Unit 4, lesson 25

Cauchy-Euler Equations

How to solve Cauchy-Euler (equidimensional) equations with a power trial or a logarithmic change of variable, how repeated and complex exponents give logarithmic factors, and why the origin is a singular point.

  • Video 25
  • Reduction of Order and Cauchy-Euler Equations

The equation x2y′′−3xy′+4y=0x^2y'' - 3xy' + 4y = 0, where y′=dy/dxy' = dy/dx, has variable coefficients, and no exponential erxe^{rx} solves it. Yet each derivative is multiplied by the matching power of xx, and that pattern points to powers of xx as solutions. This lesson shows why powers work, how the change of variable τ=ln⁡x\tau = \ln x turns the equation into one with constant coefficients, how the root cases look back in xx, and what happens at x=0x = 0, where the leading coefficient vanishes.

Key terms

Cauchy-Euler equation

Definition A second-order Cauchy-Euler equation, also called an equidimensional equation, is a linear equation

x2y′′+a xy′+b y=0x^2y'' + a\,xy' + b\,y = 0

with real constants aa and bb, where y′=dy/dxy' = dy/dx. An equation αx2y′′+βxy′+γy=0\alpha x^2y'' + \beta xy' + \gamma y = 0 with constants α≠0\alpha \ne 0, β\beta and γ\gamma takes this form after division by α\alpha. More generally, an equation of order nn is a Cauchy-Euler equation if it has the form

∑k=0nck xky(k)=0\sum_{k=0}^{n} c_k\,x^k y^{(k)} = 0

with constants ckc_k and cn≠0c_n \ne 0. The name refers to scaling. If yy solves the equation on an interval II and c≠0c \ne 0, then z(x)=y(cx)z(x) = y(cx) solves it for all xx with cxcx in II, because

xkz(k)(x)=(cx)k y(k)(cx)x^k z^{(k)}(x) = (cx)^k\,y^{(k)}(cx)

for every kk.

In plain words Each derivative comes with as many factors of xx as its order. If xx is a length, every term then has the units of yy, so measuring xx in centimeters instead of meters leaves the equation unchanged.

Example The equation x2y′′−2y=0x^2y'' - 2y = 0 is a Cauchy-Euler equation with a=0a = 0 and b=−2b = -2. Both x2x^2 and x−1x^{-1} solve it on (0,∞)(0, \infty). For the second, y′′=2x−3y'' = 2x^{-3}, so x2y′′=2x−1=2yx^2y'' = 2x^{-1} = 2y. Scaling gives nothing new here, since (cx)2=c2x2(cx)^2 = c^2x^2 and (cx)−1=c−1x−1(cx)^{-1} = c^{-1}x^{-1} are multiples of the same two solutions. The equation y′′−2y=0y'' - 2y = 0 is not equidimensional. If yy solves it, then z(x)=y(cx)z(x) = y(cx) satisfies z′′=2c2zz'' = 2c^2z, a different equation unless c=±1c = \pm 1.

Power trial solution

Definition On the interval x>0x > 0 the power xm=emln⁡xx^m = e^{m\ln x} is defined for every real number mm, and

x (xm)′=m xm,x2(xm)′′=m(m−1) xm.\begin{gathered} x\,(x^m)' = m\,x^m, \\[0.4em] x^2(x^m)'' = m(m-1)\,x^m. \end{gathered}

Substituting y=xmy = x^m into the left side of x2y′′+a xy′+b y=0x^2y'' + a\,xy' + b\,y = 0 gives (m(m−1)+am+b)xm\bigl(m(m-1) + am + b\bigr)x^m. Since xm≠0x^m \ne 0, the power xmx^m solves the equation on (0,∞)(0, \infty) exactly when mm is a root of the power characteristic equation, also called the indicial equation,

m2+(a−1) m+b=0.m^2 + (a - 1)\,m + b = 0.

