Frobenius Series Solutions

Power Series Solutions And Special Functions:Review of Power Series

Pradeep Boggarapu

Department of Mathematics

BITS PILANI K K Birla Goa Campus, Goa

September 28, 2015

Pradeep Boggarapu (Dept. of Maths) Review of Power Series September 28, 2015 1 / 65

Finding the general solution of a linear differential equation

y ′′ + P(x)y ′ + Q(x)y = R(x)

depends on determining any two linear independent solutions of thehomogeneous equation

y ′′ + P(x)y ′ + Q(x)y = 0

So far, we have a systematic procedure for constructing fundamentalsolutions (linear independent solutions of associated homogeneousequation), if the equation has constant coefficients (i.e., P(x) and Q(x) areconstant functions).

For the equations with variable coefficients we don’t have any method tofind fundamental solutions expect the case of knowing one solution wherewe use the known solution to find another solution via the formulay2(x) = v(x)y1(x) with

v(x) =

∫1

y21

e−∫P(x)dxdx


Examples: Let p, a, b, c and k are real constants,

1 Bessel’s Differential Equation: x2y ′′ + xy ′ + (x2 − p2)y = 0

2 Legender’s Differential Equation: (1− x2)y ′′ − 2xy ′ + p(p − 1)y = 0

3 Hermite Differential Equation: y ′′ − 2xy ′ + 2py = 0

4 Guass Hypergeometric Differentila Equation:

x(x − 1)y ′′ + [c − (a + b + 1)x ]y ′ − aby = 0

5 Laguerre’s Differential Equations: xy ′′ + (1− x)y ′ + py = 0

6 Airy’s Equation: y ′′ ± p2xy = 0


In most of the cases the solutions of the above differential equations arebeyond the elementary functions which are called as special functions.

Many of the special functions find applications in connection with thepartial differential equations of mathematical physics. Thare are alsoimportant in modern pure mathematics, through the theory of orthogonalexpansions.

For a larger class of linear diffrerential equations with variable coefficientssuch as above equations, we must search for solutions beyond the familiarelementary functions of calculus.

The principal tool we need is the representation of a given function by apower series.

Then, we assume that the solutions y have power series representations∑∞n=0 an(x − x0)n, and then determine the coefficients an’s so as to satisfy

the differential equation similar to the method of undeterminedcoefficients.


Review of Power Series

Power series about the point zero: It is an infinite series of the form

∞∑n=0

anxn = a0 + a1x + a2x2 + a3x3 + · · · (0.1)

where a0, a1, a2, . . . , an, . . . are real constants.

Power series about the point x0: It is an infinite series of the form

∞∑n=0

an(x − x0)n = a0 + a1(x − x0) + a2(x − x0)2 + a3(x − x0)3 + · · · (0.2)

where a0, a1, a2, . . . , an, . . . are real constants.

Examples:∞∑n=0

xn

n!= 1 +

x

1!+

x2

2!+

x3

3!+ · · · ;

∞∑n=0

xn = 1 + x1 + x2 + x3 + · · · .


Convergence of power series: The power series is said to converge at apoint x if its n-th partial sum

∑nk=0 akxk converges that is to say that the

limit L = limn→∞

n∑k=0

akxk exists.

In this case the sum of the series is the limit i.e. L =∑∞

n=0 anxn and suchpoints x are called points of convergence.

Note that x = 0 is always a point of convergence of the power series (0.1).See the following examples

∞∑n=0

xn

n!= 1 +

x

1!+

x2

2!+

x3

3!+ · · · ; (0.3)

∞∑n=0

xn = 1 + x1 + x2 + x3 + · · · ; (0.4)

∞∑n=0

n!xn = 1 + x + 2!x2 + 3!x3 + · · · . (0.5)


The first series converges for every value of x in R; second converges onlyfor |x | < 1 and the third series diverges for all x 6= 0.

The power series in x that behaves like third series are of no interest to us.

Fact: The points of convergence of a power series (0.1) (or (0.47)) forman interval. Moreover there exists 0 ≤ R ≤ ∞ such that the power series(0.1) (or (0.47)) converges for all |x | < R (resp. |x − x0| < R) anddiverges for all |x | > R (resp. |x − x0| > R). Here R is called radius ofconvergence.

In many cases the radius of convergence can be found by using thefollowing formulas,

R = limn→0

∣∣∣ anan+1

∣∣∣ or R = limn→0

1n√|an|

whenever the limits exist.

Regardless of the existence of the above limits, it is known that R alwaysexists.


Differentiation of power series: Suppose that the power series (0.1)converges for |x | < R with R > 0 and denote the sum by f (x):

f (x) =∞∑n=0

anxn = a0 + a1x + a2x2 + a3x3 + · · · .

Then f (x) is automatically is continuous and has the derivatives of allorders for |x | < R. Also,

f ′(x) =∞∑n=1

nanxn−1 = a1 + 2a2x + 3a3x2 + · · · , (0.6)

f ′′(x) =∞∑n=2

n(n − 1)xn−2 = 2a2 + 3 · 2a3x + · · · , (0.7)

and so on, and each of the resulting series converges for |x | < R.

And we can link the coefficient an to f (x) and its derivative via thefollowing formula

an =f (n)(0)

n!(0.8)


Algebra of power series: Let f (x) =∑∞

n=0 anxn and g(x) =∑∞

n=0 bnxn

be two power series with radius of convergence at least R > 0, then thesepower series can be added or subtracted termwise:

f (x)± g(x) =∞∑n=0

(an ± bn)xn = (a0 ± b0) + (a1 ± b1)x + · · · .

They can also be multiplies as they were polynomials, in the sense that

f (x)g(x) =∞∑n=0

cnxn

where cn = a0bn + a1bn−1 + · · ·+ anb0.

If f (x) = g(x) for |x | < R if and only if an = bn for all n i.e. If both seriesconverges to the same function for |x | < R if and only if they have thesame coefficients.


Power series representation of a function: Let f be a continuousfunction defined for |x | < R or (−R,R) which has derivatives of all ordersin the interval (−R,R). Can f (x) be represented by a power series aboutthe point zero? Equivalently will the following hold

f (x) =∞∑n=0

f (n)(0)

n!xn = f (0) + f ′(0)x +

f ′′(0)

2!x2 + · · · (0.9)

throughout the interval (−R,R)?

This is often true, but unfortunately it is some times false. The aboveexpansion is valid if the error term Rn(x) in Taylor’s formula:

f (x) =n∑

n=0

f (k)(0)

k!xk + Rn(x)

convegres to zero as n tends to infity.


