3.6 Cauchy–Euler Equations

INTRODUCTION

The relative ease with which we were able to find explicit solutions of linear higher-order differential equations with constant coefficients in the preceding sections does not, in general, carry over to linear equations with variable coefficients. We shall see in Chapter 5 that when a linear differential equation has variable coefficients, the best that we can usually expect is to find a solution in the form of an infinite series. However, the type of differential equation considered in this section is an exception to this rule; it is an equation with variable coefficients whose general solution can always be expressed in terms of powers of x, sines, cosines, logarithmic, and exponential functions. Moreover, its method of solution is quite similar to that for constant equations.

A photo of Augustin-Louis Cauchy.

© Colport/Alamy Stock Photo

Augustin-Louis Cauchy

Cauchy–Euler Equation

A linear differential equation of the form

where the coefficients an, an−1, …, a0 are constants, is known as a Cauchy–Euler equation. The differential equation is named in honor of two of the most esteemed mathematicians of all time, the French baron Augustin-Louis Cauchy (1789–1857) and the prolific Swiss Leonhard Euler (1707–1783). Cauchy’s name is one of 72 French mathematicians, scientists, and engineers engraved on the Eiffel Tower in Paris in recognition of their achievements. The observable characteristic of this type of differential equation is that the degree k = n, n − 1,…,1, 0 of the monomial coefficients xk matches the order k of differentiation dky/dxk:

A photo of Leonhard Euler.

© Science History Images/Alamy Stock Photo

Leonhard Euler

As in Section 3.3, we start the discussion with a detailed examination of the forms of the general solutions of the homogeneous second-order equation

(1)

The solution of higher-order equations follows analogously. Also, we can solve the nonhomogeneous equation ax2y″ + bxy′ + cy = g(x) by variation of parameters, once we have determined the complementary function yc(x) of (1).

Lead coefficient being zero at x = 0 could cause a problem.

The coefficient of d2y/dx2 is zero at x = 0. Hence, in order to guarantee that the fundamental results of Theorem 3.1.1 are applicable to the Cauchy–Euler equation, we confine our attention to finding the general solution on the interval (0, ∞). Solutions on the interval (−∞, 0) can be obtained by substituting t = −x into the differential equation. See Problems 49 and 50 in Exercises 3.6.

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, after substituting xm each term of a Cauchy–Euler equation becomes a polynomial in m times xm since

For example, by substituting y = xm the second-order equation (1) becomes

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. (2)

There are three different cases to be considered, depending on whether the roots of this quadratic equation are real and distinct, real and equal, or complex. In the last case the roots appear as a conjugate pair.

Case I: Distinct Real Roots Let m1 and m2 denote the real roots of (2) such that m1 ≠ m2. Then y1 = xm1 and y2 = xm2 form a fundamental set of solutions. Hence the general solution of (1) is

y = c1xm1 + c2xm2. (3)

EXAMPLE 1 Distinct Roots

Solve

SOLUTION

Rather than just memorizing equation (2), it is preferable to assume y = xm as the solution a few times in order to understand the origin and the difference between this new form of the auxiliary equation and that obtained in Section 3.3. Differentiate twice,

and substitute back into the differential equation:

if m2 − 3m − 4 = 0. Now (m + 1)(m − 4) = 0 implies m1 = −1, m2 = 4 and so (3) yields the general solution y = c1x−1 + c2x4. ≡

Case II: Repeated Real Roots If the roots of (2) 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 3.2. We first write the Cauchy–Euler equation in the standard form

and make the identifications P(x) = b/ax and ∫(b/ax) dx = (b/a) ln x. Thus

The general solution of (1) is then

y = c1xm1 + c2xm1 ln x. (4)

EXAMPLE 2 Repeated Roots

Solve

SOLUTION

The substitution y = xm yields

when 4m2 + 4m + 1 = 0 or (2m + 1)2 = 0. Since m1 = − is a repeated root, (4) gives the general solution y = c1 + c2 ln x. ≡

For higher-order equations, if m1 is a root of multiplicity k, then it can be shown that

are k linearly independent solutions. Correspondingly, the general solution of the differential equation must then contain a linear combination of these k solutions.