If its roots m1≠m2m_1 \ne m_2 are real, the general solution on (0,∞)(0, \infty) is y=C1xm1+C2xm2y = C_1x^{m_1} + C_2x^{m_2}. The interval (0,∞)(0, \infty) is a regular interval, because the standard-form coefficients a/xa/x and b/x2b/x^2 are continuous there, and the two powers form a fundamental set, because their Wronskian (m2−m1) xm1+m2−1(m_2 - m_1)\,x^{m_1 + m_2 - 1} is never zero there.

In plain words Multiplying by xx gives back the power lost in each differentiation, so every term is a multiple of the same power xmx^m, and the equation reduces to a quadratic for the exponent. The term m(m−1)m(m-1) comes from x2y′′x^2y'', and writing m2m^2 in its place is an easy slip.

Example Solve x2y′′−2xy′−4y=0x^2y'' - 2xy' - 4y = 0 with y(1)=0y(1) = 0 and y′(1)=5y'(1) = 5. Here a=−2a = -2 and b=−4b = -4, so the exponents solve m2−3m−4=(m−4)(m+1)=0m^2 - 3m - 4 = (m - 4)(m + 1) = 0. The general solution on (0,∞)(0, \infty) is y=C1x4+C2x−1y = C_1x^4 + C_2x^{-1}. The data give C1+C2=0C_1 + C_2 = 0 and 4C1−C2=54C_1 - C_2 = 5, so C1=1C_1 = 1, C2=−1C_2 = -1 and

y=x4−1x,x>0.y = x^4 - \frac{1}{x}, \qquad x > 0.

Substituting confirms it. The x4x^4 terms give 12−8−4=012 - 8 - 4 = 0 and the x−1x^{-1} terms give −2−2+4=0-2 - 2 + 4 = 0. The solution cannot be continued to x=0x = 0, where 1/x1/x is unbounded.

Logarithmic change of variable

Definition On x>0x > 0 set τ=ln⁡x\tau = \ln x, so x=eτx = e^{\tau}, and let Y(τ)=y(eτ)Y(\tau) = y(e^{\tau}) describe the same function in the new coordinate. Write YτY_\tau and YττY_{\tau\tau} for the first and second derivatives with respect to τ\tau. The chain rule gives

Yτ=x y′,Yττ=x2y′′+x y′,\begin{gathered} Y_\tau = x\,y', \\[0.4em] Y_{\tau\tau} = x^2y'' + x\,y', \end{gathered}

with the right sides evaluated at x=eτx = e^{\tau}. So x2y′′=Yττ−Yτx^2y'' = Y_{\tau\tau} - Y_\tau, and the Cauchy-Euler equation becomes the constant-coefficient equation

Yττ+(a−1) Yτ+b Y=0.Y_{\tau\tau} + (a - 1)\,Y_\tau + b\,Y = 0.

Its characteristic equation r2+(a−1)r+b=0r^2 + (a - 1)r + b = 0 is the power characteristic equation, and a solution emτe^{m\tau} becomes emln⁡x=xme^{m\ln x} = x^m. On x<0x < 0 the same two identities hold with τ=ln⁡∣x∣\tau = \ln\lvert x \rvert, so x=−eτx = -e^{\tau} and Y(τ)=y(−eτ)Y(\tau) = y(-e^{\tau}), and the solution formulas hold with ∣x∣\lvert x \rvert in place of xx. Neither coordinate reaches x=0x = 0, so the two half-lines are solved separately. A right side g(x)g(x) in a forced equation becomes g(eτ)g(e^{\tau}) on x>0x > 0, so forced equations can be solved in τ\tau as well.

In plain words Under τ=ln⁡x\tau = \ln x, stretching xx by a factor becomes shifting τ\tau by a constant. An equation that does not change under stretching therefore becomes one that does not change under shifts, and for these linear equations that means constant coefficients. Equal steps in τ\tau are equal ratios in xx.