By means of the procedure explained in the previous slide, it is quite easyto obtaion the following familiar expansions,

ex =∞∑n=0

xn

n!= 1 +

x

1!+

x2

2!+

x3

3!+ · · · ; (0.10)

sin x =∞∑n=0

(−1)nx2n+1

(2n + 1)!= x − x3

3!+

x5

5!− · · · ; (0.11)

cos x =∞∑n=0

(−1)2nx2n

(2n)!= 1− x2

2!+

x4

4!− · · · ; (0.12)

1

1± x=

∞∑n=0

(±1)nxn = 1± x + x2 ± x3 + · · · ; (0.13)

log (1 + x) =∞∑n=1

(−1)n−1xn

n= x − x2

2+

x3

3− x4

4+ · · · ; (0.14)

tan−1 x =∞∑n=0

(−1)nx2n+1

(2n + 1)= x − x3

3+

x5

5− · · · ; (0.15)


The function f for which the above series expansion (0.9) is valid for someneighbourhood of zero is said to be analytic at x = 0. More generally theanalyticity at any point is defined as follows.

Analytic at a point: A function f (x) with the property that a powerseries expansion of the form

f (x) =∞∑n=0

an(x − x0)n

valid in some neighbourhood of the point x0 is said to be analytic at x0.

In this case the coefficients an’s are necessarily given by

an =f (n)(x0)

n!,

and the above series is called the Taylor series of f (x) at x0.


Facts about analytic functions:

Polynomials and the functions ex , sin x and cos x are analytic at allpoints.

If f (x) and g(x) are analytic at x0, then f (x) + g(x), f (x)g(x), andf (x)/g(x) [if g(x0) 6= 0] are also analytic at x0 .

If f (x) is analytic at x0, f ′(x0) 6= 0 and f −1(x) is a continuousinverse, then f −1(x) is analytic at f (x0).

If g(x) is analytic at x0 and f (x) is analytic at g(x0), thenf ◦ g(x) = f (g(x)) is analytic at x0.

The sum of a power series is analytic at all points inside the intervalof convergence.


Series solutions of first order equations: We have repeatedlyemphasized that many interesting and important differential equationscannot be solved by any of the methods discussed so far.

And also note that solutions for equations of this kind can often be foundin terms of power series.

Our purpose in this section is to discuss that how we use power seriesrepresentation to solve a differential equation by demonstrating an examplewith a first order equations that are easy to solve by elementary methods.

Problem 1: Solve the first order differential equation y ′ = y by usingpower series representation.

Solution: We assume that the given equation has a power series solutionof the form

y =∞∑n=0

anxn = a0 + a1x + a2x2 + a3x3 + · · · (0.16)

that converges for |x | < R with R > 0.Pradeep Boggarapu (Dept. of Maths) Review of Power Series September 28, 2015 14 / 65

That is, we are assuming that the differential equation has a solution say ythat is analytic at the origin. We know that

y ′ =∞∑n=1

nanxn−1 = a1 + 2a2x + 3a3x2 + · · ·+ (n + 1)an+1xn + · · · , (0.17)

in the interval of convergence. Since y ′ = y , we have that

∞∑n=1

nanxn−1 =∞∑n=0

anxn

a1 + 2a2x + 3a3x2 + · · · = a0 + a1x + a2x2 + a3x3 + · · · .

The above both series must have the same coefficients:

a1 = a0, 2a2 = a1, 3a3 = a2, · · · (n + 1)an+1 = an, · · · .

These equations enables us to express each an in terms of a0:

a1 = a0, a2 =a12

=a02, a3 =

a23

=a0

2 · 3, · · · , an =

a0n!, · · · .


If you substitute all these values of coefficients in (0.16), we obtain ourpower series solution

y = a0

(1 +

x

1!+

x2

2!+

x3

3!+ · · ·+ xn

n!+ · · ·

)We can easily recognise that the above series as the power series

expansion of ex , so the above can be written as

y = a0ex .

, This example suggests a useful method for obtaining the power seriesexpansion of a given function:

Find the differential equation with initial condition satisfied by thefunction, and then solve this equation by power series. We consider anexample (to understand this idea).Proplem 2: Find the power series expansion of y = (1 + x)p about theorigin (where p is a constant) by using the method described in the above.Use the result to show that

√2 = 1 +

1

2· 1

2+

1 · 32 · 4

· 1

22+

1 · 3 · 52 · 4 · 6

· 1

23+ · · · .


Part 1: First find a differential equation satisfied by y :

(1 + x)y ′ = py , y(0) = 1

As before we assume that the above equation has a power series solution

y =∞∑n=0

anxn = a0 + a1x + a2x2 + a3x3 + · · · (0.18)

with positive radius R > 0 of convergence. It follows that

y ′ = a1 + 2a2x + 3a3x2 + · · ·+ (n + 1)an+1xn + · · · ,xy ′ = a1x + 2a2x2 + · · ·+ nanxn + · · · ,py = pa0 + pa1x + pa2x2 + · · ·+ panxn + · · · .

Since the d.e. is (1 + x)y ′ = py , the sum of the first two series must equalthe third, so equating the coeffiecients of the successive powers of x gives

a1 = pa0, 2a2 + a1 = pa1, 3a3 + 2a2 = pa2, . . . ,

(n + 1)an+1 + nan = pan, . . .


The initial condition y(0) = 1 implies that a0 = 1, so

a1 = p, a2 =a1(p − 1)

2=

p(p − 1)

2,

a3 =a2(p − 2)

3=

p(p − 1)(p − 2)

2 · 3, . . . ,

an =p(p − 1)(p − 2) · · · (p − n + 1)

n!, . . . .

With these coefficints, the solution (0.18) becomes

y =1 + px +p(p − 1)

2!x2 +

p(p − 1)(p − 2)

3!x3 + · · ·

+p(p − 1)(p − 2) · · · (p − n + 1)

n!xn + · · · . (0.19)

To conclude that (0.19) actually is the desired solution, it suffices toobserve that the series converges for |x | < R for some R > 0.


We can show that R = 1 by using the ratio tests or by using first of theformulas given previously

R = limn→∞

∣∣∣ anan+1

∣∣∣ = limn→0

∣∣∣n + 1

p − n

∣∣∣ = 1.

On comparing the two solutions y = (1 + x)p and (0.19), and using thefact that the initial value problem has only one solution, we have

(1 + x)p =1 + px +p(p − 1)

2!x2 +

p(p − 1)(p − 2)

3!x3 + · · ·

+p(p − 1)(p − 2) · · · (p − n + 1)

n!xn + · · · (0.20)

for |x | < 1. This expansion is called the binomial series and this formulageneralises the binomial theorem to the case of an arbitrary exponent.


