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Solution of linear differential equations by power series. Solutions about ordinary points and singular points.

Introduction. Not every differential equation can be solved — a solution may not exist. There may be no function that satisfies the differential equation. If a solution does exist it may not be possible to express it in closed form in terms of the elementary functions familiar from calculus (i.e. the algebraic, trigonometric, exponential, etc. functions). There are differential equations of great importance in higher mathematics and engineering that cannot be solved in terms of elementary functions. In such cases one must turn to approximate methods such as power series. The solutions to many differential equations are expressible in terms of a power series of the form

1) y = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ....... .

We will now consider methods for finding solutions in terms of a power series. We remark, however, that an equation may be solvable but yet not solvable in terms of a series of the form 1). In such a case we must seek a solution in some other form. One such form is

2) y = x^{c}(a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ....... )

which is a generalization of 1), since 2) gives 1) when c = 0. For example, if a particular differential equation had the solution

an assumption of a solution of type 1) would not give the solution but an assumption of a
solution of type 2) would give it since 3) is a special case of 2) with c = 1/2, a_{0} = a_{1} = .... = 1. A
series of type 2) is called a Frobenius type series.

Solution of linear equations by power series

Def. Ordinary point, singular point. Given a linear differential equation with polynomial coefficients

a point x = x_{0} is called an ordinary point if b_{0}(x_{0})
0. If b_{0}(x_{0}) = 0 the point is called a
singular point.

Theorem 1. Let x = 0 be an ordinary point of the linear equation

1) b_{0}(x)y" + b_{1}(x)y' + b_{2}(x)y = 0

with polynomial coefficients. Then there is a solution of the form

2) y = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + .......

containing two arbitrary coefficients a_{0} and a_{1} which converges inside a circle with center at x = 0
that extends out to the nearest singular point. Solution 2) is a solution of 1) but it is not the only
solution. The complete solution of 1) is given in the following theorem:

Theorem 2. Let x = 0 be an ordinary point of the linear equation

1) b_{0}(x)y" + b_{1}(x)y' + b_{2}(x)y = 0

with polynomial coefficients. Then this equation can be solved in series about point x = 0 as

3) y = A{series in x} + B{series in x}

in which A and B are arbitrary constants and the two “series in x” are two different series to be determined. The two series are linearly independent and both are convergent in a region surrounding x = 0.

The solution 3) is the complete solution of 1).

Theorem 3. The complete solution of the linear equation

4) b_{0}(x)y" + b_{1}(x)y' + b_{2}(x)y = G(x)

consists of its complementary function plus any particular solution. The complementary function of 4) consists of the complete solution of the homogeneous equation

b_{0}(x)y" + b_{1}(x)y' + b_{2}(x)y = 0

**********************************************************************

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Methods of solution

Method of Undetermined Coefficients. The Method of Undetermined Coefficients is the most common method of solving a differential equation by power series. Suppose we wish to solve the equation with polynomial coefficients

about an ordinary point x = 0 using a power series. We proceed as follows:

Step 1. Assume a solution of the form

2) y = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + .......

where a_{0}, a_{1}, .... ,a_{n-1 }are arbitrary constants and the remaining a’s are constants to be determined.

Step 2. Substitute the assumed series 2) into the differential equation 1). This substitution will yield an equation of form

3) A_{0}(a_{0}, a_{1, }a_{3, ... }) + A_{1}(a_{0}, a_{1, }a_{3 ... }) x + A_{2}(a_{0}, a_{1, }a_{3 ... }) x^{2} + A_{3}(a_{0}, a_{1, }a_{3 ... }) x^{3} + .......
0

This equation represents an identity and for this power series to vanish identically over any interval each coefficient must be zero. Thus

A_{0}( a_{0}, a_{1, }a_{3 ... }) = 0

A_{1}(a_{0}, a_{1, }a_{3 . .. }) = 0

4) A_{2}(a_{0}, a_{1, }a_{3 ... }) = 0

..............................

..............................

A_{n}(a_{0}, a_{1, }a_{3 ... }) = 0

..............................

Step 3. From equations 4) we can compute the values of the a’s._{ } We will now give an example
that will clarify details of the procedure.

Technique for finding solutions near an ordinary point by the Method of Undetermined Coefficients.

Problem 1. Solve the equation

1) (1 - x^{2})y" - 6xy' - 4y = 0

near the ordinary point x = 0.

Solution. This equation has singular points at x = 1 and x = -1. Thus there will be a solution

2) y = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + .......

valid for |x| < 1.

