
2 
= uy
1
into the DE to get  y
2
’ = u’y
1
+ uy
1
’  and
y
2
” = u”y
1
+ 2u’y
1
’ + uy
1
”.
Then u”y
1
+ 2u’y
1
’ + uy
1
” + p(u’y
1 
+ uy
1
’) + quy
1 
= 0
? u”y
1
+ u’(2y
1
’ + py
1
) + u(y
1
” + py
1
’ + qy
1
) = 0
? u”y
1 
+ u’(2y
1
’ + py
1
) = 0
Let v = u’
Advanced Engineering Mathematics             2. Second-order Linear ODEs 8
? Ex. Find a basis of solution for the DE
x
2
y” – xy’ + y = 0
Solution.
One solution is y
1
= x
Since ?vdx = u can’t be a constant, y
1
and y
2
form a basis 
of solution.        
Advanced Engineering Mathematics             2. Second-order Linear ODEs 9
? An initial value problem now consists of y’’ + py’ + qy = 0  
and two initial conditions  y(x
0
) = k
0
,  y’(x
0
) = k
1
.
The conditions are used to determine the two arbitrary 
constants c
1
and c
2
in the general solution.
? Ex.
Solve the initial value problem
y’’ – y = 0,  y(0) = 4 ,  y’(0) = -2
y = c
1
e 
x
+ c
2
e
-x
y’ = c
1
e 
x
– c
2
e
-x
y(0) = c
1
+ c
2
= 4  c
1
= 1
y’(0) = c
1
– c
2
= -2              c
2
= 3
? solution  y = e 
x
+ 3e 
–x
.
? ?
? Problems of Section 2.1.
Advanced Engineering Mathematics             2. Second-order Linear ODEs 10
2.2 Homogeneous linearODEs with constant coefficients
y’’ + ay’+ by =0,
where coefficients a and b are constant.
Solution.
Assume y=e
?x
and substituting it into the original DE to get
( ?
2
+ a ? + b) e
?x
=0
The equation ?
2
+a ?+b= 0 is called the characteristic 
equation of the DE. Its roots are                                          
and                                          .
Then the solutions are and .
Consider three cases:
case 1. two distinct real roots if a
2
–4b >0.
case 2. a real double root if a
2
–4b =0.
case 3. complex conjugate roots if a
2
–4b<0.