Part2: Now we substitute x = −12 and p = −1

2 in the power series (0.20)to get the required indentity.

(1− 1

2

)− 12

= 1 +(− 1

2

)·(− 1

2

)+

(− 1

2

)(− 3

2

)2!

·(− 1

2

)2+

(− 1

2

)(− 3

2

)(− 5

2

)3!

·(− 1

2

)3+ · · · .

After simplying the above you get the required identity

√2 = 1 +

1

2· 1

2+

1 · 32 · 4

· 1

22+

1 · 3 · 52 · 4 · 6

· 1

23+ · · · .


Problem 3: Express sin−1 x in the form of a power series∑

anxn bysolving y ′ = (1− x2)−1/2 in two ways. (Hint: Remember the binomialseries.) Use this result to obtain the formula

π

6=

1

2+

1

2· 1

3 · 23+

1 · 32 · 4

· 1

5 · 25+

1 · 3 · 52 · 4 · 6

· 1

7 · 27+ · · · .

Solution:Part 1: We first find an intial value problem satisfied by y = sin−1 x :

y ′ = (1− x2)−1/2, y(0) = 0

As before we assume that the above equation has a power series solution

y =∞∑n=0

anxn = a0 + a1x + a2x2 + a3x3 + · · · (0.21)

with positive radius R > 0 of convergence. It follows that

y ′ = a1 + 2a2x + 3a3x2 + 4a4x3 + · · ·+ (n + 1)an+1xn + · · · (0.22)


We know that

(1 + x)p =1 + px +p(p − 1)

2!x2 +

p(p − 1)(p − 2)

3!x3 + · · ·

+p(p − 1)(p − 2) · · · (p − n + 1)

n!xn + · · · .

Now we replace p by −1/2 and x by −x2 in the above to get(1− x2

)−1/2= 1 +

1

2x2 +

1 · 32 · 4

· x4 +1 · 3 · 52 · 4 · 6

· x6 + · · ·

+1 · 3 · 5 · · · (2n − 1)

2 · 4 · 6 · · · (2n)· x2n + · · · .

Since the d.e. is y ′ = (1− x2)−1/2, so equating the coeffiecients of thesuccessive powers of x on both series gives

2a2 = 4a4 = a2n = · · · = 0; a1 = 1; 3a3 =1

2; 5a5 =

1 · 32 · 4

;

7a7 =1 · 3 · 52 · 4 · 6

; . . . ; (2n + 1)a2n+1 =, . . .1 · 3 · 5 · · · (2n − 1)

2 · 4 · 6 · · · (2n)


Further simplification implies

2a2 = 4a4 = a2n = · · · = 0; a1 = 1; a3 =1

2· 1

3; a5 =

1 · 32 · 4

· 1

5;

a7 =1 · 3 · 52 · 4 · 6

· 1

7; . . . ; a2n+1 =, . . .

1 · 3 · 5 · · · (2n − 1)

2 · 4 · 6 · · · (2n)· 1

2n + 1

Since y(0) = 0, a0 = 0, now substitute all the values of coefficients inpower series solution (0.21) and the initial value problem has only onesolution which means the power series solution is nothing but sin−1 x ,therefore

sin−1 x = x +1

2· 1

3x3 +

1 · 32 · 4

· 1

5x5 +

1 · 3 · 52 · 4 · 6

· 1

7x7 + · · · .

If x = 12 then sin−1 x = π

6 and hence in view of the above we get therequired identity

π

6=

1

2+

1

2· 1

3 · 23+

1 · 32 · 4

· 1

5 · 25+

1 · 3 · 52 · 4 · 6

· 1

7 · 27+ · · · .


Problem 4: Find the power series solution of the initial value problem

(1 + x)y ′ = 1; y(0) = 0,

and also find solution of the same by using variable separable method toget the follwoing identity

loge 2 =∞∑n=1

1

n· 1

2n=

1

2+

1

2· 1

22+

1

3· 1

23+

1

4· 1

24+ · · · .

Problem 5: Find the power series solutions of the each of the followingfirst order differentail equations:

(1) y ′ − y = 0 (2) y ′ = ex2y (3) y ′ − xy = 0

(4) (1− x)y ′ = y (5) y ′ − y = x2 (6) y + xy = 1 + x

(7) (1 + x2)y ′ = 0.


Second order linear equations. Ordinary points:We have seen the power series solutions for first order linear equations.We now turn our attention to the general homogeneous second orderlinear equation

y ′′ + P(x)y ′ + Q(x)y = 0 (0.23)

As we know, it is occasionally possible to solve such an equation in termsof familiar elementary functions.

For instance, when P(x) and Q(x) are constants and in few other cases aswell.

For most of the equations having the greatest significance in pure andapplied mathematics can only be solved by power series.

The central fact about the equation (0.23) is that the behaviour of thecoefficients P(x) and Q(x) near a point x0 determines the behaviour itssolutions near this point.


In this section we confine ourselves to the case in which P(x) and Q(x)are “well behave” in the sense of being analytic at x0 that is the casewhere x0 is an ordinary point.

Ordinary point : If each of P(x) and Q(x) has a power series expansionvalid in some neighbourhood of a point x0, then x0 is called an ordinarypoint of the equation (0.23). In this case all solutions of equation (0.23)will have power series expansion valid in a neighbourhood of x0.

In other words, if the coefficient funtions P(x) and Q(x) of equation(0.23) are analytic at a point x0 then its all solutions are also analytic atthis point.

Any point that is not ordinary point of (0.23) is called singular point. Wewill see a bit more about singular points after few slides.


Examples: Let p, a, b, c and k are real constants, find out all ordinary andsigular points of the following equations.

1 Bessel’s Differential Equation: x2y ′′ + xy ′ + (x2 − p2)y = 0

2 Legender’s Differential Equation: (1− x2)y ′′ − 2xy ′ + p(p − 1)y = 0


4 Guass Hypergeometric Differentila Equation:

x(x − 1)y ′′ + [c − (a + b + 1)x ]y ′ − aby = 0




We now illustrate the above we said by an example of familiar equations

y ′′ + y = 0 (0.24)

The coefficients P(x) = 0 and Q(x) = 1 are analytic at all the points, weexpect the solutions are also analytic at all the points.

