The δ Operator as Applied toSeries Solutions of Differential Equations

J.B. McLeodDepartment of Mathematics

University of Pittsburgh, Pittsburgh, PA 15260

1 The Frobenius Method and Its Disadvan-

tage

The usual method of solving a differential equation in series is that called byPiaggio the method of Frobenius. To see its advantages, let us start by workingout a simple example by this method.

Let us solve

(2x+ x3)d2y

dx2− dy

dx− 6xy = 0.

We try as a solution

y =∞∑

n=0

anxc+n,

where c is not necessarily an integer and is so chosen that xc is truly the firstterm in the series, so that a0 6= 0. Formally,

dy

dx=∞∑

n=0

an(c+ n)xc+n−1,

d2y

dx2=∞∑

n=0

an(c+ n)(c+ n− 1)xc+n−2.

Substituting this in the differential equation, we have

∞∑n=0

an{2(c+n)(c+n−1)xc+n−1+(c+n)(c+n−1)xc+n+1−(c+n)xc+n−1−6xc+n+1} = 0,

i.e.,∑

an{(c+ n)[2(c+ n)− 3]xc+n−1 + [(c+ n)(c+ n− 1)− 6]xc+n+1} = 0.

1

For the last equation to hold, the coefficients of all powers of x must be zero.The lowest power is xc−1, and the coefficient of this gives

a0c(2c− 3) = 0,

which is called the Indicial Equation. Since a0 6= 0, this yields

c = 0 or c =3

2.

The coefficient of xc gives

a1(c+ 1){2(c+ 1)− 3} = 0,

so that

a1 = 0 or c = −1 or c =1

2.

The values of c contradict those given by the Indicial Equation, and so wechoose a1 = 0.

For higher powers, the coefficient of xc+m−1 gives

am(c+m)(2c+ 2m− 3) + am−2[(c+m− 2)(c+m− 3)− 6] = 0, (1)

so that every coefficient is some multiple of the coefficient two before it. Buta1 = 0, and so a3 = a5 = −−− = 0. And

a2n = −(c+ 2n− 2)(c+ 2n− 3)− 6

(c+ 2n)(2c+ 4n− 3)a2n−2

= −(c+ 2n)(c+ 2n− 5)

(c+ 2n− 5)(c+ 2n)(2c+ 4n− 3) a2n−2

= − c+ 2n− 5

2c+ 4n− 3a2n−2

since c+ 2n 6= 0 b y the Indicial Equation.

If c = 0, this gives

a2n = −2n−54n−3

a2n−2

= (−1)2 (2n−5)(2n−2−5)(4n−3)(4n−4−3)

a2n−4

= · · · = (−1)n (2n−5)(2n−2−5)(4n−3)(4n−4−3)

a2n−4

so that

y = a0{1 + 3x2 + 35x4 − 1

15x6 + · · ·+ (−1)n (2n−5)(2n−7)···(−1)(−3)

(4n−3)(4n−7)···5.1x2n + · · ·}

= a0y0,

2

say, where y0, is a particular solution and a0 may be regarded as an arbitraryconstant.

If c = 32, we obtain similarly y = a0y1, say, and the general solution of theequation is y = Ay0 +By1, for arbitrary constants A,B.

This whole process is formal. It is justified inside the circle of convergenceof the resulting infinite series, for we know that a power series can be differen-tiated term-by-term as often as we please inside its circle of convergence. It istherefore necessary finally to investigate the convergence of the infinite series,and this is best done by the Ratio Test. It is easily verified that for both y0, y1

the radius of convergence is√

2.

This is the so-called Frobenius Method. There are two major disadvantagesin its use. In the first place, the algebra, even in as simple an example as theone above, is inclined to be heavy, and there is certainly no suggestion thatthe solution can be written down at sight. In the second place, the methodmay break down completely either because the recurrence relationship in thecoefficients an is not two-term, as (1) is , and so becomes unmanageable, orbecause the resulting power series is not convergent, and unfortunately theoriginal differential equation gives no indication of whether such a break-downwill occur, so that we may find after some heavy algebra that all our work iswasted.

It is however possible to transform the differential equation so as to enableus to decide at once whether a series solution can be obtained or not, andwhat its range of validity will be; and if there is a valid series solution, we areenabled to write it down at sight.

The necessary transformation depends on the δ operator.