Step 1. Obtain basic equation. We wish to write down the equation that is obtained when 2) is
substituted into 1). We can do this by inspection. Simply apply 1) to the general term a_{n}x^{n} of 2)
i.e substitute the general term a_{n}x^{n} of 2) into 1).

Let us write 1) as

3) y" - x^{2}y" - 6xy' - 4y = 0

Then substitution of a_{n}x^{n} into 3) gives

y"
n(n - 1)a_{n}x^{n -2}

x^{2}y"
n(n - 1)a_{n}x^{n }

6xy'
6na_{n}x^{n}

4y
4a_{n}x^{n}

The equation obtained by substituting 2) into 1) is then

Combining terms we get

We now factor the coefficient in the second series to get

Step 2. Do index shift. We wish now to do an index shift on 4). We wish to alter 4) into an equivalent equation through a device called an index shift. The object of the index shift is to make the exponents of x in both series the same.

We now shift the index in the second series by replacing n everywhere with (n - 2). The n = 0 in the summation sign then becomes n - 2 = 0, or n = 2. With this index shift 4) becomes

We now note that the equation

corresponds to the equation

A_{n}(a_{0}, a_{1, }a_{3 ... }) = 0

discussed above. Equation 5) represents an identity and in order the series to vanish identically the coefficients in the series must be zero.

Step 3. Get recurrence relation. Let us now write out 6) for different values of n. For n = 0 and n = 1 the second series has not yet started and we get contributions only from the first series.

We have

As expected, we see that a_{0} and a_{1} are arbitrary. We can rewrite the last equation above as

Equation 8) is called a recurrence relation. It gives a_{n} in terms of preceding a’s. In this
particular case each a is determined by the a with a subscript two lower than its own. Thus all
the a’s are ultimately expressible in terms of either a_{0} or a_{1} according to whether the a has an even
or odd subscript.

Step 4. Set up columns and use multiplication device to obtain an expression for the a’s. We now obtain expressions for the values of the a’s utilizing a multiplication device that we shall shortly explain.

Let us list the iterated instances of 8) in two columns (two columns because the subscripts in 8) differ by two) listing n = 2, 4, 6, .... in the first column and n = 3, 5, 7, .... in the second column.

We now employ our multiplication device on the first column to obtain an expression for a_{2k}.
The device consists simply of this: Multiply all equations of the column together and then
perform cancellations and simplifications on the result. Multiplying the equations together we
get

Cancellation and simplification in 9) gives

10) a_{2k} = (k + 1)a_{0} k = 1, 2, 3, ....

We now do the same thing with the second column. Multiplying the equations together we get

which, on simplification, gives

Step 5. Substitute the expressions for the coefficients into the assumed series 2) to obtain the solution. We now substitute the expressions for the coefficients into series 2). Because of the nature of the expressions for the a’s the final solution will be written as the sum of two series, one with even subscripts and the other with odd subscripts. The solution will be of the form

Substituting the values of a_{2k} and a_{2k+1} into 13) we get as the general solution

In this example the equation was homogeneous and of second order. Raising the order introduces nothing except additional labor. A nonhomogeneous equation with right member G(x) that is an analytic function is no worse to handle than a homogeneous equation.

Problem 2. Solve the equation

1) y" + (x - 1)^{2}y' - 4(x - 1)y = 0

about the ordinary point x = 1.

Solution. To solve an equation about a point x = a means to obtain a solution valid in a region surrounding the point a with the solution expressed in powers of (x - a).

We first translate the axes, setting x - 1 = v. Thus when v = 0, x = 1. With this translation 1) becomes

In a pure translation, x - a = v, the following holds

a relevant fact in transforming 1) into 2).

We now solve 2) about point v = 0.

As usual we assume a solution

3) y = a_{0} + a_{1}v + a_{2}v^{2} + a_{3}v^{3} + ....... .

The equation obtained by substituting 3) into 2) is

[obtained by inspection by substituting the general term a_{n}v^{n} of 3) into 2)].

Collecting like terms we get

We now do a shift of index from n to (n - 3) in the second series to obtain

We now give the expressions for the coefficients:

or

which is the recurrence relation.

a_{0} and a_{1} are arbitrary constants and a_{2} = 0.

The a’s fall into three groups, those coming from a_{0}, those coming from a_{1} and those coming
from a_{2.} We use three columns:

Using our multiplication device we multiply the equations of the first column together and obtain

As for the a’s that are determined by a_{1}, we note that a_{4} = a_{1}/4 and all the rest are zero. Since a_{2}
= 0, all a’s determined by it are zero.