So we seek a solution of the form

y =∞∑n=0

anxn = a0 + a1x + a2x2 + a3x3 + · · · (0.25)

Differentiating the above, yields

y ′(x) = a1 + 2a2x + 3a3x2 + · · ·+ (n + 1)an+1xn + · · · ,y ′′(x) = 2a2 + 3 · 2a3x + · · ·+ (n + 1)(n + 2)an+2xn + · · · , (0.26)

Since y ′′ + y = 0, on adding the above two series term by term andequating to zero, we get

(2a2 + a0) + (2 · 3a3 + a1)x + (3 · 4a4 + a2)x2 + (4 · 5a5 + a3)x3

+ · · ·+ [(n + 1)(n + 2)an+2 + an]xn + · · · = 0;


now equating the coefficients of successive powers of x gives

2a2 + a0 = 0; 2 · 3a3 + a1 = 0; 3 · 4a4 + a2 = 0;

4 · 5a5 + a3 = 0; · · · ; (n + 1)(n + 2)an+2 + an = 0; . . . .

By means of these equations we can express an in terms of a0 or a1according as n is even or odd:

a2 = −a02

; a3 = − a12 · 3

; a4 = − a23 · 4

=a0

2 · 3 · 4;

a5 = − a34 · 5

=a1

2 · 3 · 4 · 5; . . . .

With these coefficients, (0.25) becomes

y = a0 + a1x − a02

x2 − a12 · 3

x3 +a0

2 · 3 · 4x4 +

a12 · 3 · 4 · 5

x5 − · · ·

= a0

(1− x2

2!+

x4

4!− · · ·

)+ a1

(x − x3

3!+

x5

5!− · · ·

)(0.27)

where y(0) = a0 and y ′(0) = a1.Pradeep Boggarapu (Dept. of Maths) Review of Power Series September 28, 2015 29 / 65

Let y1(x) and y2(x) denote the two series in parentheses. We have shownformally that (0.27) satisfies (0.24) for any two constants a0 and a1.

In particular, by choosing a0 = 1 and a1 = 0 we see that y1 satisfies thisequation, and the choice a0 = 0 and a1 = 1 shows that y2 also satisfies theequation.

Note that the two series defining y1 and y2 converges for all x ,furthermore, y1 and y2 are linearly independent as neither series is aconstant multiple of the other.

We therefore see that (0.27) is the general solution of (0.24), and that anyparticular solution is obtained by specifying the values of y(0) = a0 andy ′(0) = a1.

In the above example two series in parentheses are easily recognizable asthe expansions of cos x and sin x , so (0.27) can be written in the form

y = a0 cos x + a1 sin x .


We now apply the method of these example to establish the followinggeneral theorem about the nature of solutions near ordinary points.

Theorem 0.1.

Let x0 be an ordinary point of the differential equation

y ′′ + P(x)y ′ + Q(x)y = 0, (0.28)

and let a0 and a1 be arbitrary constants. Then there exists a uniquefunction y(x) that is analytic at x0, is a solution of equation (0.28) in acertain neighborhood of this point, and satisfies the initial conditionsy(x0) = a0 and y ′(x0) = a1. Furthermore, if the power series expansions ofP(x) and Q(x) are valid on an interval |x − x0| < R, R > 0, then thepower series expansion of this solution is also valid on the same interval.

For the proof of this theorem we refer to our text book. We solve theLegendre’s equation by the same procedure used to solve the first example.


Legendre’s differential equation:

(1− x2)y ′′ − 2xy ′ + p(p + 1)y = 0, (0.29)

where p is a constant.

It is clear that the coefficient functions

P(x) =−2x

1− x2and Q(x) =

p(p + 1)

1− x2

are analytic at the origin. The origin is therefore an ordinary point and weexpect a solution of the form

y =∞∑n=0

anxn = a0 + a1x + a2x2 + a3x3 + · · · .

Since y ′ =∑

(n + 1)an+1xn, we get the following expansions for theindvidual terms on the left side of equation (0.29):



(1− x2)y ′′ − 2xy ′ + p(p + 1)y = 0, (0.29)

where p is a constant. It is clear that the coefficient functions

P(x) =−2x

1− x2and Q(x) =

p(p + 1)

1− x2

are analytic at the origin.

The origin is therefore an ordinary point and weexpect a solution of the form

y =∞∑n=0

anxn = a0 + a1x + a2x2 + a3x3 + · · · .

Since y ′ =∑




(1− x2)y ′′ − 2xy ′ + p(p + 1)y = 0, (0.29)

where p is a constant. It is clear that the coefficient functions

P(x) =−2x

1− x2and Q(x) =

p(p + 1)

1− x2

are analytic at the origin. The origin is therefore an ordinary point and weexpect a solution of the form

y =∞∑n=0

anxn = a0 + a1x + a2x2 + a3x3 + · · · .

Since y ′ =∑



y ′′ =∞∑n=0

(n + 1)(n + 2)an+2xn,

−x2y ′′ =∞∑n=2

−(n − 1)nanxn,

−2xy ′ =∞∑n=1

−2nanxn,

and

p(p + 1)y =∞∑n=0

p(p + 1)anxn.

By equation (0.29), the sum of the these series is required to be zero, sothe coefficients of xn must be zero for every n:

(n + 1)(n + 2)an+2 − (n − 1)nan − 2nan + p(p + 1)an = 0.


With a little manipulation, this becomes

an+2 = −(p − n)(p + n + 1)

(n + 1)(n + 2)an.

This recursion formula enables us to express an in terms of a0 or a1according as n is even or odd:

a2 = −p(p + 1)

1 · 2a0,

a3 = −(p − 1)(p + 2)

2 · 3a1,

a4 = −(p − 2)(p + 3)

3 · 4a2 =

p(p − 2)(p + 1)(p + 3)

4!a0,

a5 = −(p − 3)(p + 4)

4 · 5a3 =

(p − 1)(p − 3)(p + 2)(p + 4)

5!a1,

a6 = −(p − 4)(p + 5)

5 · 6a4 = −p(p − 2)(p − 4)(p + 1)(p + 3)(p + 5)

6!a0


a7 = −(p − 5)(p + 6)

6 · 7a5

= −(p − 1)(p − 3)(p − 5)(p + 2)(p + 4)(p + 6)

7!a1,

and so on. By inserting these coefficients into y =∑

anxn, we obtain

y = a0[1− p(p + 1)

1 · 2x2 +

p(p − 2)(p + 1)(p + 3)

4!x4

− p(p − 2)(p − 4)(p + 1)(p + 3)(p + 5)

6!x6 + · · ·

]+ a1

[x − (p − 1)(p + 2)

2 · 3x3 +

(p − 1)(p − 3)(p + 2)(p + 4)

5!x5

− (p − 1)(p − 3)(p − 5)(p + 2)(p + 4)(p + 6)

7!x7 + · · ·

](0.30)

as our solution of (0.29).


When p is not an integer, each series in bracket has radius of convergesR = 1.

