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.

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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.

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