We thus have for y

Substituting v = x - 1 into 7) we obtain the general solution of 1):

Summary of procedure. Ordinary point.

1. Assume a solution of the form

y = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + .......

2. Substitute assumed solution into differential equation to get basic equation L(y) = 0.

3. Collect in like powers of x

4. Shift index

6. Derive recurrence relation.

7. Set up columns and use multiplication devise to obtain a_{n}. Choose a_{0} = 1.

8. Write solution.

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Regular and irregular singular points. Let x = x_{0} be a singular point of the equation

1) b_{0}(x)y" + b_{1}(x)y' + b_{2}(x)y = 0

with polynomial coefficients. Then b_{0}(x_{0}) = 0, which implies that b_{0}(x_{0}) has a factor (x - x_{0}) to
some power. Let us now divide 1) by b_{0}(x), putting 1) into the form

2) y" + p(x)y' + q(x)y = 0

where p(x) = b_{1}(x)/b_{0}(x) and q(x) = b_{2}(x)/b_{0}(x). Here p(x) and q(x) are rational functions where
at least one and possibly both have denominators containing the factor (x - x_{0}). If the
denominator of p(x) does not contain the factor (x - x_{0}) to a power higher than one, and if
the denominator of q(x) does not contain the factor (x - x_{0}) to a power higher than two, then
x = x_{0} is called a *regular singular point *of 1). If a singular point is not a regular singular
point it is called an *irregular singular point*.

Singular points of equations of higher order are classified in a similar way. For example, the
singular point x = x_{0} of the equation

3) y"' + p_{1}(x)y" + p_{2}(x)y' + p_{3}(x)y = 0

is called regular if the factor (x - x_{0}) does not appear in the denominator of p_{1}(x) to a power
higher than one, of p_{2}(x) to a power higher than two and of p_{3}(x) to a power higher than three. If
the singular point is not regular, it is called irregular.

Example 1. Classify the singular points, in the finite plane, of the equation

x^{4}(x^{2} + 1)(x - 3)y" + x^{3}y' + (x + 1)y = 0

Solution. We have

The regular singular points are at x = i, -i, 3. The irregular singular points are at x = 0.

Theorem 4. If x = a is a regular singular point of the differential equation

b_{0}(x)y" + b_{1}(x)y' + b_{2}(x)y = 0

a solution of the form

4) y = (x - a)^{c }[a_{0} + a_{1}(x - a) + a_{2}(x - a)^{2} + a_{3}(x - a)^{3} + ....... ]

always exists.

If x = a is an irregular singular point, solutions of the form 4) may or may not exist.

We will deal with the problem of finding solutions near regular singular points. The problem of finding solutions near irregular singular points is substantially more difficult and we won’t deal with it.

Technique for finding solutions in the vicinity of a regular singular point by the Method of Undetermined Coefficients.

We will consider the problem of finding solutions of the equation

1) b_{0}(x)y" + b_{1}(x)y' + b_{2}(x)y = 0

about a singular point x = 0. If we wish to obtain a solution about a point x = a, we first translate the origin to that point by a change of variable. We thus reduce the problem of finding a solution about a point x = a to one of finding a solution about the point x = 0.

Let us denote the left member of 1) by

L(y) = b_{0}(x)y" + b_{1}(x)y' + b_{2}(x)y .

We will find the notation useful in referring to the left member.

Solving equation 1) in the vicinity of a regular singular point involves solving an equation called an indicial equation that arises at a certain point in the procedure. Once we have the roots of the indicial equation subsequent procedure depends on the nature of the roots of the indicial equation.

We will now outline of the procedure up to the point where we obtain the indicial equation. The procedure after that is best explained by example.

Step 1. Assume solution. Assume a solution of the form

2) y = x^{c}[a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ....... ] = a_{0}x^{c} + a_{1}x^{c+1 }+ a_{2}x^{c+2 } + a_{3}x^{c+3 } + .......

where a_{0}
0.

Step 2. Substitute assumed solution into differential equation. Substitute the assumed series 2) into the differential equation 1), collect like terms, and do an index shift in such a way as to bring all the exponents of x down to the smallest one present. This will yield an equation of form

3) L(x) = A_{0}(c, a_{0 })x^{m} + A_{1}(c, a_{0}, a_{1}) x^{m+1 } + A_{2}(a_{0}, a_{1, }a_{3 ... }) x^{m+2} + A_{3}(a_{0}, a_{1, }a_{3 ... }) x^{m+3 } +
.......
0

This equation is an identity and because of that all of the coefficients A_{0}(c, a_{0 }), A_{1}(c, a_{0}, a_{1}),
A_{2}(a_{0}, a_{1, }a_{3 ... }), ....... must vanish.