This is most easily seen by using the recursion formula: for the first series,this formul (with n replaced by 2n)∣∣∣∣a2n+2x2n+2

a2nx2n

∣∣∣∣ =

∣∣∣∣(p − 2n)(p + 2n + 1)

(2n + 1)(2n + 2)

∣∣∣∣|x2| → |x2|

as n→∞, and similarly for the second series.

The fact that the each series has positive radius of convergence justifiesthe operations we have performed and shows that the power series (0.30)is a valid solution of (0.29) for every choice of constants a0 and a1.

Each bracketed series is a particular solution and since it is clear that thefunctions defined by these series are linearly independent, (0.30) is thegeneral solution of (0.29)




3 Chebyshev’s equation: (1− x2)y ′′ − xy + p2y = 0′



Regular Singular Points: We recall the point x0 is a singular point of thedifferentail equation

y ′′ + P(x)y ′ + Q(x)y = 0, (0.31)

if one (or both) of the coeffiecient functions P(x) and Q(x) fails to beanalytic at x0.

In this case the theorem and method of the previous section do not apply.If we wish to study the solution of the above differential equations near x0,then new ideas are necessary.

At this moment we can not say that the equation has (or has no) asolution near the point x0.

But if we were in a situation that the coefficient functions are not so farthat to be analytic at the point x0 in the sense a mild modifications of thecoefficient functions are analytic, would we be able to find the solutions ofequation (0.31)?

The answer is yes partially. We have some singular points called RegularSingular Points near which we can find the solutions.


What are Regular singular points: A singular point x0 is said to beregular if the functions (x − x0)P(x) and (x − x0)2Q(x) are analytics atx0. If the singular point is not regular then it is called irregular.

Roughly speaking, at regular point x0, the sigularity of P(x) and Q(x) cannot be worse than 1/(x − x0) and 1/(x − x0)2 respectively.Example:1. x2y ′′ + pxy ′ + rxy = 0 and 2. x3y ′′ + px2y ′ + sin xy = 0, In bothexamples the point x = 0 is regular point.

3. x3(1− x2)y ′′ + 2xy ′ − 2y = 0, here x = 0 is irregular singular but notregular singular, x = ±1 are regular singular points and any other point isan ordinary point.

Let p, a, b, c and k are real constants, find out all ordinary points, regularand irregular sigular points of the following equations.

Legender’s Differential Equation: (1− x2)y ′′ − 2xy ′ + p(p − 1)y = 0x = ±1 are the regular singular point of the above equations all others areordinary points.


Guass Hypergeometric Differentila Equation:

x(x − 1)y ′′ + [c − (a + b + 1)x ]y ′ − aby = 0

x = 0, 1 are the regular singular points and all other points are ordinaypoints.

Bessel’s Differential Equation: x2y ′′ + xy ′ + (x2 − p2)y = 0x = 0 is only the regular singular point and any other points are ordinay.

Hermite Differential Equation: y ′′ − 2xy ′ + 2py = 0There are no singular points. All points are ordinary points.

Laguerre’s Differential Equations: xy ′′ + (1− x)y ′ + py = 0x = 0 is regular singular and others are ordinary point.


How do we solve the differential equation

y ′′ + P(x)y ′ + Q(x)y = 0, (0.32)

near x = x0, if it has regular singularity at x = x0.

In this case we will find a solutions y which is of the form

y = (x − x0)m[a0 + a1(x − x0)2 + a2(x − x0)2 + a3(x − x0)3 + · · · ]= a0(x − x0)m + a1(x − x0)m+1 + a2(x − x0)m+2 + a3(x − x0)m+3 + · · ·

(0.33)

where the coefficients a0 6= 0, a1, a2, . . . and the exponent m are unknown.The above series is called Frobenius series.

We will find the unknown coefficients and the exponent so that y is asolution the equation (0.42), the exponent m may be a negative integer, afraction or even an irrational real number.

Question: What is the guarantee that the solution of the above formexists?


We answere this question later after we get enough familiar with this kindof solutions. We will see some example to get familiar with those.

We consider the equation

2x2y ′′ + x(2x + 1)y − y = 0. (0.34)

If we write the equation in standard form

y ′′ +12 + x

xy ′ +

−1/2

x2y = 0

then we see at once that xP(x) = 12 + x and x2Q(x) = −1

2 are analytic atx = 0, so x = 0 is a regular singular point.

We want to find a solution near this point. We now introduce ourassumed Frobenius series solution

y = xm(a0 + a1x + a2x2 + a3x3 + · · · )= a0xm + a1xm+1 + a2xm+2 + a3xm+3 + · · · , (0.35)

with a0 6= 0,Pradeep Boggarapu (Dept. of Maths) Review of Power Series September 28, 2015 42 / 65

and its derivatives are given by

y ′ = a0mxm−1 + a1(m + 1)xm + a2(m + 2)xm+1 + a3(m + 3)xm+2 + · · ·

and

y ′′ = a0m(m − 1)xm−2 + a1(m + 1)mxm−1 + a2(m + 2)(m + 1)xm

+ a3(m + 3)(m + 2)xm+1 + · · · .

To find the coefficients in (0.35), we proceed in essentially the same wayas in the case of an ordinary point, with the significant difference that nowwe must find the appropriate value (or values) of the exponents m.

When we substitute three series y , y ′ and y ′′ in the equation (0.34), thenthe common factor xm−2 of the series y , y ′/x and y ′′/x2 gets cancel, theresult is


a0m(m−1)+a1(m+1)mx +a2(m+2)(m+1)x2+a3(m+3)(m+2)x3+· · ·

+

(1

2+ x

)[a0m + a1(m + 1)x + a2(m + 2)x2 + a3(m + 3)x3 + · · · ]

− 1

2[a0 + a1x + a2x2 + a3x3 + · · · ] = 0

By inspection, we combine correspoding powers of x and equate thecoefficient of each power of x to zero.

This yields the following system of equations:

a0[m(m − 1) +

1

2m − 1

2

]= 0,

a1[(m + 1)m +

1

2(m + 1)− 1

2

]+ a0m = 0, (0.36)

a2[(m + 2)(m + 1) +

1

2(m + 2)− 1

2

]+ a1(m + 1) = 0,

· · · .


Since a0 6= 0, it follows from first of the above equations that

m(m − 1) +1

2m − 1

2= 0.

This is called the indicial equation of the differential euqation (0.34). Itsroots are

m1 = 1 and m2 = −1

2

These are only possible values for the exponent m in (0.35).

For each of these values of m, we now use the remaining equations of(0.36) to calculate a1, a2, a3, . . . in terms of a0. For m1 = 1, we obtain

a1 = − a0

2 · 1 + 12 · 2−

12

= −2

5a0

a2 = − 2a1

3 · 2 + 12 · 3−

12

= −2

7a1 =

4

35a0,

· · · .