Case III: Conjugate Complex Roots If the roots of (2) are the conjugate pair m1 = α + iβ, m2 = α − iβ, where α and β > 0 are real, then a solution is y = C1xα+iβ + C2xα−iβ. But when the roots of the auxiliary equation are complex, as in the case of equations with constant coefficients, we wish to write the solution in terms of real functions only. We note the identity

xiβ = (eln x)iβ = eiβ ln x,

which, by Euler’s formula, is the same as

Adding and subtracting the last two results yields

xiβ + x−iβ = 2 cos(β ln x) and xiβ − x−iβ = 2i sin(β ln x),

respectively. From the fact that y = C1xα+iβ + C2xα−iβ is a solution for any values of the constants, we see, in turn, for C1 = C2 = 1 and C1 = 1, C2 = −1 that

are also solutions. Since W(xα cos(β ln x), xα sin(β ln x)) = βx2α−1 ≠ 0, β > 0, on the interval (0, ∞), we conclude that

y1 = xα cos(β ln x) and y2 = xα sin(β ln x)

constitute a fundamental set of real solutions of the differential equation.

Hence the general solution of (1) is

y = xα[c1 cos(β ln x) + c2 sin(β ln x)]. (5)

EXAMPLE 3 An Initial-Value Problem

Solve the initial-value problem 4x2y″ + 17y = 0, y(1) = −1, y′(1) = −.

SOLUTION

The y′ term is missing in the given Cauchy–Euler equation; nevertheless, the substitution y = xm yields

4x2y″ + 17y = xm(4m(m − 1) + 17) = xm(4m2 − 4m + 17) = 0

when 4m2 − 4m + 17 = 0. From the quadratic formula we find that the roots are m1 = + 2i and m2 = − 2i. With the identifications α = and β = 2, we see from (5) that the general solution of the differential equation on the interval (0, ∞) is

y = [c1 cos(2 ln x) + c2 sin(2 ln x)].

By applying the initial conditions y(1) = −1, y′(1) = 0 to the foregoing solution and using ln 1 = 0 we then find, in turn, that c1 = −1 and c2 = 0. Hence the solution of the initial-value problem is y = −cos (2 ln x). The graph of this function, obtained with the aid of computer software, is given in FIGURE 3.6.1. The particular solution is seen to be oscillatory and unbounded as x → ∞. ≡

The curve is graphed on an x y coordinate plane. It starts at the point (0, 1), goes down and slightly to the right to an approximate point (1, negative 1), and goes up to the right to an approximate point (5, 2). Then, it goes down and to the right through an approximate point (10, 0), and reaches an approximate low point (25, negative 5). From there, the curve goes up and to the right through an approximate point (45, 0), and continues up to the right smoothly, and ends at the approximate point (105, 10).

FIGURE 3.6.1 Graph of solution of IVP in Example 3

The next example illustrates the solution of a third-order Cauchy–Euler equation.

EXAMPLE 4 Third-Order Equation

Solve

SOLUTION

The first three derivatives of y = xm are

so that the given differential equation becomes

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 on the interval (0, ∞) is

y = c1x−2 + c2 cos(2 ln x) + c3 sin(2 ln x). ≡

Nonhomogeneous Equations

The method of undetermined coefficients as described in Section 3.4 does not carry over, in general, to linear differential equations with variable coefficients. Consequently, in the following example the method of variation of parameters is employed.

EXAMPLE 5 Variation of Parameters

Solve x2y″ − 3xy′ + 3y = 2x4ex.

SOLUTION

Since the equation is nonhomogeneous, we first solve the associated homogeneous equation. From the auxiliary equation (m − 1)(m − 3) = 0 we find yc = c1x + c2x3. Now before using variation of parameters to find a particular solution yp = u1y1 + u2y2, recall that the formulas = W1/W and = W2/W, where W1, W2, and W are the determinants defined on page 141, and were derived under the assumption that the differential equation has been put into the standard form y″ + P(x)y′ + Q(x)y = f(x). Therefore we divide the given equation by x2, and from

we make the identification f(x) = 2x2ex. Now with y1 = x, y2 = x3 and

we find and

The integral of the latter function is immediate, but in the case of we integrate by parts twice. The results are u1 = − x2ex + 2xex − 2ex and u2 = ex. Hence

yp = u1y1 + u2y2 = (− x2ex + 2xex − 2ex) x + exx3 = 2x2ex − 2xex.

Finally the general solution is y = yc + yp = c1x + c2x3 + 2x2ex − 2xex. ≡

A Generalization

The second-order differential equation

(6)

is a generalization of equation (1). Note that (6) reduces to (1) when x0 = 0. We can solve Cauchy–Euler equations of the form given in (6) exactly as we did in (1), namely, by seeking solutions y = (x − x0)m and using

See Problems 39–42 in Exercises 3.6.

