Applied Mathematics Letters 25 (2012) 408–411

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Applied Mathematics Letters

journal homepage: www.elsevier.com/locate/aml

A procedure for solving some second-order linear ordinary

differential equations
M. Saravi ∗
Department of Mathematics, Islamic Azad University, Nour Branch, Nour, Iran

article info abstract

Article history: The purpose of this work is to introduce the concept of pseudo-exactness for second-order
Received 12 May 2011 linear ordinary differential equations (ODEs), and then to try to solve some specific ODEs.
Received in revised form 5 August 2011 © 2011 Elsevier Ltd. All rights reserved.
Accepted 9 September 2011

Keywords:
Pseudo-exact
Linear ODEs
Closed form solution

1. Introduction

Many of the differential equations that describe physical phenomena are linear differential equations, and among these,
the second-order differential equation is the most common and the most important special case. They arise in the fields of
mechanics, heat, electricity, aerodynamics, stress analysis, and so on. In general, a second-order linear differential equation
with variable coefficients cannot be solved directly, and in most cases this is impossible. Therefore, the problem of finding
the solution of a second-order homogeneous ODE

Py′′ + Qy′ + Ry = 0, (1)

where y = f (x) is an analytical solution of (1) and P , Q , R are known functions of x with no restrictions placed on their
nature, arises frequently in science and engineering. In practice, equations of the form (1) do not usually have closed form
solutions; and even when they do, it may be very difficult to find them. Hence the exploitation of analytical methods for
solving such equations becomes a main subject of considerable interest in most textbooks [1–3]. In efforts to overcome this
problem several methods been introduced [1], but all were for specific forms or needed to have a precondition. Probably
the most well-known and widely used algorithm for finding a solution is the method of reduction of order [1–3]; and if the
equation is exact, methods are known. In the next section, we introduce a new concept for a second-order linear ODE, and
using this concept we try to find closed form solutions of (1) if they exist, and then we extend the consideration to finding
a particular solution of the corresponding non-homogeneous case of (1).

2. The definition and procedure

Definition. We call Eq. (1) a pseudo-exact equation if

P′′ − Q′ + R = 1. (2)

∗ Tel.: +98 9111213090; fax: +98 1226252752.

E-mail addresses: masoud@saravi.info, msaravi2001@yahoo.com.

0893-9659/$ – see front matter © 2011 Elsevier Ltd. All rights reserved.
doi:10.1016/j.aml.2011.09.024
 M. Saravi / Applied Mathematics Letters 25 (2012) 408–411 409

If Eq. (1) is not exact, it can be made exact by multiplying it by an appropriate integrating factor µ(x). Thus we require that
µ(x) be such that
µ(x)Py′′ + µ(x)Qy′ + µ(x)Ry = 0, (3)

is exact. Since, (3) is exact then we will have

(µ(x)P)′′ − (µ(x)Q)′ + Rµ(x) = 0.

This leads to

Pµ′′ + (2P′ − Q)µ′ + P′′ − Q′ + R µ = 0.

 
(4)

This equation is known as the adjoint equation. It plays a very important role in the advanced theory of differential
equations [1]. In general the problem of solving the adjoint differential equation is as difficult as that of solving the original
equation. But if we choose R = 1 it can be seen that (4) is a pseudo-exact equation and we may solve it. Therefore, we will
have two options: change the original equation to a pseudo-exact form, or find µ(x) in (4) with R = 1.
How can we change Eq. (1) to a pseudo-exact equation? By a simple procedure Eq. (1) can be inverted into a pseudo-exact
form.

Procedure. Divide Eq. (1) by R and rewrite (1) as

P ′′ Q
y + y′ + y = 0. (5)
R R
P Q
Now if we choose R
= p and R
= q, then Eq. (5) can be written as

py′′ + qy′ + y = 0. (6)

Differentiating (6) leads to

py′′′ + (q + p′ )y′′ + (q′ + 1)y′ = 0. (7)

Now if we let y′ = z, then we obtain

pz′′ + (q + p′ )z′ + (q′ + 1)z = 0, (8)

which is a pseudo-exact equation. Eq. (8) may be solved easily if Eq. (1) has closed form solutions.
The idea of a pseudo-exact equation can be extended to non-homogeneous equations. Consider

y′′ + py′ + qy = 0, (9)

′ ′
which is a pseudo-exact equation. Hence q − p = 1. That is, q = p + 1. Therefore (9) becomes

y′′ + py′ + (1 + p′ )y = 0. (10)

Using Eq. (10) we write

x
y′′ + py′ + (1 + p′ )y = e− a pdx . (11)
x
It is not difficult to show that the particular solution of Eq. (11) is given by yp = e− a pdx .
This idea can also be extended to any second-order linear ODE in the form of
x
y′′ + py′ + qy = (q − p′ )e− a pdx , (12)
x
where its homogeneous equation is not pseudo-exact, and it can easily be shown that its particular solution is yp = e− a pdx .
For example, yp = sin2 x is a particular equation of

y′′ − 2 cot x.y′ + (2 cot2 x + 1)y = − sin2 x.