2 The δ Operator and its Properties

The usual differential operator is D = ddx. We know from experience that it isparticularly suited to dealing with exponentials, i.e., powers of ex, as in linearequations with constant coefficients. Since we are now interested in powers ofx, it is suggested that the relevant operator be

d

d(log x)=

d1xdx

= xd

dx,

and so we write

δ ≡ xd

dx.

3

Property I.δxa = axa.

Proof.

δxa = xd

dxxa = x · axa−1 = axa.

Property II. If f(δ) is a polynomial in δ, then

f(δ)xa = f(a)xa.

Proof. This follows by repeated application of Prop. I.(There is of course a strong resemblance to properties of D when operating onexponentials.)Property III.

xnDn = δ(δ − 1)(δ − 2) · · · (δ − n+ 1).

(It is this property which enables us to transform a differential equation in Dto one in δ.)

Proof. The proof is induction. It is true for n = 1 by definition of δ.Suppose it is true for n = r, so that xrDr = δ(δ − 1) · · · (δ − r + 1). Then

Dr+1 = D{x−rδ(δ − 1) · · · (δ − r + 1)}= −rx−(r+1)δ(δ − 1) · · · (δ − r + 1) + x−rDδ(δ − 1) · · · (δ − r + 1)= x−(r+1)(−r)δ(δ − 1) · · · (δ − r + 1) + x−(r+1)δδ(δ − 1) · · · (δ − r + 1)= x−(r+1)(δ − r)(δ)(δ − 1) · · · (δ − r + 1)= x−(r+1)δ(δ − 1) · · · (δ − r + 1)(δ − r),

since plainly δ − r must commute with a polynomial in δ. This completes theproof.

3 A Formal Solution using δ.

Let us return to the example

(2x+ x3)d2y

dx2− dy

dx− 6xy = 0.

To transform this equation, we multiply by x and then use Prop. III above,when the equation becomes:

{(2 + x2)δ(δ − 1)− δ − 6x2)y = 0.

Taking the terms in x to one side, this is

{2δ(δ − 1)− δ}y = x2δ(δ − 1) + 6}y,δ(2δ − 3)y = x2(−δ + 3)(δ + 2)y,

F (δ)y = x2G(δ)y,

4

say.

I now claim that, having made this simple transformation on the differentialequation, all the information we require about it can be written down at sight.The truth of the information we obtain can be checked with the example asworked out in †1 above, although of course a more general justification will begiven later.

First, the indicial equation is just F (c) = 0, i.e.,

c(2c− 3) = 0,c = 0 or c = 3

2,

which checks with †1.

Secondly, for either of these values of c, the corresponding series solution is

xc +G(c)

F (c+ 2)xc+2 +

G(c)G(c+ 2)

F (c+ 2)F (c+ 4)xc+4 + · · ·+

+G(c)G(c+ 2) · · ·G(c+ 2n− 2)

F (c+ 2)F (c+ 4) · · ·F (c+ 2n)xc+2n + · · · ,

and that this is the same as the series in †1 can be checked by noting that theratio of two successive coefficients is

G(c+ 2n− 2)

F (c+ 2n)= −(c+ 2n− 2− 3)(c+ 2n− 2 + 2)

(c+ 2n)[2(c+ 2n)− 3]= − c+ 2n− 5

2c+ 4n− 3,

corresponding to †1. (It will be noticed that there is an increment of 2 in theseries, i.e., we have xc, xc+2, xc+4, · · ·; this is due to the presence of x2 in thedifferential equation in terms of †.)

Finally, the radius of convergence of this series is given by

| x2 |<| limc→∞

F (c)

G(c)|= 2,

again checking with †1.

4 The Justification of the δ Method.

The δ method applies to any differential equation which can be cast into theform

F (δ)y = xhG(δ)y, (2)

where F,G are polynomials in δ and h is some constant, and this is the mostgeneral form we shall consider.

5

Suppose a satisfies F (z) = 0.

Guided by what we saw †3, we try y = xa + y1 as a solution of (2). Then

F (δ){xa + y1} = xhG(δ){xa + y1},

andF (δ)xa = F (a)xa = 0.

HenceF (δ)y1 = xhG(a)xa + xhG(δ)y1.