And for m = −12 , we obtain

a1 =12a0

12

(− 1

2

)+ 1

2 ·12 −

12

= −a0

a2 = −12a1

32 ·

12 + 1

2 ·32 −

12

= −1

2a1 =

1

2a0,

· · · .We therefore have the following two Frobenius series solutions in each ofwhich we have put a0 = 1:

y1 = x(

1− 2

5x +

4

35x2 + · · ·

)(0.37)

y2 = x−1/2(

1− x +1

2x2 + · · ·

). (0.38)

These solutions are clearly linear independent for x > 0, so the generalsolution of (0.34) on this interval is

y = c1x(

1− 2

5x +

4

35x2 + · · ·

)+ c2x−1/2

(1− x +

1

2x2 + · · ·

)Pradeep Boggarapu (Dept. of Maths) Review of Power Series September 28, 2015 46 / 65

In view of the above example, it is easy to see that the indicial equation ofthe more general diffierential equation (0.42) is

m(m − 1) + mp0 + q0 = 0. (0.39)

where p0 and q0 are constant terms of the power series expansions of(x − x0)P(x) and (x − x0)2Q(x) near the point x = x0 respectively.

In our example, the inditial equation had two distinct real roots leading tothe two independent series solutions (0.37) and (0.38).

It is natural to expect such a result whenever the indicial equation hasdistinct real roots m1 and m2. This turns out to be true if the differencebetween m1 and m2 is not an integer.

If, however, this difference is an integer, then it often (but not always)happens that one of two expected series solution does not exist.

In this case it is necessary just as in the case m1 = m2 to find a secondsolution by other methods. In the next section we investigate thesedifficulties in greater detaills.


Regular Singular Points (Continued): We recall the point x0 is aregular singular point of the differentail equation

y ′′ + P(x)y ′ + Q(x)y = 0, (0.40)

if one (or both) of the coeffiecient functions P(x) and Q(x) fails to beanalytic at x0 and if the functions (x − x0)P(x) and (x − x0)2Q(x) areanalytics at x0.

The type of solutions we are aiming at for (0.40) is a “quasi power series”or Frobenius series of the form

y = (x − x0)m[a0 + a1(x − x0)2 + a2(x − x0)2 + · · · ]= a0(x − x0)m + a1(x − x0)m+1 + a2(x − x0)m+2 + · · · . (0.41)

We might have the following questions in our minds.

Question 1: What are the reasons behind the definition of a regular point?

Question 2: Where do we get the idea that series of the form (0.41) mightbe suitable solutions for the equation (0.40) near the regular singular pointx = x0?



y ′′ + P(x)y ′ + Q(x)y = 0, (0.40)









y ′′ + P(x)y ′ + Q(x)y = 0, (0.40)









y ′′ + P(x)y ′ + Q(x)y = 0, (0.40)








Now let us try to answer the above questions.

To simplify matters, we may assume that the singular point x0 is locatedat the origin. If it is not, then we can always move it to the origin bychanging the independent variable from x to x − x0.

We know that the general form of a function analytic at x = 0 is

a0 + a1x + a2x2 + a3x3 + · · · .

As a consequense, the origin will certainly be a singular point of(0.40) if

P(x) = · · ·+ b−2x2

+b−1

x+ b0 + b1x + b2x2 + · · ·

andQ(x) = · · ·+ c−2

x2+

c−1x

+ c0 + c1x + c2x2 + · · · ,

and at least one of the coefficients with negative subscripts is nonzero.





a0 + a1x + a2x2 + a3x3 + · · · .


P(x) = · · ·+ b−2x2

+b−1

x+ b0 + b1x + b2x2 + · · ·

andQ(x) = · · ·+ c−2

x2+

c−1x

+ c0 + c1x + c2x2 + · · · ,






a0 + a1x + a2x2 + a3x3 + · · · .


P(x) = · · ·+ b−2x2

+b−1

x+ b0 + b1x + b2x2 + · · ·

andQ(x) = · · ·+ c−2

x2+

c−1x

+ c0 + c1x + c2x2 + · · · ,






a0 + a1x + a2x2 + a3x3 + · · · .


P(x) = · · ·+ b−2x2

+b−1

x+ b0 + b1x + b2x2 + · · ·

andQ(x) = · · ·+ c−2

x2+

c−1x

+ c0 + c1x + c2x2 + · · · ,



Consider the differential equation y ′′ +p

xby ′ +

q

xcy = 0, where p and q

are nonzero real numbers and b and c are positive integers. It clear thatx = 0 is an irregular singular point if b > 1 or c > 2.

(a) If b = 2 and c = 3, show that there is only one possible value of mfor which there might exist a Frobenius series solution.

(b) Show that b = 1 and c ≤ 2 if and only if m satisfies a quadraticequation and hence we can hope for two Frobenius series solutions,corresponding to the roots of this equation.

The above exercise shows that two independent Frobenius series solutionsare possible only if the expression for P(x) (and Q(x)) given in the lastslide does not contain, more than the first term (resp. more than twoterms) to the left of the constant term b0 (resp. c0).

An equivalent statement is that xP(x) and x2Q(x) must be analytic at theorigin. Accordint to the definition, this is precisely what is meant by sayingthat the singular point x = 0 is regular



xby ′ +

q









xby ′ +

q









xby ′ +

q








We now attempt to answer the second question. The simple differentialequation that we can solve completely near a singular point is the Eulerequation:

x2y ′′ + pxy ′ + qy = 0.

If this is written in the form

y ′′ +p

xy ′ +

q

x2y = 0, (0.42)

so that P(x) = p/x and Q(x) = q/x2, the it clear that the origin is aregular point whenever the constants p and q are not both zero.

The general solutions of this kind equation provide a very suggestivebridge to the general case, we briefly recall the detials.

The key to find the solutions is that changing the independent variablefrom x to z = log x transforms the above equation into an equation withconstant coefficents.



x2y ′′ + pxy ′ + qy = 0.


y ′′ +p

xy ′ +

q

x2y = 0, (0.42)






x2y ′′ + pxy ′ + qy = 0.


y ′′ +p

xy ′ +

q

x2y = 0, (0.42)






x2y ′′ + pxy ′ + qy = 0.


y ′′ +p

xy ′ +

q

x2y = 0, (0.42)





The transformed equation is clearly

d2y

dz2+ (p − 1)

dy

dz+ qy = 0

whose auxiliary equation is m2 + (p − 1)m + q = 0.

If the roots of the auxiliary equation are m1 and m2, the we know that thefollowing are independent solutions:

em1z and em2z if m1 6= m2;

em1z and zem1z if m1 = m2.

