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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.
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FAQs on Second order Linear Ordinary Differential Equations - PowerPoint Presentation - Calculus for MAT

1. What is a second order linear ordinary differential equation?
Ans. A second order linear ordinary differential equation is a differential equation that involves the second derivative of an unknown function with respect to the independent variable, along with the first derivative and the unknown function itself. It can be written in the form: $$\frac{{d^2y}}{{dx^2}} + p(x)\frac{{dy}}{{dx}} + q(x)y = r(x)$$ where y is the unknown function, x is the independent variable, and p(x), q(x), and r(x) are given functions.
2. How do you solve a second order linear ordinary differential equation?
Ans. To solve a second order linear ordinary differential equation, we can use various methods such as the method of undetermined coefficients, variation of parameters, or by using Laplace transforms. The specific method used depends on the form of the given equation and the given initial or boundary conditions.
3. What are the applications of second order linear ordinary differential equations in engineering?
Ans. Second order linear ordinary differential equations find numerous applications in engineering. Some common applications include modeling damped and undamped oscillations in mechanical systems, analyzing electrical circuits, studying fluid flow in pipes or channels, and analyzing the behavior of vibrating systems.
4. Can a second order linear ordinary differential equation have complex roots?
Ans. Yes, a second order linear ordinary differential equation can have complex roots. This occurs when the characteristic equation associated with the differential equation has complex roots. Complex roots indicate that the solution to the differential equation will involve complex numbers, representing oscillatory or decaying behavior.
5. What are the boundary conditions for solving a second order linear ordinary differential equation?
Ans. The boundary conditions for solving a second order linear ordinary differential equation depend on the specific problem being addressed. Common boundary conditions include specifying the values of the unknown function and its derivatives at certain points or specifying the values of the function and/or its derivatives at the boundaries of a certain interval. These conditions are used to obtain a unique solution to the differential equation.
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Second order Linear Ordinary Differential Equations - PowerPoint Presentation | Calculus for MAT

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

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

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