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.
The equation x2y′′−3xy′+4y=0, where y′=dy/dx, has variable coefficients, and no exponential erx solves it. Yet each derivative is multiplied by the matching power of x, and that pattern points to powers of x as solutions. This lesson shows why powers work, how the change of variable τ=lnx turns the equation into one with constant coefficients, how the root cases look back in x, and what happens at x=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′′+axy′+by=0
with real constants a and b, where y′=dy/dx. An equation αx2y′′+βxy′+γy=0 with constants α=0, β and γ takes this form after division by α. More generally, an equation of order n is a Cauchy-Euler equation if it has the form
k=0∑nckxky(k)=0
with constants ck and cn=0. The name refers to scaling. If y solves the equation on an interval I and c=0, then z(x)=y(cx) solves it for all x with cx in I, because
xkz(k)(x)=(cx)ky(k)(cx)
for every k.
In plain words Each derivative comes with as many factors of x as its order. If x is a length, every term then has the units of y, so measuring x in centimeters instead of meters leaves the equation unchanged.
Example The equation x2y′′−2y=0 is a Cauchy-Euler equation with a=0 and b=−2. Both x2 and x−1 solve it on (0,∞). For the second, y′′=2x−3, so x2y′′=2x−1=2y. Scaling gives nothing new here, since (cx)2=c2x2 and (cx)−1=c−1x−1 are multiples of the same two solutions. The equation y′′−2y=0 is not equidimensional. If y solves it, then z(x)=y(cx) satisfies z′′=2c2z, a different equation unless c=±1.
Power trial solution
Definition On the interval x>0 the power xm=emlnx is defined for every real number m, and
x(xm)′=mxm,x2(xm)′′=m(m−1)xm.
Substituting y=xm into the left side of x2y′′+axy′+by=0 gives (m(m−1)+am+b)xm. Since xm=0, the power xm solves the equation on (0,∞) exactly when m is a root of the power characteristic equation, also called the indicial equation,
m2+(a−1)m+b=0.
If its roots m1=m2 are real, the general solution on (0,∞) is y=C1xm1+C2xm2. The interval (0,∞) is a regular interval, because the standard-form coefficients a/x and b/x2 are continuous there, and the two powers form a fundamental set, because their Wronskian (m2−m1)xm1+m2−1 is never zero there.
In plain words Multiplying by x gives back the power lost in each differentiation, so every term is a multiple of the same power xm, and the equation reduces to a quadratic for the exponent. The term m(m−1) comes from x2y′′, and writing m2 in its place is an easy slip.
Example Solve x2y′′−2xy′−4y=0 with y(1)=0 and y′(1)=5. Here a=−2 and b=−4, so the exponents solve m2−3m−4=(m−4)(m+1)=0. The general solution on (0,∞) is y=C1x4+C2x−1. The data give C1+C2=0 and 4C1−C2=5, so C1=1, C2=−1 and
y=x4−x1,x>0.
Substituting confirms it. The x4 terms give 12−8−4=0 and the x−1 terms give −2−2+4=0. The solution cannot be continued to x=0, where 1/x is unbounded.
Logarithmic change of variable
Definition On x>0 set τ=lnx, so x=eτ, and let Y(τ)=y(eτ) describe the same function in the new coordinate. Write Yτ and Yττ for the first and second derivatives with respect to τ. The chain rule gives
Yτ=xy′,Yττ=x2y′′+xy′,
with the right sides evaluated at x=eτ. So x2y′′=Yττ−Yτ, and the Cauchy-Euler equation becomes the constant-coefficient equation
Yττ+(a−1)Yτ+bY=0.
Its characteristic equation r2+(a−1)r+b=0 is the power characteristic equation, and a solution emτ becomes emlnx=xm. On x<0 the same two identities hold with τ=ln∣x∣, so x=−eτ and Y(τ)=y(−eτ), and the solution formulas hold with ∣x∣ in place of x. Neither coordinate reaches x=0, so the two half-lines are solved separately. A right side g(x) in a forced equation becomes g(eτ) on x>0, so forced equations can be solved in τ as well.
In plain words Under τ=lnx, stretching x by a factor becomes shifting τ 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 are equal ratios in x.
Example For x2y′′+xy′+y=0 we have a=1 and b=1, so the new equation is Yττ+Y=0, with solutions cosτ and sinτ. Back in x, the general solution on (0,∞) is
y=C1cos(lnx)+C2sin(lnx).
The zeros of cosτ are evenly spaced, π apart. The zeros of cos(lnx) are the points x=eπ/2+kπ for integers k, such as 0.2079, 4.8105 and 111.32, so each is eπ≈23.14 times the one before. Toward x=0 they crowd together, and every nonzero solution has infinitely many zeros in (0,1). The constant-coefficient root cases are explained in Second-Order Constant-Coefficient Equations.
Repeated and complex exponents
Definition Let a and b be real, and let m1 and m2 be the roots of m2+(a−1)m+b=0. On x>0 the general solution has one of three forms.
Distinct real roots give y=C1xm1+C2xm2.
A repeated root, which occurs when (a−1)2=4b and equals m=(1−a)/2, gives y=xm(C1+C2lnx).