Reduction to Constant Coefficients

The similarities between the forms of solutions of Cauchy–Euler equations and solutions of linear equations with constant coefficients is not just a coincidence. For example, when the roots of the auxiliary equations for and are distinct and real, the respective general solutions are

and (7)

In view of the identity the second solution given in (7) can be expressed in the same form as the first solution:

where This last result illustrates the fact that any Cauchy–Euler equation can always be rewritten as a linear differential equation with constant coefficients by means of the substitution and its equivalent . The idea is to solve the new differential equation in terms of the variable t, using the methods of the previous sections, and once the general solution is obtained, resubstitute and

This method illustrated in the last example requires the use of the Chain Rule of differentiation.

EXAMPLE 6 Changing the DE to Constant Coefficients

Solve

SOLUTION

With the substitution t = ln x, it follows from the Chain Rule and Product Rule that

Substituting the foregoing derivatives in the original DE and simplifying yields

or

Since the last equation has constant coefficients, its auxiliary equation is or Thus

Then by undetermined coefficients we try a particular solution of the form This assumption leads to so and . Using we get By resubstituting and we see that the general solution of the original differential equation on the interval is ≡

See Problems 43–48 in Exercises 3.6.

Solutions for x < 0

In the preceding discussions, we have solved Cauchy–Euler equations for One way of solving a Cauchy–Euler equation for is to change the independent variable by means of the substitution (which implies and using the Chain Rule:

and

See Problems 49 and 50 in Exercises 3.6.

3.6 Exercises Answers to selected odd-numbered problems begin on page ANS-6.

In Problems 1–18, solve the given differential equation.

  1. x2y″ − 2y = 0
  2. 4x2y″ + y = 0
  3. xy″ + y′ = 0
  4. xy″ − 3y′ = 0
  5. x2y″ + xy′ + 4y = 0
  6. x2y″ + 5xy′ + 3y = 0
  7. x2y″ − 3xy′ − 2y = 0
  8. x2y″ + 3xy′ − 4y = 0
  9. 25x2y″ + 25xy′ + y = 0
  10. 4x2y″ + 4xy′ − y = 0
  11. x2y″ + 5xy′ + 4y = 0
  12. x2y″ + 8xy′ + 6y = 0
  13. 3x2y″ + 6xy′ + y = 0
  14. x2y″ − 7xy′ + 41y = 0
  15. x3y‴ − 6y = 0
  16. x3y‴ + xy′ − y = 0
  17. xy(4) + 6y‴ = 0
  18. x4y(4) + 6x3y‴ + 9x2y″ + 3xy′ + y = 0

In Problems 19–24, solve the given differential equation by variation of parameters.

  1. xy″ − 4y′ = x4
  2. 2x2y″ + 5xy′ + y = x2 − x
  3. x2y″ − xy′ + y = 2x
  4. x2y″ − 2xy′ + 2y = x4ex
  5. x2y″ + xy′ − y = ln x
  6. x2y″ + xy′ − y =

In Problems 25–30, solve the given initial-value problem. Use a graphing utility to graph the solution curve.

  1. x2y″ + 3xy′ = 0, y(1) = 0, y′(1) = 4
  2. x2y″ − 5xy′ + 8y = 0, y(2) = 32, y′(2) = 0
  3. x2y″ + xy′ + y = 0, y(1) = 1, y′(1) = 2
  4. x2y″ − 3xy′ + 4y = 0, y(1) = 5, y′(1) = 3
  5. xy″ + y′ = x,y(1) = 1, y′(1) = −
  6. x2y″ − 5xy′ + 8y = 8x6, y = 0, y′ = 0

In Problems 31 and 32, solve the given boundary-value problem.

  1. xy″ − 7xy′ + 12y = 0, y(0) = 0, y(1) = 0
  2. x2y″ − 3xy′ + 5y = 0, y(1) = 0, y(e) = 1

In Problems 33–38, find a homogeneous Cauchy–Euler differential equation whose general solution is given.

  1. y = c1x4 + c2x−2
  2. y = c1 + c2x5

In Problems 39–42, use the substitution y = (x − x0)m to solve the given equation.

  1. (x + 3)2y″ − 8(x + 3)y′ + 14y = 0
  2. (x − 1)2y″ − (x − 1)y′ + 5y = 0
  3. (x + 2)2y″ + (x + 2)y′ + y = 0
  4. (x − 4)2y″ − 5(x − 4)y′ + 9y = 0

In Problems 43–48, use the substitution t = ln x to transform the given Cauchy–Euler equation to a differential equation with constant coefficients. Solve the original equation by solving the new equation using the procedures in Sections 3.3–3.5.