Now y1 is in the nature of an error term, so that we certainly hope that thethird term in the last equation is negligible compared with the other two. Andif we neglect the third term, we can satisfy the first two by choosing

y1 =G(a)

F (a+ h)xa+h,

for

F (δ) { G(a)

F (a+ h)xa+h} = G(a)xa+h.

We therefore try

y1 =G(a)

F (a+ h)xa+h + y2,

and obtain that

F (δ)y2 =G(a)G(a+ h)

F (a+ h)xa+2h + xhG(δ)y2.

This similarly leads to trying

y2 =G(a)G(a+ h)

F (a+ h)F (a+ 2h)xa+2h + y3,

and so on, so that we become morally certain that the solution is

xa +G(a)

F (a+ h)xa+h + · · ·+ G(a)G(a+ h) · · ·G[a+ (n− 1)h]

F (a+ h)F (a+ 2h) · · ·F (a+ nh)xa+nh + · · · .

This will in fact be a solution inside the circle of convergence of the series, forit can then be differentiated term-by-term and substituted in the differentialequation (in its δ form) and the terms will cancel on either side of the equationfrom the very way in which we have built up the solution.

Thus, granted convergence, which we must now discuss, we have a seriessolution for each root of F (z) = 0 in ascending powers of xh (which may ofcourse mean descending powers of x in h is negative.)

6

We shall assume until further notice that F (z) = 0 does not have any equalroots, since this would imply that two solutions which we would normallyexpect to be linearly independent become the same. And we shall also assumethat, for an root a, it is impossible that F (a+nh) = 0 for some positive integern, since this would imply the occurrence of zeros in the denominators of theterms of the infinite series. These difficulties will be dealt with later.

5 The Convergence of the δ Solution

We shall assume that h > 0. This is no restriction, because if h < 0, then werewrite the equation as

G(δ)y = x−hF (δ)y,

and now −h > 0.

The ratio of two successive terms is

G[a+ (n− 1)h]

F (a+ nh)xh,

so that the series is convergent for

|xh| < | limn→∞

F (a+ nh)

G[a+ (n− 1)h]|.

Suppose F (δ) is of degree q, with leading term f0δp, and that G(δ) is of degree

q, with leading term g0δq. Then

limn→∞

F (a+ nh)

G[a+ (n− 1)h]= lim

n→∞

fnhp

0

g0(nh)q= lim

n→∞

f0

g0

(nh)p−q.

The subsequent argument depends on whether p><q.

p > q: Then the series is convergent for |xh| < ∞, i.e., everywhere. We thushave p solutions of the differential equation, corresponding to the p solutionsof F (z) = 0, and this is as we expect since the differential equation is of orderp, since p > q; and every solution converges everywhere.

p < q: Then the series is convergent for |xh| < 0, i.e., nowhere, except possiblyand trivially at x = 0 itself. We must now consider the equation in the form

G(δ)y = x−hF (δ)y. (3)

if b is a solution of G(z) = 0, then a solution of (3), granted convergence, is

xb +F (b)

G(b− h)xb−h +

F (b)F (b− h)

G(b− h)G(b− 2h)xb−2h + · · · ,

7

the ratio of two successive terms being

F [b− (n− 1)h]

G(b− nh)x−h,

so that the series converges for

|x−h| < | limn→∞

G(a− nh)

F [a− (n− 1)h]| =∞,

i.e., for |x| > 0, i.e., for all x except x = 0.

We thus have q solutions (in descending powers of x) of the differentialequation, corresponding to the q solutions of G(z) = 0, and this is as weexpect since the differential equation is of order q; and again every solutionconverges everywhere except at x = 0.

p = q: In this case

| limn→∞

F (a+ nh)

G[a+ (n− 1)h]| = |f0

g0

| = `,

say, and we have p solutions (in ascending powers of x), derived from the proots of F (z) = 0, each being valid for |xh| < `; and we have q(= p) solutions(in descending powers of x), derived from the q roots of G(z) = 0, each beingvalid for |xh| > `.

To sum up, then except on one special circle (which may, in particularcases, have zero or infinite radius), we can find the correct number of solutionsto suit the equation, and the equation may be regarded as solved. In practice,the exceptional circle is one on which singularities of the solution occur, sothat we could not possibly expect to find valid power series expressions there.

6 An Example

x2 d2y

dx2+ x

dy

dx+ (x2 − v2)y = 0.

