Differential equations, a tourist’s guide | DE1
Differential equations, a tourist’s guide | DE1

Introduction:
The same relative ease with which we were able to find explicit solutions of
higher-order linear differential equations with constant coefficients in the
preceding sections does not, in general, carry over to linear equations with variable
coefficients. However, the type of differential equation that we consider in this
section is an exception to this rule; it is a linear equation with variable coefficients
whose general solution can always be expressed in terms of powers of x, sines,
cosines, and logarithmic functions. Moreover, its method of solution is quite
similar to that for constant-coefficient equations in that an auxiliary equation must
be solved.

Cauchy-Euler Equation:
A linear differential equation of the form
𝑎𝑛𝑥𝑛
𝑑𝑛𝑦
𝑑𝑥2
+ 𝑎𝑛−1𝑥𝑛−1
𝑑𝑛−1𝑦
𝑑𝑥𝑛−1
+ ⋯ + 𝑎1𝑥
𝑑𝑦
𝑑𝑥
+ 𝑎0𝑦 = 𝑔 𝑥 ,
where the coefficients 𝑎𝑛, 𝑎𝑛−1, . . . , 𝑎0 are constants, is known as a Cauchy-Euler
equation. The observable characteristic of this type of equation is that the degree 𝑘 =
𝑛, 𝑛 − 1, . . . , 1, 0 of the monomial coefficients 𝑥𝑘 matches the order k of
differentiation 𝑑𝑘
𝑦/𝑑𝑥𝑘
:
Note the following properties of these equations
Any solution will be on a subset of (−∞, 0) or (0, ∞).
The powers of x must match the order of the derivatives.

Method of Solution
We try a solution of the form 𝑦 = 𝑥𝑚, where 𝑚 is to be determined. Analogous to
what happened when we substituted 𝑒𝑚𝑥
into a linear equation with constant
coefficients, when we substitute 𝑥𝑚, each term of a Cauchy-Euler equation
becomes a polynomial in 𝑚 times 𝑥𝑚
, since
𝑎𝑘𝑥𝑘
𝑑𝑘
𝑦
𝑑𝑥𝑘
= 𝑎𝑘𝑥𝑘
𝑚(𝑚 − 1)(𝑚 − 2) ⋯ (𝑚 − 𝑘 + 1)𝑥𝑚−𝑘
= 𝑎𝑘𝑚(𝑚 − 1)(𝑚 − 2) ⋯ (𝑚 − 𝑘 + 1)𝑥𝑚.
For example, when we substitute 𝑦 = 𝑥𝑚, the second-order equation becomes

𝑎𝑥2
𝑑2𝑦
𝑑𝑥2
+ 𝑏𝑥
𝑑𝑦
𝑑𝑥
+ 𝑐𝑦 = 𝑎𝑚(𝑚 − 1)𝑥𝑚 + 𝑏𝑚𝑥𝑚 + 𝑐𝑥𝑚
= (𝑎𝑚(𝑚 − 1) + 𝑏𝑚 + 𝑐)𝑥𝑚.
Thus 𝑦 = 𝑥𝑚 is a solution of the differential equation whenever 𝑚 is a
solution of the auxiliary equation
𝑎𝑚(𝑚 − 1) + 𝑏𝑚 + 𝑐 = 0 or 𝑎𝑚2 + (𝑏 − 𝑎)𝑚 + 𝑐 = 0

First order Cauchy-Euler
Note that
𝑎1𝑥
𝑑𝑦
𝑑𝑥
+ 𝑎0𝑦 = 0 ⇒
𝑑𝑦
𝑑𝑥
=
𝑎0𝑦
𝑎1𝑥

𝑑𝑦
𝑦
=
𝑎0
𝑎1
1
𝑥
𝑑𝑥
We can separate the variables as seen, and solve for 𝑦 = 𝑥𝑚.
Example:
We make the following substitution: 𝑥 = 𝑒𝑡.Then the derivatives will be
𝑦′ =
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
=
𝑑𝑦
𝑑𝑡
𝑒𝑡
= 𝑒−𝑡
𝑑𝑦
𝑑𝑡
,
𝑦′′ =
𝑑
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
𝑑
𝑑𝑥
𝑒−𝑡
𝑑𝑦
𝑑𝑡
=
𝑑
𝑑𝑡
𝑑𝑥
𝑑𝑡
𝑒−𝑡
𝑑𝑦
𝑑𝑡
=
−𝑒−𝑡 𝑑𝑦
𝑑𝑡
+ 𝑒−𝑡 𝑑2
𝑦
𝑑𝑡2
𝑒𝑡
= 𝑒−2𝑡
𝑑2
𝑦
𝑑𝑡2

𝑑𝑦
𝑑𝑡
.