Complex roots α±iβ with β>0 give the real general solution below.
y=C1xαcos(βlnx)+C2xαsin(βlnx)
The factor lnx is the factor τ of the repeated-root solution τemτ, and the complex case comes from the real and imaginary parts of xα+iβ=xαeiβlnx. The pairs have Wronskians (m2−m1)xm1+m2−1, x2m−1 and βx2α−1, never zero on x>0, so each pair is a fundamental set there. On x<0 every formula holds with ∣x∣ in place of x.
In plain words The root cases are the familiar ones for constant coefficients, written in τ=lnx and translated back. A factor τ becomes lnx, and eατcosβτ becomes xαcos(βlnx).
Example Solve x2y′′−3xy′+4y=0 with y(1)=1 and y′(1)=0. The exponents solve m2−4m+4=(m−2)2=0, a repeated root m=2, so y=x2(C1+C2lnx) on (0,∞). Then y(1)=C1=1, and the product rule gives
y′=2x(C1+C2lnx)+C2x,
so y′(1)=2C1+C2=0 and C2=−2. The solution and its derivatives are
y=x2(1−2lnx),y′=−4xlnx,y′′=−4(lnx+1).
In x2y′′−3xy′+4y the terms with x2lnx have coefficients −4, 12 and −8, and the terms with x2 alone have −4 and 4. Both groups cancel, so the equation holds on (0,∞), the half-line that contains the initial point. For complex exponents take x2y′′−xy′+5y=0. Its exponents solve m2−2m+5=0, so m=1±2i, and the general solution on x>0 is
y=C1xcos(2lnx)+C2xsin(2lnx).
Singular point at the origin
Definition The leading coefficient x2 vanishes at x=0, so the equation can be divided by it only for x=0. In standard form
y′′+xay′+x2by=0,
the coefficients are continuous on (−∞,0) and on (0,∞). Each half-line is a regular interval, so data y(x0)=y0 and y′(x0)=y1 with x0=0 determine exactly one solution on the half-line that contains x0. When a and b are not both zero, at least one coefficient is unbounded near 0, and the origin is a singular point. The existence and uniqueness theorem does not apply on any interval that contains 0. A solution found on a half-line extends across 0 only if the extended function is twice differentiable at 0 and satisfies the original equation there, and data given at x=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=0, and whether data at x=0 pick out one solution, is decided by the original equation, not by the formula.
Example The solution y=x2(1−2lnx) found above has y→0 and y′=−4xlnx→0 as x→0+, so it seems to arrive smoothly at the origin. Any differentiable extension therefore has y(0)=0 and y′(0)=0. But the difference quotient
hy′(h)−y′(0)=−4lnh
grows without bound as h→0+, so no extension has a second derivative at 0, in line with y′′=−4(lnx+1)→∞. Uniqueness can fail as well. The solutions of x2y′′−2y=0 on x>0 are C1x2+C2x−1, and on x<0 they have the same form. A solution on an interval containing 0 is bounded near 0, so C2=0 on each side, and the existence of y′′(0) forces the same C1 on both sides. The solutions on an interval containing 0 are therefore exactly the functions y=Cx2. Every one of them has y(0)=0 and y′(0)=0, so these data have infinitely many solutions, while data with y(0)=1 have none.
Check yourself
Find the general solution of x2y′′+5xy′+4y=0 on x>0 and on x<0.
Show the answer
The exponents solve m(m−1)+5m+4=m2+4m+4=(m+2)2=0, a repeated root m=−2. On x>0 the general solution is
y=x−2(C1+C2lnx).
On x<0 replace x by ∣x∣. Since ∣x∣−2=x−2, this gives y=x−2(C1+C2ln∣x∣), with constants that are independent of those on x>0. Every nonzero solution is unbounded near 0, so none of them crosses the origin.
A classmate substitutes y=xm into x2y′′+3xy′−3y=0 and gets m2+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>0.
Show the answer
The term x2y′′ contributes m(m−1)xm, not m2xm. The correct equation is
m(m−1)+3m−3=m2+2m−3=0,
which factors as (m+3)(m−1)=0. So y=C1x+C2x−3 on x>0. For x−3, for example, the three terms are 12x−3, −9x−3 and −3x−3, which add to zero. If instead m is a root of the classmate’s equation, substitution gives (m2+2m−3)xm=−mxm, which is not zero, because the roots m=(−3±21)/2 are not zero.
Show that the problem x2y′′−2xy′+2y=0, y(0)=0, y′(0)=1 has infinitely many solutions on (−1,1), and that with y(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)=0, so x and x2 are solutions. Both are polynomials, so y=x+Cx2 is defined on all of R, and with y′=1+2Cx and y′′=2C
x2(2C)−2x(1+2Cx)+2(x+Cx2)=0,
since the x2 terms give 2C−4C+2C=0 and the x terms give −2+2=0. Every C gives y(0)=0 and y′(0)=1. On each half-line the general solution is C1x+C2x2, which tends to 0 as x→0, so every solution on (−1,1) has y(0)=0 by continuity, and the data y(0)=1 cannot be met. There is no contradiction, because the theorem needs the standard form
y′′−x2y′+x22y=0
with coefficients continuous on an interval containing the initial point, and these coefficients are not defined at x=0. For data given at a point x0=0 the theorem does apply on the half-line containing x0.
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