This is Bessel’s equation of order v. It transforms to

{δ(δ − 1) + δ + x2 − v2}y = 0,

i.e.,(δ2 − v2)y = −x2y,

8

so thatF (δ) ≡ δ2 − vz = (δ − v)(δ + v), G(δ) ≡ −1.

The roots of the indicial equation are ±v, and the ascending solutions areconvergent, everywhere since F is of a higher degree than G. The solutioncorresponding to the root +v is

xv +(−1)

2(2v + 2)xv+2 +

(−1)2

2 · 4 · (2v + 2)(2v + 4)xv+4 + · · ·+

+(−1)n

2 · 4 · · · · · 2n · (2v + 2) · · · (2v + 2n)xv+2n + · · · ,

while the other is given by replacing v with −v. These are constant multiplesof the Bessel functions Jv, J−v.

It should be noted that we must suppose v non-integral. For, if v = 0, wehave equal roots; while if v is any other integer, we have the difficulty that oneof the two series will have zeros in the denominators of its term.

7 The Case of Equal Roots

Let us write

S(a) ≡ xa +G(a)

F (a+ h)xa+h +

G(a)G(a+ h)

F (a+ h)F (a+ 2h)xa+2h + · · · ,

giving the solution of the differential equation corresponding to a solution a ofF (z) = 0. if a is a double root of F (z) = 0, we can consider this as the limitas ε→ 0 of the case where a, a+ ε are roots of F (z) = 0. Then S(a), S(a+ ε)are solutions of the differential equation, and so also therefore is

S(a+ ε)− S(a)

ε

Letting ε → 0, this suggests that, if F (z) = 0 has a double root a, thenS(a), S ′(a) are solutions of the differential equation.

The rigorous proof of this comes in two parts. In (a), we show that theinfinite series for S ′(a) has the same radius of convergence as that for S(a),and this, especially at a first reading, may well be omitted. In (b), we showthat S ′(a) is truly a solution of the differential equation.

(a) We write

S(a) ≡∞∑

n=0

Cnxa+nh

9

and we shall suppose, merely for simplicity in writing, that h = 1. Then

S ′(a) = log x S(a) +∞∑

n=0

∂Cn

∂axa+n (4)

provided that the second term on the right-hand side is convergent uniformlyin a. In fact we shall show that the limit as n→∞ of the ratio of successiveterms in this series is the same as that for S(a), and, as we have already seen,this is independent of a; and it will be clear from the various proofs that allthis is so uniformly in a. Hence we will have proved that the last series in (4)is convergent uniformly in a for x inside the circle of convergence of S(a), andso S ′(a) has the same circle of convergence as S(a).

To prove that the limit as n → ∞ of the ratio of successive terms in thelast series in (4) is the same as for S(a), we note that

Cn =G(a) · · ·G(a+ n− 1)

F (a+ 1) · · ·G(a+ n),

so that, differentiating logarithmically, we have

1

Cn

∂Cn

∂a=

n−1∑v=0

G′(a+ v)

G(a+ v)−

n∑v=1

F ′(a+ v)

F (a+ v)

SupposeF (t) ≡ f0(t− d1) · · · (t− dp),G(t) ≡ g0(t− e1) · · · (t− eq)

so thatF ′(t)

F (t)=

p∑r=1

1

t− dr

,G′(t)

G(t)=

q∑r=1

1

t− er

,

1

Cn

∂Cn

∂a=

q∑r=1

n−1∑v=0

1

a+ v − er

−p∑

r=1

n∑v=1

1

a+ v − dr

.

Now,n−1∑v=0

1

a+ v − er

− log n→ Er as n→∞,

for some constant Er, and

n∑v=1

1

a+ v − dr

− log n→ Dr,

and so1

Cn

∂Cn

∂n− (q − p) log n→ A,

10

say. Then

limn→∞

∂Cn

∂a∂Cn−1

∂a

= limn→∞

Cn

Cn−1

× (q − p) log n+ A

(q − p) log (n− 1) + A

= limn→∞

Cn

Cn−1

,

which shows that the required limits of ratios of successive terms are the same.