But in this work we focus on homogeneous equations and leave the non-homogeneous case to the reader. In the next section,
by considering some examples we can obtain a better understanding about the application of pseudo-exact form equations.

3. Examples

This section deals with some tests on a few examples that been considered in several ODE textbooks. Here we resolved
them by using the above procedure and one can thus compare the methods.
410 M. Saravi / Applied Mathematics Letters 25 (2012) 408–411

Example 1. Consider

(1 − x cot x)y′′ − xy′ + y = 0.

The general solution of this equation is y = c1 x + c2 sin x, and is taken from [1]. The author in this book assumed that y1 = x,
and then found y2 by method of reduction of order. But if we apply the procedure introduced here, first we differentiate,
and then put y′ = z. We obtain

(1 − x cot x)z ′′ − cot x(1 − x cot x).z ′ = 0.

That is, z ′′ − cot x.z ′ = 0.
It is not difficult to show that z = c1 + c2 cos x. Therefore, y′ = c1 + c2 cos x. By a simple integration we obtain
y = c1 x + c2 sin x + c3 . It is easy to show that c3 = 0, unless we are dealing with a non-homogeneous case.

Example 2. Consider

x(x − 2)u′′ − 2(x − 1)u′ + 2u = 0, 0 < x < 2.

This equation was chosen from [2] and its general solution is u = c1 x2 + c2 (x − 1).
The author did not solve it. He just asked the reader to show that u1 = x2 and u2 = x − 1 are two solutions of this
′′′
equation. But if we follow the above procedure we come to 12 x(x − 2)u = 0. That is, u′′′ = 0. Integrating three times,
successively, we obtain u = c1 x2 + c2 x + c3 . By a simple substitution, we come to c3 = −c2 .

Example 3. Consider

(1 + x2 )y′′ − 4xy′ + 6y = 0.
This equation was chosen from [3]. The author offered to solve it by a power series method. Here we inverted it to a pseudo-
exact form, i.e.,
1 1 1
(1 + x2 )z ′′ − xz ′ + z = 0.
6 3 3
x2
Obviously, z1 = x, and by method of reduction z2 = x2 − 1. Thus, y1 ′ = x, whence y1 = 2
+ c1 , and y2 ′ = x2 − 1, that is
x3
y2 = 3
− x + c2 . Substituting in the given equation leads to c1 = 6, and c2 = 0. Therefore, the general solution will be
 2  3
x3
   
x 1 x
y = b1 − + b2 − x = a0 (1 − 3x ) + a1 x −
2
,
2 6 3 3

where a0 = − 16 b1 and a1 = −b2 , respectively.

We end this section by considering the Hermite equation.

Example 4. The Hermite equation is given by

y′′ − 2xy′ + 2λy = 0. (13)

It is clear that for λ = 1, y1 (x) = x. Let λ = 2, that is,

y′′ − 2xy′ + 4y = 0. (14)

Writing this equation as

1 ′′ 1
y − xy′ + y = 0,
4 2
and differentiating, we can write

1 ′′ 1 1
z − xz ′ + z = 0, (15)
4 2 2
x2
where z = y′ . Obviously z2 (x) = x; hence y2 (x) = 2
+ c. Substituting in (13) leads to

y2 (x) = 1 − 2x2 .

If we choose λ = 3 in (12), we get

y′′ − 2xy′ + 6y = 0.
 M. Saravi / Applied Mathematics Letters 25 (2012) 408–411 411

In the same manner we obtain

1 ′′ 1 1
z − xz ′ + z = 0,
6 3 3
which has the same solution as (14). That is, z3 (x) = 1 − 2x2 . Since z3 (x) = y′ 3 (x), then it is easy to show that y3 (x) = x − 32 x3 .
If we carry through this implementation we can obtain the following results:
4
y4 (x) = 1 − 4x2 + x4 ,
3
4 4 5
y5 (x) = x − x3 + x ,
3 15
..
.
This implementation can also be used to solve the Legendre equation.

4. Conclusions

The results from all the examples in this work show the efficiency of this procedure. In spite of the fact that in Examples 1–
3, we did not know that the given equation has a closed form solution, with this procedure such solutions may be found. In
Example 4 it was shown that the Hermite equation can also be solved by this procedure. The author hopes to extend this
method to all second-order linear ODEs, or, working on (12), to get a general solution of (1). Research on this matter is one
of my future goals.

References

[1] William E. Boyce, Richard C. Diprima, Elementary Differential Equations and Boundary Value Problems, 8th ed., John Wiley & Son, 2005.
[2] Merle C. Potter, Jack Goldberg, Mathematical Methods, 2nd ed., Prentice-Hall International Inc., 1987.
[3] Erel D. Rainville, Philipe A. Bedint, Elementary Differential Equations, Macmillan Publishing Co., 1974.