Since ez = x , the corresponding pairs of solutions are

xm1 and xm2 if m1 6= m2;

xm1 and xm1 log x if m1 = m2. (0.43)



d2y

dz2+ (p − 1)

dy

dz+ qy = 0










d2y

dz2+ (p − 1)

dy

dz+ qy = 0









The most general differential equation with regular singular point at theorigin is simply the Euler equation with the constant p and q replaced bypower series:

y ′′+1

x(p0 + p1x + p2x2 + · · · )y ′+

1

x2(q0 + q1x + q2x2 + · · · )y = 0. (0.44)

If the transition from (0.42) to (0.44) is accomplished by replaceing theconstants by power series. then it is natural to guess that thecorresponding transition from (0.43) to the solutions of (0.44) might beaccomplished by replacing power functions xm by Frobenius series in x .

We therefore expect that equation (0.44) will have two independentFrobenius series solutions in x or perhaps one of Frobenius series form andone of the form

y = xm log x(a0 + a1x + a2x2 + · · · )

where we assume that x > 0. We will show that these are very goodguesses.



y ′′+1

x(p0 + p1x + p2x2 + · · · )y ′+

1

x2(q0 + q1x + q2x2 + · · · )y = 0. (0.44)



y = xm log x(a0 + a1x + a2x2 + · · · )




y ′′+1

x(p0 + p1x + p2x2 + · · · )y ′+

1

x2(q0 + q1x + q2x2 + · · · )y = 0. (0.44)



y = xm log x(a0 + a1x + a2x2 + · · · )



Our work so far was mainly directed at motivation and technique. We nowconfront the theoretical side of the problem of solving the general secondorder linear equation

y ′′ + P(x)y ′ + Q(x)y = 0 (0.45)

near the regular singular point x = 0.

The ideas developed above suggest that we attempt a formal calculationsof any solutions of (0.45) that have the Frobenius series form

y = xm(a0 + a1x + a2x2 + · · · ), (0.46)

where a0 6= 0 and m is number to be determined.

For reasons already explained, we cofine our attention to the the intervalx > 0. The behaviour of solution on the interval x < 0 can be studied bychanging the variable to t = −x and solving the resulting equation fort > 0.



y ′′ + P(x)y ′ + Q(x)y = 0 (0.45)



y = xm(a0 + a1x + a2x2 + · · · ), (0.46)





y ′′ + P(x)y ′ + Q(x)y = 0 (0.45)



y = xm(a0 + a1x + a2x2 + · · · ), (0.46)




Our hypothesis is that xP(x) and x2Q(x) are analytic at x = 0, andtherefore have power series expansions

xP(x) =∞∑n=0

pnxn and x2Q(x) =∞∑n=0

qnxn (0.47)

which are valid on an interval |x | < R for R > 0.

Just as in the example of previous section, we must find the possiblevalues of m in (0.46) and then, for each acceptable m, we must calculatethe corresponding coefficients a0, a1, a2, . . ..

If we write (0.46) in the form

y = xm∞∑n=0

anxn =∞∑n=0

anxm+n,

then differentiation yields

y ′ =∞∑n=0

an(m + n)xm+n−1 and



xP(x) =∞∑n=0


qnxn (0.47)




y = xm∞∑n=0

anxn =∞∑n=0

anxm+n,


y ′ =∞∑n=0




xP(x) =∞∑n=0


qnxn (0.47)




y = xm∞∑n=0

anxn =∞∑n=0

anxm+n,


y ′ =∞∑n=0



y ′′ =∞∑n=0

an(m + n)(m + n − 1)xm+n−2

= xm−2∞∑n=0

an(m + n)(m + n − 1)xn.

The terms P(x)y ′ and x2Q(x) in (0.45) can now be written as

P(x)y ′ =1

x

( ∞∑n=0

pnxn

)[ ∞∑n=0

an(m + n)xm+n−1]

= xm−2( ∞∑

n=0

pnxn

)[ ∞∑n=0

an(m + n)xn

]

= xm−2∞∑n=0

[ n∑k=0

pn−kak(m + k)

]xn

= xm−2∞∑n=0

[ n−1∑k=0

pn−kak(m + k) + p0an(m + n)

]xn


y ′′ =∞∑n=0

an(m + n)(m + n − 1)xm+n−2

= xm−2∞∑n=0

an(m + n)(m + n − 1)xn.

The terms P(x)y ′ and x2Q(x) in (0.45) can now be written as

P(x)y ′ =1

x

( ∞∑n=0

pnxn

)[ ∞∑n=0

an(m + n)xm+n−1]

= xm−2( ∞∑

n=0

pnxn

)[ ∞∑n=0

an(m + n)xn

]

= xm−2∞∑n=0

[ n∑k=0

pn−kak(m + k)

]xn

= xm−2∞∑n=0

[ n−1∑k=0

pn−kak(m + k) + p0an(m + n)

]xn


and

Q(x)y =1

x2

( ∞∑n=0

qnxn

)( ∞∑n=0

anxm+n

)= xm−2

( ∞∑n=0

qnxn

)( ∞∑n=0

anxn

)

= xm−2∞∑n=0

[ n∑k=0

qn−kak

]xn = xm−2

∞∑n=0

[ n−1∑k=0

qn−kak + q0an

]xn.

When these expressions for y ′′,P(x)y ′,Q(x)y are inserted in (0.45) andthe common factor xm−2 is canceled, then the diffierential equationbecomes

∞∑n=0

{an[(m + n)(m + n − 1) + (m + n)p0 + q0]

+n−1∑k=0

ak [(m + k)pn−k + qn−k ]

}xn = 0;

and equating to zero the coefficient of xn yields the following recursionformula for the an:


and

Q(x)y =1

x2

( ∞∑n=0

qnxn

)( ∞∑n=0

anxm+n

)= xm−2

( ∞∑n=0

qnxn

)( ∞∑n=0

anxn

)

= xm−2∞∑n=0

[ n∑k=0

qn−kak

]xn = xm−2

∞∑n=0

[ n−1∑k=0

qn−kak + q0an

]xn.


∞∑n=0

{an[(m + n)(m + n − 1) + (m + n)p0 + q0]

+n−1∑k=0


}xn = 0;



and

Q(x)y =1

x2

( ∞∑n=0

qnxn

)( ∞∑n=0

anxm+n

)= xm−2

( ∞∑n=0

qnxn

)( ∞∑n=0

anxn

)

= xm−2∞∑n=0

[ n∑k=0

qn−kak

]xn = xm−2

∞∑n=0

[ n−1∑k=0

qn−kak + q0an

]xn.