  1. x2y″ + 9xy′ − 20y = 0
  2. x2y″ − 9xy′ + 25y = 0
  3. x2y″ + 10xy′ + 8y = x2
  4. x2y″ − 4xy′ + 6y = ln x2
  5. x2y″ − 3xy′ + 13y = 4 + 3x
  6. x3y‴ − 3x2y″ + 6xy′ − 6y = 3 + ln x3

In Problems 49 and 50, use the substitution t = −x to solve the given initial-value problem on the interval (−∞, 0).

  1. 4x2y″ + y = 0, y(−1) = 2, y′(−1) = 4
  2. x2y″ − 4xy′ + 6y = 0, y(−2) = 8, y′(−2) = 0

Mathematical Models

Contributed Problem

Pierre Gharghouri, Professor Emeritus, Jean-Paul Pascal, Associate Professor, Department of Mathematics Ryerson University, Toronto, Canada

  1. Temperature of a Fluid A very long cylindrical shell is formed by two concentric circular cylinders of different radii. A chemically reactive fluid fills the space between the concentric cylinders as shown in green in FIGURE 3.6.2. The inner cylinder has a radius of 1 and is thermally insulated, while the outer cylinder has a radius of 2 and is maintained at a constant temperature . The rate of heat generation in the fluid due to the chemical reactions is proportional to , where is the temperature of the fluid within the space bounded between the cylinders defined by Under these conditions the temperature of the fluid is defined by the following boundary-value problem:

    .

    Find the temperature distribution within the fluid.

    The cylindrical shell is a hollow cylinder formed by two concentric cylinders. The radius of the inner cylinder is labeled 1. The radius of the outer cylinder is labeled 2. The outer cylinder is maintained at a temperature T subscript 0.

    FIGURE 3.6.2 Cylindrical shell in Problem 51

  2. In Problem 51, find the minimum and maximum values of on the interval defined by Why do these values make intuitive sense?
  3. Bending of a Circular Plate In the analysis of the bending of a uniformly loaded circular plate, the equation w(r) of the deflection curve of the plate can be shown to satisfy the third-order differential equation

    , (8)

    where q and D are constants. Here r is the radial distance from a point on the circular plate to its center.

    1. Use the method of this section along with variation of parameters as given in (10) of Section 3.5 to find the general solution of equation (8).
    2. Find a solution of (8) that satisfies the boundary conditions

      where is the radius of the plate. [Hint: The condition is correct. Use this condition to determine one of the constants in the general solution of (8) found in part (a).]

  4. In the engineering textbook where the differential equation in Problem 53 was found, the author states that the differential equation is readily solved by integration. True, but you have to realize that equation (8) can be written in the form given next.
    1. Verify that equation (8) can be written in the alternative form

      . (9)

    2. Solve equation (9) using only integration with respect to r. Show your result is equivalent to the solution obtained in part (a) of Problem 53.

Discussion Problems

  1. Give the largest interval over which the general solution of Problem 42 is defined.
  2. Find a Cauchy–Euler differential equation of lowest order with real coefficients if it is known that 2 and 1 − i are two roots of its auxiliary equation.
  3. The initial conditions y(0) = y0, y′(0) = y1, apply to each of the following differential equations:

    For what values of y0 and y1 does each initial-value problem have a solution?

  4. What are the x-intercepts of the solution curve shown in Figure 3.6.1? How many x-intercepts are there in the interval defined by 0 < x < ?

Computer Lab Assignments

In Problems 59–62, solve the given differential equation by using a CAS to find the (approximate) roots of the auxiliary equation.

  1. 2x3y‴ − 10.98x2y″ + 8.5xy′ + 1.3y = 0
  2. x3y‴ + 4x2y″ + 5xy′ − 9y = 0
  3. x4y(4) + 6x3y‴ + 3x2y″ − 3xy′ + 4y = 0
  4. x4y(4) − 6x3y‴ + 33x2y″ − 105xy′ + 169y = 0

In Problems 63 and 64, use a CAS as an aid in computing roots of the auxiliary equation, the determinants given in (10) of Section 3.5, and integrations.

  1. x3y‴ − x2y″ − 2xy′ + 6y = x2
  2. x3y‴ + 2x2y″ − 8xy′ + 12y = x−4