(b) We can regard (a) as having dismissed questions of convergence, andthese will not be mentioned again. We have

F (a+ n)Cn = G(a+ n− 1)Cn−1,

so that

F ′(a+ n)Cn + F (a+ n)∂Cn

∂a

= G′(a+ n− 1)Cn−1 +G(a+ n− 1)∂Cn−1

∂a. (5)

Then

{F (δ)− xG(δ)}S ′(a)

= {F (δ)− xG(δ)} log x S(a) +∞∑

n=0

∂Cn

∂axa+n = log x {Fδ)− xG(δ)}S(a) + {F ′(δ)− xG′(δ)}S(a)+

+ {F (δ)− xG(δ)}∞∑

n=0

∂Cn

∂axa+n, (6)

the proof of this step being given below in (c),

= 0 + {F ′(δ)− xG′(δ)}S(a) + {F (δ)− xG(δ)}∞∑

n=0

∂Cn

∂axa+n,

since S(a) is a solution of the differential equation.

To prove that the two remaining terms in the last line combine to give zero,we consider coefficients of the various powers of x. The coefficient of xa+n is

F ′(a+ n)Cn −G′(a+ n− 1)Cn−1 + F (a+ n)∂Cn

∂a−G(a+ n− 1)

∂Cn−1

∂a,

which vanishes by (5). The coefficient of xa is exceptional, being just

F ′(a)C0 + F (a) ∂C0

∂a

= 0 because F (a) = F ′(a) = 0.

11

(c) The step in (6) above is a particular case of the property

F (δ){log x · V } = log x F (δ)V + F ′(δ)V,

where V is some function of x.

It will be sufficient to prove this for F (δ) ≡ δn, where n is a positive integer.Then

δ ≡ d

d(log x),

so thatδn{log x · v} = dn

{d(log x)}n + n d(log x)d(log x)

· dn−1y{d(log x)}n−1 ,

applying Leibniz’s formula with log x as the variable in place of the usual x,

= log xδnv + nδn−1v,

as required.

8 An Example

x2 d2y

dx2− 3x

dy

dx+ (4− x

10y = 0.

This transforms into

{δ(δ − 1)− 3δ + 4− x

10})y = 0,

(δ − 2)2y =x

10y.

Since F (δ) is of higher degree than G(δ), the ascending series are convergenteverywhere, but the indicial equation has the repeated root 2. The series S(a)is this particular example is

xa +110xa+1

(a− 1)2+

( 110

)nxa+n

(a− 1)2a2 · · · (a+ n− 2)2+ · · · ,

and the first solution is obtained by putting a = 2 in this.

The second solution is obtained by differentiating S(a) with respect to aand then putting a = 2 again. We have

S ′(a) = log x S(a)+∞∑

n=1

( 110

)nxa+n

(a− 1)2a2 · · · (a+ n− 2)2(−2){ 1

a− 1+

1

a+· · ·+ 1

a+ n− 2},

and so the second solution is given by putting a = 2 in this.

It should be remarked that if the indicial equation has a root a repeatedthree times, then S(a), S ′(a), S ′′(a) will all be solutions, and so on for highermultiplicities.

12

9 The Case Where Coefficients in the Series

Solution become Infinite

This case occurs, assuming again for simplicity that h = 1, if a is a root ofF (z) = 0 and F (a + m) = 0 for some positive integer m; for this will meanthat the coefficient Cm in S(a), and also all subsequent coefficients, becomeinfinite. S(a) thus becomes meaningless.

We can then consider as a solution of the differential equation, not S(a),but [F (t + m)S(t)]t=a. This will be a solution, because [F (t + m)]t=a is aconstant as far as x is concerned, and the awkward infinite coefficients areremoved. But, in this solution, all coefficients of lesser powers x than xa+m

disappear, because they have a factor F (a+m), and so the series starts witha term in xa+m and is in fact just a multiple of the series for the root a+m ofF (z) = 0.

Two normally independent solutions have thus become identified, and wehave to look for another independent solution. To do this, we substitute F (t+m) S(t) in the differential equation, obtaining

F (t+m){F (δ)−xG(δ)}∑

Cnxt+n = F (t+m)

∑{F (t+n)Cnx

t+n−xG(t+n)Cnxt+n}.

Since F (t+n)Cn = G(t+n−1)Cn−1, the terms in the last sum cancel in pairs,leaving only F (t)C0x

t. Hence

{F (δ)− xG(δ)}F (t+m)S(t) = F (t)F (t+m)C0xt.

But F (t)F (t+m) has a double root t = a, and so, formally at least, both [F (t+m)S(t))]t=a and [ ∂

∂t{F (t + m)S(t)}]t=a satisfy the differential equation. (This

can be verified rigorously as for the case of equal roots.) This compensates forthe previously mentioned lost solution.