Example For x2y′′+xy′+y=0x^2y'' + xy' + y = 0 we have a=1a = 1 and b=1b = 1, so the new equation is Yττ+Y=0Y_{\tau\tau} + Y = 0, with solutions cos⁡τ\cos\tau and sin⁡τ\sin\tau. Back in xx, the general solution on (0,∞)(0, \infty) is

y=C1cos⁡(ln⁡x)+C2sin⁡(ln⁡x).\begin{aligned} y &= C_1\cos(\ln x) \\[0.4em] &\quad + C_2\sin(\ln x). \end{aligned}

The zeros of cos⁡τ\cos\tau are evenly spaced, π\pi apart. The zeros of cos⁡(ln⁡x)\cos(\ln x) are the points x=eπ/2+kπx = e^{\pi/2 + k\pi} for integers kk, such as 0.20790.2079, 4.81054.8105 and 111.32111.32, so each is eπ≈23.14e^{\pi} \approx 23.14 times the one before. Toward x=0x = 0 they crowd together, and every nonzero solution has infinitely many zeros in (0,1)(0, 1). The constant-coefficient root cases are explained in Second-Order Constant-Coefficient Equations.

Repeated and complex exponents

Definition Let aa and bb be real, and let m1m_1 and m2m_2 be the roots of m2+(a−1)m+b=0m^2 + (a - 1)m + b = 0. On x>0x > 0 the general solution has one of three forms.

  • Distinct real roots give y=C1xm1+C2xm2y = C_1x^{m_1} + C_2x^{m_2}.
  • A repeated root, which occurs when (a−1)2=4b(a - 1)^2 = 4b and equals m=(1−a)/2m = (1 - a)/2, gives y=xm(C1+C2ln⁡x)y = x^m(C_1 + C_2\ln x).
  • Complex roots α±iβ\alpha \pm i\beta with β>0\beta > 0 give the real general solution below.
y=C1xαcos⁡(βln⁡x)+C2xαsin⁡(βln⁡x)\begin{aligned} y &= C_1x^{\alpha}\cos(\beta\ln x) \\[0.4em] &\quad + C_2x^{\alpha}\sin(\beta\ln x) \end{aligned}

The factor ln⁡x\ln x is the factor τ\tau of the repeated-root solution τemτ\tau e^{m\tau}, and the complex case comes from the real and imaginary parts of xα+iβ=xαeiβln⁡xx^{\alpha + i\beta} = x^{\alpha}e^{i\beta\ln x}. The pairs have Wronskians (m2−m1) xm1+m2−1(m_2 - m_1)\,x^{m_1 + m_2 - 1}, x2m−1x^{2m - 1} and β x2α−1\beta\,x^{2\alpha - 1}, never zero on x>0x > 0, so each pair is a fundamental set there. On x<0x < 0 every formula holds with ∣x∣\lvert x \rvert in place of xx.

In plain words The root cases are the familiar ones for constant coefficients, written in τ=ln⁡x\tau = \ln x and translated back. A factor τ\tau becomes ln⁡x\ln x, and eατcos⁡βτe^{\alpha\tau}\cos\beta\tau becomes xαcos⁡(βln⁡x)x^{\alpha}\cos(\beta\ln x).

Example Solve x2y′′−3xy′+4y=0x^2y'' - 3xy' + 4y = 0 with y(1)=1y(1) = 1 and y′(1)=0y'(1) = 0. The exponents solve m2−4m+4=(m−2)2=0m^2 - 4m + 4 = (m - 2)^2 = 0, a repeated root m=2m = 2, so y=x2(C1+C2ln⁡x)y = x^2(C_1 + C_2\ln x) on (0,∞)(0, \infty). Then y(1)=C1=1y(1) = C_1 = 1, and the product rule gives

y′=2x(C1+C2ln⁡x)+C2 x,\begin{aligned} y' &= 2x(C_1 + C_2\ln x) \\[0.4em] &\quad + C_2\,x, \end{aligned}

so y′(1)=2C1+C2=0y'(1) = 2C_1 + C_2 = 0 and C2=−2C_2 = -2. The solution and its derivatives are

y=x2(1−2ln⁡x),y′=−4xln⁡x,y′′=−4(ln⁡x+1).\begin{gathered} y = x^2(1 - 2\ln x), \\[0.4em] y' = -4x\ln x, \\[0.4em] y'' = -4(\ln x + 1). \end{gathered}