Putting this into the original Euler equation gives:
As it can be seen, we obtain the linear equation with constant coefficients. The
corresponding characteristic equation has the form:
𝑘2 + 𝐴 − 1 𝑘 + 𝐵 = 0.
Now we can determine the roots of the characteristic equation and write the general
solution for the function 𝑦(𝑡). Then we can easily return to the function 𝑦(𝑥) taking
into account that
𝑦 𝑡 = 𝑦 ln 𝑥 .
𝑒2𝑡
𝑒−2𝑡
𝑑2
𝑦
𝑑𝑡2

𝑑𝑦
𝑑𝑡
+ 𝐴𝑒𝑡
𝑒−𝑡
𝑑𝑦
𝑑𝑡
+ 𝐵𝑦 = 0, ⇒
𝑑2
𝑦
𝑑𝑡2

𝑑𝑦
𝑑𝑡
+ 𝐴
𝑑𝑦
𝑑𝑡
+ 𝐵𝑦 = 0,

𝑑2
𝑦
𝑑𝑡2
+ (𝐴 − 1)
𝑑𝑦
𝑑𝑡
+ 𝐵𝑦 = 0.

Second order Cauchy-Euler
We now assume we are searching for solutions of the form 𝑦 = 𝑥𝑚
. Sure enough,
𝑎𝑥2
𝑑2
𝑦
𝑑𝑥2
+ 𝑏𝑥
𝑑𝑦
𝑑𝑥
+ 𝑐𝑦 = 0 ⇒ 𝑥𝑚
𝑎𝑚2
+ (𝑏 − 𝑎)𝑚 + 𝑐 = 0
Second Way of Solving an Euler Equation
In the second method we look for a solution of the equation in the form of the power
function 𝑦 = 𝑥𝑘,, where 𝑘 is an unknown number. It follows from here that
𝑑𝑦
𝑑𝑥
= 𝑘𝑥𝑘−1,
𝑑2𝑦
𝑑𝑥2
= 𝑘 𝑘 − 1 𝑥𝑘−2.

Substituting into the differential equation gives the following result:
𝑥2𝑘 𝑘 − 1 𝑥𝑘−2 + 𝐴𝑥𝑘𝑥𝑘−1 + 𝐵𝑥𝑘 = 0, ⇒ 𝑘 𝑘 − 1 𝑥𝑘 + 𝐴𝑘𝑥𝑘 + 𝐵𝑥𝑘 = 0,
⇒ [𝑘(𝑘 − 1) + 𝐴𝑘 + 𝐵]𝑥𝑘 = 0.
As 𝑥𝑘 ≠ 0, then
𝑘(𝑘 − 1) + 𝐴𝑘 + 𝐵 = 0, ⇒ 𝑘2 + (𝐴 − 1)𝑘 + 𝐵 = 0.
We get the same characteristic equation as in the first way. After finding the roots,
one can write the general solution of the differential equation.

Third order Equation
Solve 𝑥3 𝑑3𝑦
𝑑𝑥3 + 5𝑥2 𝑑2𝑦
𝑑𝑥2 + 7𝑥
𝑑𝑦
𝑑𝑥
+ 8𝑦 = 0.
Solution
The first three derivatives of 𝑦 = 𝑥𝑚
are
𝑑𝑦
𝑑𝑥
= 𝑚𝑥𝑚−1
,
𝑑2
𝑦
𝑑𝑥2
= 𝑚(𝑚 − 1)𝑥𝑚−2
,
𝑑3
𝑦
𝑑𝑥3
= 𝑚(𝑚 − 1)(𝑚 − 2)𝑥𝑚−3
,
so the given differential equation becomes
𝑥3
𝑑3
𝑦
𝑑𝑥3
+ 5𝑥2
𝑑2
𝑦
𝑑𝑥2
+ 7𝑥
𝑑𝑦
𝑑𝑥
+ 8𝑦 = 𝑥3𝑚(𝑚 − 1)(𝑚 − 2)𝑥𝑚−3 + 5𝑥2𝑚(𝑚 − 1)𝑥𝑚−2 + 7𝑥𝑚𝑥𝑚−1 + 8𝑥𝑚
= 𝑥𝑚
(𝑚(𝑚 − 1)(𝑚 − 2) + 5𝑚(𝑚 − 1) + 7𝑚 + 8)
= 𝑥𝑚 𝑚3 + 2𝑚2 + 4𝑚 + 8 = 𝑥𝑚(𝑚 + 2) 𝑚2 + 4 = 0.
In this case we see that 𝑦 = 𝑥𝑚 will be a solution of the differential equation for
𝑚1 = −2, 𝑚2 = 2𝑖, and 𝑚3 = −2𝑖. Hence the general solution is 𝑦 = 𝑐1𝑥−2 +
𝑐2cos(2ln 𝑥) + 𝑐3sin(2ln 𝑥).

