Differential equations
Section 2: Second order differential equations
Notes and Examples
These notes contain subsections on
 Classification of differential equations
 The auxiliary equation method
 Finding complementary functions
 Finding particular integrals
 Finding general and particular solutions
Classification of differential equations
It is important to understand the classification of differential equations
because the different methods of solution apply to particular types of
equation. For example, you have learnt two methods (separation of variables,
and integrating factors) which are applicable only to first order linear
differential equations. In this chapter you will be looking at a method (the
auxiliary equation method) which can be applied to linear differential
equations of any order, but only if it has constant coefficients.
Uses of the different methods can be illustrated using the following flow chart.
NO
NO
A substitution
may be
possible
Is the
differential
equation
linear?
YES
Does it have
constant
coefficients?
Are the
variables
separable?
YES
Method of
separation of
variables
NO
YES
NO
Auxiliary
equation method
Is the
equation
first order?
A substitution
may be
possible
YES
Integrating factor
method
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Some differential equations may be solved by more than one method.
However, many differential equations cannot be solved by analytical methods.
There are numerical methods which can be used to obtain solutions to any
degree of accuracy required.
The auxiliary equation method
The auxiliary method can be applied to any linear differential equation with
constant coefficients. Most of the work in this chapter is applied to second
order differential equations, but the method can be used for first order
equations, and also for third and higher order equations (which you are not
required to cover).
The example below illustrates how three different methods can be used to
solve a first order differential equation.
Example 1
Find the general solution of the differential equation
dy
2  3y  0 .
dx
Solution 1: separation of variables
dy
2  3y  0
dx
1
3
 y dy    2 dx
ln y   32 x  c
ye
 32 x  c
 Ae
 32 x
Solution 2: using an integrating factor
dy 3
 y0
dx 2
3 dx
3x
Integrating factor  e 2  e 2
3x
e2
dy 3 32 x
 e y0
dx 2
 
3x
d
ye 2  0
dx
3x
ye 2  A
y  Ae
 32 x
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Solution 3: the auxiliary equation method
2  3  0
The auxiliary equation is:
   32
The general solution is
y  Ae
 32 x
The example above demonstrates that the auxiliary equation method is the
quickest and easiest for this type of equation.
However, remember that all three methods cannot be applied to all first order
differential equations. If there were a function of x on the right-hand side of the
equation, then the auxiliary equation method could be used, or an integrating
factor, but separation of variables probably could not.
You can practice applying the auxiliary equation method to first order
differential equations using the interactive questions First order differential
equations.
Finding complementary functions
In the case of a first order differential equation with constant coefficients, the
auxiliary equation is linear and has one root , resulting in a complementary
function of the form y  Ae x .
In the case of a second order differential equation with constant coefficients,
the auxiliary equation is quadratic and therefore has either two distinct real
roots, one repeated root, or two complex roots.
For two distinct real roots 1 and 2 , the complementary function has the form
y  Ae x  Be x .
1
2
For one repeated root , the complementary function has the form
y  ( Ax  B)e x .
The case in which the auxiliary equation has complex roots will be covered in
section 3.
The method can also be extended to third and higher order equations. In the
case of a third order equation, for example, the auxiliary equation will be cubic
and will have three roots, so that the complementary function has three parts
to it. Similarly the complementary function for a fourth order equation would
have four parts.
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Finding particular integrals
The worked examples in the textbook illustrate all the types of particular
integral that you will meet.
The table below illustrates all these forms of particular integrals.
Function f(x)
Constant term
Linear function
Exponential function involving e px
Function involving sin px and / or cos px
Particular integral
c
ax + b
ke px
a cos px  b sin px
Notice that in Example 3.5.1, although the right-hand side of the equation
involves only sin 2x, the trial form of the particular integral involves both sin 2x
and cos 2x. Similarly, in Example 3.5.2, although the right-hand side is a term
in just x, the trial form of the particular integral includes a constant term as
well as a term in x.
An important special case is shown in Example 3.5.3. In this case, the form
which you would expect the particular integral to take ( ae3 x ) is the same as
part of the complementary function. This is a problem, since this means that
dy
dy
any particular integral satisfies the homogeneous equation
 6  9y  0 ,
dx
dx
dy
dy
 6  9 y  e3 x . In such
and therefore does not satisfy the whole equation
dx
dx
3x
cases, a particular integral of the form axe is usually used. However, in this
case, this form is also the same as part of the complementary function (since
the auxiliary equation has a repeated root) so instead a particular integral of
the form ax 2e3 x is used.
Finding general and particular solutions
You can see from the worked examples in the textbook that solving a
d2 y
dy
differential equation of the form a 2  b  cy  f ( x) is quite a lengthy
dx
dx
process. However, although it is easy to make careless errors, the steps
involved are all quite straightforward.
Here is a summary of the process.
 Write down and solve the quadratic auxiliary equation.
 Use the solutions of the auxiliary equation to write down the
complementary function in the appropriate form, with unknown
constants A and B.
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 Look at the function f(x) on the right-hand side of the differential
equation. Check whether it is of the same form as any part of the
complementary function, and then write down the appropriate form of
the particular integral, which will involve one or two unknown
constants.
 Differentiate the particular integral twice, and substitute into the
original differential equation. Then equate coefficients to find the
values of the unknown constants in the particular integral.
 Write down the general solution of the differential equation by adding
the particular integral to the complementary function.
 If a particular solution is required, substitute the given conditions to
find the values of the unknown constants A and B. (This may involve
solving a pair of simultaneous equations).
Here is a further example.
Example 2
Find the particular solution of the differential equation
d² y
dy
 2  y  sin x
dx²
dx
dy
in the case where y = 0 and
= 0 when x = 0.
dx
Solution
The auxiliary equation is:
 2  2  1  0
(  1)²  0
  1
The complementary function is:
y  ( A  Bx)e  x
The particular integral is y  a sin x  b cos x
dy
 a cos x  b sin x
dx
d² y
  a sin x  b cos x
dx ²
Substituting into the differential equation:
a sin x  b cos x  2(a cos x  b sin x)  a sin x  b cos x  sin x
(a  2b  a)sin x  (b  2a  b) cos x  sin x
2b sin x  2a cos x  sin x
Equating coefficients of cos x: a  0
Equating coefficients of sin x: b   12
The particular integral is y   12 cos x
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The general solution is:
y  ( A  Bx)e  x  12 cos x
When x = 0, y = 0  0  A  12  A  12
dy
 Be x  ( A  Bx)e x  12 sin x
dx
dy
When x = 0,
= 0  0B A  B 
dx
1
2
The particular solution is y  12 (1  x)e  x  12 cos x
For additional practice, try the interactive questions Second order
differential equations (general solutions) and Second order differential
equations (particular solutions). All of these questions deal with equations
in which the auxiliary equation has two distinct roots.
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