In x2y′′−3xy′+4yx^2y'' - 3xy' + 4y the terms with x2ln⁡xx^2\ln x have coefficients −4-4, 1212 and −8-8, and the terms with x2x^2 alone have −4-4 and 44. Both groups cancel, so the equation holds on (0,∞)(0, \infty), the half-line that contains the initial point. For complex exponents take x2y′′−xy′+5y=0x^2y'' - xy' + 5y = 0. Its exponents solve m2−2m+5=0m^2 - 2m + 5 = 0, so m=1±2im = 1 \pm 2i, and the general solution on x>0x > 0 is

y=C1xcos⁡(2ln⁡x)+C2xsin⁡(2ln⁡x).\begin{aligned} y &= C_1x\cos(2\ln x) \\[0.4em] &\quad + C_2x\sin(2\ln x). \end{aligned}

Singular point at the origin

Definition The leading coefficient x2x^2 vanishes at x=0x = 0, so the equation can be divided by it only for x≠0x \ne 0. In standard form

y′′+ax y′+bx2 y=0,y'' + \frac{a}{x}\,y' + \frac{b}{x^2}\,y = 0,

the coefficients are continuous on (−∞,0)(-\infty, 0) and on (0,∞)(0, \infty). Each half-line is a regular interval, so data y(x0)=y0y(x_0) = y_0 and y′(x0)=y1y'(x_0) = y_1 with x0≠0x_0 \ne 0 determine exactly one solution on the half-line that contains x0x_0. When aa and bb are not both zero, at least one coefficient is unbounded near 00, and the origin is a singular point. The existence and uniqueness theorem does not apply on any interval that contains 00. A solution found on a half-line extends across 00 only if the extended function is twice differentiable at 00 and satisfies the original equation there, and data given at x=0x = 0 may have no solution, exactly one, or infinitely many. In the language of series solutions this singular point is a regular singular point, the kind treated in Regular Singular Points and Frobenius.

In plain words The power formulas live on one side of the origin. Whether a solution can cross x=0x = 0, and whether data at x=0x = 0 pick out one solution, is decided by the original equation, not by the formula.

Example The solution y=x2(1−2ln⁡x)y = x^2(1 - 2\ln x) found above has y→0y \to 0 and y′=−4xln⁡x→0y' = -4x\ln x \to 0 as x→0+x \to 0^+, so it seems to arrive smoothly at the origin. Any differentiable extension therefore has y(0)=0y(0) = 0 and y′(0)=0y'(0) = 0. But the difference quotient

y′(h)−y′(0)h=−4ln⁡h\frac{y'(h) - y'(0)}{h} = -4\ln h

grows without bound as h→0+h \to 0^+, so no extension has a second derivative at 00, in line with y′′=−4(ln⁡x+1)→∞y'' = -4(\ln x + 1) \to \infty. Uniqueness can fail as well. The solutions of x2y′′−2y=0x^2y'' - 2y = 0 on x>0x > 0 are C1x2+C2x−1C_1x^2 + C_2x^{-1}, and on x<0x < 0 they have the same form. A solution on an interval containing 00 is bounded near 00, so C2=0C_2 = 0 on each side, and the existence of y′′(0)y''(0) forces the same C1C_1 on both sides. The solutions on an interval containing 00 are therefore exactly the functions y=Cx2y = Cx^2. Every one of them has y(0)=0y(0) = 0 and y′(0)=0y'(0) = 0, so these data have infinitely many solutions, while data with y(0)=1y(0) = 1 have none.