Distinct Real Roots:
Let 𝑚1 and 𝑚2 denote the real roots of (1) such that 𝑚1 ≠ 𝑚2. Then 𝑦1 = 𝑥𝑚1 and
𝑦2 = 𝑥𝑚2 form a fundamental set of solutions. Hence the general solution is
𝑦 = 𝑐1𝑥𝑚1 + 𝑐2𝑥𝑚2.
Example: Distinct Roots
Solve 𝑥2 𝑑2𝑦
𝑑𝑥2 − 2𝑥
𝑑𝑦
𝑑𝑥
− 4𝑦 = 0

Solution
Rather than just memorizing equation (1), it is preferable to assume 𝑦 = 𝑥𝑚 as the
solution a few times to understand the origin and the difference between this new
form of the auxiliary equation. Differentiate twice,
𝑑𝑦
𝑑𝑥
= 𝑚𝑥𝑚−1,
𝑑2𝑦
𝑑𝑥2
= 𝑚(𝑚 − 1)𝑥𝑚−2,
and substitute back into the differential equation:
𝑥2
𝑑2𝑦
𝑑𝑥2
− 2𝑥
𝑑𝑦
𝑑𝑥
− 4𝑦 = 𝑥2 ⋅ 𝑚(𝑚 − 1)𝑥𝑚−2 − 2𝑥 ⋅ 𝑚𝑥𝑚−1 − 4𝑥𝑚
= 𝑥𝑚(𝑚(𝑚 − 1) − 2𝑚 − 4) = 𝑥𝑚 𝑚2 − 3𝑚 − 4 = 0
if 𝑚2 − 3𝑚 − 4 = 0. Now (𝑚 + 1)(𝑚 − 4) = 0 implies 𝑚1 = −1, 𝑚2 = 4, so 𝑦 =
𝑐1𝑥−1
+ 𝑐2𝑥4

Repeated Real Roots:
If the roots of (1) are repeated (that is, 𝑚1 = 𝑚2 ), then we obtain only one solution
namely, 𝑦 = 𝑥𝑚1.
When the roots of the quadratic equation 𝑎𝑚2
+ (𝑏 − 𝑎)𝑚 + 𝑐 = 0 are equal, the
discriminant of the coefficients is necessarily zero. It follows from the quadratic
formula that the root must be 𝑚1 = −(𝑏 − 𝑎)/2𝑎.
Now we can construct a second solution 𝑦2, using (5) of Section 4.2. We first write
the Cauchy-Euler equation in the standard form
𝑑2𝑦
𝑑𝑥2
+
𝑏
𝑎𝑥
𝑑𝑦
𝑑𝑥
+
𝑐
𝑎𝑥2
𝑦 = 0

And make the identifications 𝑃(𝑥) = 𝑏/𝑎𝑥 and 𝑓(𝑏/𝑎𝑥)𝑑𝑥 = (𝑏/𝑎)ln 𝑥. Thus
𝑦2 = 𝑥𝑚1
𝑒−(
𝑏
𝑎
)ln
𝑥
𝑥2𝑚1
𝑑𝑥
= 𝑥𝑚1 𝑥−
𝑏
𝑎 ⋅ 𝑥−2𝑚1 𝑑𝑥 ← 𝑒−(𝑏/𝑎)ln
𝑥
= 𝑒ln
𝑥−𝑏/𝑎
= 𝑥−𝑏/𝑎
= 𝑥𝑚1 𝑥−
𝑏
𝑎 ⋅ 𝑥
𝑏−𝑎
𝑎 𝑑𝑥 ← −2𝑚1 = (𝑏 − 𝑎)/𝑎
= 𝑥𝑚1
𝑑𝑥
𝑥
= 𝑥𝑚1 ln
𝑥.
The general solution is then
𝑦 = 𝑐1𝑥𝑚1 + 𝑐2𝑥𝑚1 ln 𝑥 .

Conclusion
The development of the solution set of certain ordinary differential equations still
remains the object of research, with attractive problems and high applicability in the
phenomena of nature. It is evident the difficulty encountered by students to establish
a relation of interest with the area of calculation, particularly in Differential
Equations, perhaps because they do not know the wide field of application that these
equations make available. In the light of the above, it is expected that this work may
significantly awaken other research on Cauchy-Euler Equation in order to minimize
the lags between mathematical abstraction and its practice.

References:
1. Boyce, William E.; DiPrima, Richard C. Equações Diferenciais Elementares e Problemas de
Valores de Contorno. 7. ed. Rio de Janeiro: LTC Editora, 2002.
2. Abunahman, Sergio Antonio. Equações Diferenciais. Rio de Janeiro: Livros Técnicos e Científicos,
1979.
3. Bronson, Richard. Moderna Introdução às Equações Diferencias. São Paulo: Mcgraw-Hill do
Brasil, 1976.
4. Zill, Dennis G; Cullen, Michael R. Equações Diferenciais. São Paulo: Makron Books Ltda, 2001,
v.1.
5. JR, Wylie, C. R. Matematicas Superiores para Ingenieria. New York: Mcgraw-Hill, 1969.
6. BOYER, Carl, B. Historia da Matemática. São Paulo: Edgard Blucher, 1974.

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