∞∑n=0

{an[(m + n)(m + n − 1) + (m + n)p0 + q0]

+n−1∑k=0


}xn = 0;



an[(m + n)(m + n − 1) + (m + n)p0 + q0]

+n−1∑k=0

ak [(m + k)pn−k + qn−k ] = 0. (0.48)

On writting this out for the succussive values of n, we get

a0[m(m − 1) + mp0 + q0] = 0,

a1[(m + 1)m + (m + 1)p0 + q0] + a0(mp1 + q1) = 0

a2[(m + 2)(m + 1) + (m + 2)p0 + q0]

+ a0(mp2 + q2)a1[(m + 1)p1 + q1] = 0,

· · ·

an[(m + n)(m + n − 1) + (m + n)p0 + q0]

+ a0(mpn + qn) + · · ·+ an−1[(m + n − 1)p1 + q1] = 0;

· · ·


an[(m + n)(m + n − 1) + (m + n)p0 + q0]

+n−1∑k=0

ak [(m + k)pn−k + qn−k ] = 0. (0.48)

On writting this out for the succussive values of n, we get

a0[m(m − 1) + mp0 + q0] = 0,

a1[(m + 1)m + (m + 1)p0 + q0] + a0(mp1 + q1) = 0

a2[(m + 2)(m + 1) + (m + 2)p0 + q0]

+ a0(mp2 + q2)a1[(m + 1)p1 + q1] = 0,

· · ·

an[(m + n)(m + n − 1) + (m + n)p0 + q0]

+ a0(mpn + qn) + · · ·+ an−1[(m + n − 1)p1 + q1] = 0;

· · ·


If we put f (m) = m(m − 1) + mp0 + q0, then these equations become

a0f (m) = 0,

a1f (m + 1) + a0(mp1 + q1) = 0,

a2f (m + 2) + a0(mp2 + q2) + a1[(m + 1)p1 + q1] = 0,

· · ·anf (m + n) + a0(mpn + qn) + · · ·+ an−1[(m + n − a)p1 + q1] = 0,

· · · .

Since a0 6= 0, we conclude from the first of these equations that f (m) = 0or, equivalently, that

m(m − 1) + mp0 + q0 = 0. (0.49)

This is the indicial equation, and its roots m1 and m2 which are possiblevalues of m in our assumed equations are called the exponents of thedifferential equation (0.45) at the regular singular point x = 0.



a0f (m) = 0,

a1f (m + 1) + a0(mp1 + q1) = 0,

a2f (m + 2) + a0(mp2 + q2) + a1[(m + 1)p1 + q1] = 0,


· · · .


m(m − 1) + mp0 + q0 = 0. (0.49)




a0f (m) = 0,

a1f (m + 1) + a0(mp1 + q1) = 0,

a2f (m + 2) + a0(mp2 + q2) + a1[(m + 1)p1 + q1] = 0,


· · · .


m(m − 1) + mp0 + q0 = 0. (0.49)



The remaining equations give a1 in terms of a0, a2 in terms of a0 and a1,and so on. The an’s, are therefore determined in terms of a0 for eachchoice of m unless f (m + n) = 0 for some positive integer n, in which casethe process breaks off.

Thus, if m1 = m2 + n for some integer n ≥ 1 , the choice m = m1 gives aformal solution but in general m = m2 does not, sincef (m2 + n) = f (m1) = 0.

If m1 = m2 we also obtain only one formal solution. In all other caseswhere m1 and m2 are real numbers, this procedure yields two independentformal solutions.

It is possible, of course, for m1 and m2 to be conjugate complex numbers,but we do not discuss this case because an adequate treatment would leadus too far into complex analysis.

These ideas are formulated more precisely in the following theorem.


























Theorem A. Assume that x = 0 is a regular singular point of thedifferential equation (0.45) and that the power series expansions (0.47) ofxP(x) and x2Q(x) are valid on an interval lxl < R with R > 0. Let theindicial equation (0.49) have real roots m1 and m2 with m2 ≤ m1. Thenequation (0.45) has at least one solution

y1 = xm1

∞∑n=0

anxn (a0 6= 0) (0.50)

on the interval 0 < x < R, where the an’s are determined in terms of a0 bythe recursion formula (0.48) with m replaced by m1, and the series∑

anxn converges for |x | < R. Furthermore, if m1 −m2 is not zero or apositive integer, then equation (0.45) has a second independent solution

y2 = xm2

∞∑n=0

anxn (a0 6= 0) (0.51)

on the same interval, where in this case an’s are determined in terms of a0by formula (0.48) with m replaced by m2 and again the series

∑anxn

converges for |x | < R.Pradeep Boggarapu (Dept. of Maths) Review of Power Series September 28, 2015 61 / 65

In view of what we have already done, the proof of this theorem can becompleted by showing that in each case the series

∑anxn converges on

the interval |x | < R which we are not going to prove now.

If you are interested the you can be find this remaing proof in theAppendix A of Chapter 5 in the text book.

Theorem A unfortunately fails to answer the question of how to find asecond solution when th defference m1 −m2 is zero or positive integer.

In order to convey an idear of the possibilities here, we distinguish threecases.

Case A. If m1 = m2, there can not exist a second Frobenius seriessolution.

The other two cases deal with the situation when m1 −m2 = n0 orm1 = m2 + n0 for some positive integer n0. In veiw of recursion formula(0.48) we have that

anf (m2 + n) = −a0(mpn + qn)− · · · − an−1[(m + n− 1)p1 + q1]. (0.52)




















































Since f (m2 + n0) = f (m1) = 0, the above recursion formula will not helpus to find an0 . The next two case deal with this problem.

Case B. If the right side of (0.52) is not zero for n = n0. Asf (m + n0) = 0 is zero there is no possible way of continuing thecalculation of the coefficient an0 and further coefficients and hence therecan not exist a second Frobenius series solution.

Case C. If the right side of (0.52) happens to be zero for n = n0 where asf (m + n0) = 0 then an0 is unrestricted and can be assigned any valuewhatever.

In particular, we can put an0 = 0 and continue to compute the coefficientswithout any further difficulties. Hence in this case there does exist asecond Frobenius series solution.

In Case A and Case B, there can not exist second Frobenious seriessolution, so we need to find a second solution in any other way.


























Possiblly we can use the first Frobenius series solutiony1 = xm1(a0 + a1x + a2x2 + · · · )to find second solution y2 via the formula

y2(x) = v(x)y1(x) with v(x) =

∫1

y21

e−∫P(x)dxdx or

v ′ =1

y21

e−∫P(x)dx


Thank you for your attention


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Frobenius Series Solutions