Check yourself

  1. Find the general solution of x2y′′+5xy′+4y=0x^2y'' + 5xy' + 4y = 0 on x>0x > 0 and on x<0x < 0.
Show the answer

The exponents solve m(m−1)+5m+4=m2+4m+4=(m+2)2=0m(m - 1) + 5m + 4 = m^2 + 4m + 4 = (m + 2)^2 = 0, a repeated root m=−2m = -2. On x>0x > 0 the general solution is

y=x−2(C1+C2ln⁡x).y = x^{-2}(C_1 + C_2\ln x).

On x<0x < 0 replace xx by ∣x∣\lvert x \rvert. Since ∣x∣−2=x−2\lvert x \rvert^{-2} = x^{-2}, this gives y=x−2(C1+C2ln⁡∣x∣)y = x^{-2}(C_1 + C_2\ln\lvert x \rvert), with constants that are independent of those on x>0x > 0. Every nonzero solution is unbounded near 00, so none of them crosses the origin.

  1. A classmate substitutes y=xmy = x^m into x2y′′+3xy′−3y=0x^2y'' + 3xy' - 3y = 0 and gets m2+3m−3=0m^2 + 3m - 3 = 0. Find the error, show that a root of this equation does not give a solution, and find the general solution on x>0x > 0.
Show the answer

The term x2y′′x^2y'' contributes m(m−1)xmm(m - 1)x^m, not m2xmm^2x^m. The correct equation is

m(m−1)+3m−3=m2+2m−3=0,\begin{aligned} &m(m - 1) + 3m - 3 \\[0.4em] &\quad = m^2 + 2m - 3 = 0, \end{aligned}

which factors as (m+3)(m−1)=0(m + 3)(m - 1) = 0. So y=C1x+C2x−3y = C_1x + C_2x^{-3} on x>0x > 0. For x−3x^{-3}, for example, the three terms are 12x−312x^{-3}, −9x−3-9x^{-3} and −3x−3-3x^{-3}, which add to zero. If instead mm is a root of the classmate’s equation, substitution gives (m2+2m−3)xm=−m xm(m^2 + 2m - 3)x^m = -m\,x^m, which is not zero, because the roots m=(−3±21)/2m = (-3 \pm \sqrt{21})/2 are not zero.

  1. Show that the problem x2y′′−2xy′+2y=0x^2y'' - 2xy' + 2y = 0, y(0)=0y(0) = 0, y′(0)=1y'(0) = 1 has infinitely many solutions on (−1,1)(-1, 1), and that with y(0)=1y(0) = 1 instead it has none. Why does this not contradict the existence and uniqueness theorem?
Show the answer

The exponents solve m2−3m+2=(m−1)(m−2)=0m^2 - 3m + 2 = (m - 1)(m - 2) = 0, so xx and x2x^2 are solutions. Both are polynomials, so y=x+Cx2y = x + Cx^2 is defined on all of R\mathbb{R}, and with y′=1+2Cxy' = 1 + 2Cx and y′′=2Cy'' = 2C

x2(2C)−2x(1+2Cx)+2(x+Cx2)=0,\begin{aligned} &x^2(2C) - 2x(1 + 2Cx) \\[0.4em] &\quad + 2(x + Cx^2) = 0, \end{aligned}

since the x2x^2 terms give 2C−4C+2C=02C - 4C + 2C = 0 and the xx terms give −2+2=0-2 + 2 = 0. Every CC gives y(0)=0y(0) = 0 and y′(0)=1y'(0) = 1. On each half-line the general solution is C1x+C2x2C_1x + C_2x^2, which tends to 00 as x→0x \to 0, so every solution on (−1,1)(-1, 1) has y(0)=0y(0) = 0 by continuity, and the data y(0)=1y(0) = 1 cannot be met. There is no contradiction, because the theorem needs the standard form

y′′−2x y′+2x2 y=0y'' - \frac{2}{x}\,y' + \frac{2}{x^2}\,y = 0

with coefficients continuous on an interval containing the initial point, and these coefficients are not defined at x=0x = 0. For data given at a point x0≠0x_0 \ne 0 the theorem does apply on the half-line containing x0x_0.

How this lesson connects