10 An Example

x2 d2y

dx2+ x

dy

dx+ (x2 − 1)y = 0.

(This is Bessel’s equation of order 1.) It transforms to

(δ2 − 1)y = −x2y.

In the usual notation, S(a) is

xa+

(−1)xa+2

(a+1)(a+3)+···+ (−1)nxa+2n

(a+1)(a+3)2···(a+2n−1)2(a+2n+1)+···..

13

One solution is given by putting a = 1 in this. The second solution wouldnormally be given by putting a = −1, but this produces zeros in the denomi-nators of the terms of S(a), because of the factor a+ 1. We therefore multiplyS(a) by a + 1 (it is not necessary to multiply by the full F (a + m)” as in †9,provided that what we multiply by is sufficient to remove the awkward factor),and differentiate the result by a, obtaining

log x(a+ 1), S(a) + xa + xa+2

(a+3)2+ · · ·+

+ (−1)nxa+2n

(a+3)2···(a+2n−1)2(a+2n+1)(−2)

{1

a+3+ · · ·+ 1

a+2n−1+ 1

21

a+2n+1

}+ · · · .

Putting a = −1 in this, we obtain the second solution.

11 Simplifications

Simplifications occur if coefficients vanish, so that an infinite series solutionbecomes a polynomial. And this occurs (again assuming h = 1) if G(a+m) = 0for a a root of F (z) = 0 and m some positive integer. Since there is the furtherpossibility that in addition F (a+r) = 0 for some positive integer r, it is perhapsworthwhile summarizing some conceivable situations.

(a) F(a+m) = 0;noG(a+r) = 0:This case is covered in †9, and leads to the usual series solution commencing

with xa+m, i.e. S(a+m), together with the solution

∂

∂t{F (t+m)S(t)}]t=a.

(b) G(a+m) = 0;noF(a+r) = 0:The coefficients in S(a) after a certain point vanish, and so the solution

S(a) simplifies to a polynomial.(c) G(a+m) = 0; F(a+r) = 0 (r ≥ m+ 1):

Factors in the denominator of S(a) now vanish from the same (or a later)term than factors in the numerator vanish. The coefficients in S(a) are thusnon-zero up to and including the coefficient of xa+m, and are thereafter zero upto and including the coefficient of xa+r−1. (This set of zeros may be void if r =m+ 1.) The coefficient of xa+r is indeterminate, because both numerator anddenominator are zero, and this means (in terms of the Frobenius method) thatthe recurrence relation for this particular coefficient becomes 0 = 0, so thatthe coefficient is an arbitrary constant. Choosing the constant to be zero, wefind that one solution is a polynomial containing terms in xa, xa+1, · · · , xa+m.

A second solution is plainly the infinite series solution S(a+ r).

14

There would thus seem to be the possibility of an extra solution, since thearbitrary constant which appeared in the first solution could have been chosenother than zero. But the arbitrary constant multiplies a series commencingwith xa+r, in fact S(a+ r), and so to choose the arbitrary constant other thanzero does not introduce any fresh solutions.

noindent Example.

δ(δ − 5)y = x(δ − 3)y.

Two solutions to this equation will be a polynomial containing terms inx0, x1, x2, x3, and an infinite series commencing with x5.(d) G(a+m) = 0; F (a+ r) = 0 r < m+ 1) :

This is only a slightly modified form of (a). For the fact that G(a+m) = 0merely means that S(a+ r) is a polynomial instead of an infinite series, whilethe other solution is as before

[∂

∂t{F (t+ r)S(t)}]t=a.

Example.δ(δ − 2)y = x(δ − 2)y.

By the previous analysis, this should give a polynomial solution “S(a+ r)”which both starts and finishes with x2, and so consists of x2 alone, and also asolution involving log x and a power series; and this can be verified by solvingthe equation directly. Put

(δ − 2)y = z.

Thenδz = xz,xDz = xz,

z = Aex, A an arbitrary constant,(δ − 2)y = Aex,

x dydx− 2y = Aex,

1x2

dydx− 2y

x3 = Aex

x3 ,

y = Bx2 + Ax2∫ x et

t3dt.

The solution x2 has now made its appearance. To show that the other solutioninvolves log x, expand et in a power series and integrate term-by-term.

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