Step 3. Compute roots of indicial equation. The coefficient of the lowest power of x, A_{0}(c, a_{0
}), will be of the form

A_{0}(c, a_{0 }) = f(c)a_{0}

where f(c) is an algebraic expression of the second degree in c. Because 3) is an identity this
coefficient must vanish. Since a_{0}
0, the only way for it to vanish is for f(c) to be zero. Thus
the coefficient A_{0}(c, a_{0 }) will vanish only at those values of c for which f(c) = 0. The equation

4) f(c) = 0

is called the indicial equation. So we now compute its roots c_{1} and c_{2}. Solutions of our
equation will only occur at those values of c.

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Theorem 5. Let x = 0 be a regular singular point of the equation

1) y" + p(x)y' + q(x)y = 0

where p and q are rational functions of x. Let

r(x) = xP(x)

s(x) = x^{2}Q(x)

Now let

a = r(0)

b = s(0)

i.e. the values of r(x) and s(x) at x = 0. Then the indicial equation is given by

c^{2} + (a - 1)c + b = 0

Theorem 6. Let x = 0 be a regular singular point of the equation

b_{0}(x)y" + b_{1}(x)y' + b_{2}(x)y = 0 .

Then the equation has a general solution either of the form

or of the form

where c_{1} and c_{2} are roots of the indicial equation and A and B are arbitrary constants.

The infinite series that occur in the above general solutions converge in at least the annular region of the complex plane bounded by two circles centered at x = 0, one of arbitrarily small radius, the other extending to the singular point nearest x = 0.

The form of the general solution of the equation depends on the nature of the roots c_{1} and c_{2} of
the indicial equation according to the following three cases:

Case 1. The roots c_{1} and c_{2} are distinct and the difference c_{2} - c_{1 } is not an integer. The
general solution is of the form

y = Ay_{1} + By_{2}

where A and B are arbitrary constants and y_{1} and y_{2} are particular solutions given by

The a_{i} coefficients are determined by substituting c = c_{1} into the recurrence relation and the b_{i}
coefficients are determined by substituting c = c_{2} into the recurrence relation.

Case 2. The roots c_{1} and c_{2} are equal. The general solution is of the form

y = Ay_{1} + By_{2}

where A and B are arbitrary constants and y_{1} and y_{2} are particular solutions given by

where the variable c in the a’s is carried through to the end,

a'_{n} = ∂a_{n} /∂c,

and y_{1} and y_{2} are evaluated at c = c_{1} = c_{2}.

Case 3. The roots c_{1} and c_{2} are distinct and the difference c_{2} - c_{1 } is an integer. The general
solution may be of the general form

or of the form

depending on the equation.

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Case 1. Roots of indicial equation differ by non-integer.

Problem. Solve the equation

1) 2xy" + (1 + x)y' - 2y = 0

about the point x = 0.

Solution. This equation has a regular singular point at x = 0.

Step 1. Assume solution. Assume a solution of the form

2) y = x^{c}[a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ....... ] = a_{0}x^{c} + a_{1}x^{c+1 }+ a_{2}x^{c+2 } + a_{3}x^{c+3 } + .......

where a_{0}
0.

Step 2. Substitute assumed solution into differential equation. Substitute the assumed series 2) into the differential equation 1), collect like terms, and do an index shift in such a way as to bring all the exponents of x down to the smallest one present.

Substitution of 2) into 1) gives

or

We now shift the index to bring all exponents down to the smallest one present by replacing all n’s in the second summation by (n -1)

Step 3. Compute roots of indicial equation. The total coefficient of each power of x in 4) must vanish. The equations for determining c and the a’s are

5) n = 0: c(2c - 1)a_{0} = 0

n
1: (n + c)(2n + 2c - 1)a_{n} + (n + c - 3)a_{n-1 } = 0

The indicial equation is then given by

c(2c - 1) = 0

Its roots are c_{1} = 0 and c_{2} = 1/2. At these values of c and only at these values will solutions to the
equation exist since only at these values will the n = 0 coefficient be zero. We note that the
difference c_{2} - c_{1} is not an integer. We are thus dealing with the nonintegral case. In this case
our method will always give two linearly independent solutions of the form 2), one for each root
of c.

Computation of solution corresponding to the first root.

Step 4. Derive recurrence relation for first root. We will do the root c = 1/2 first. Substituting c = 1/2 into the second equation of 5) we obtain

(n + 1/2)(2n + 1 - 1)a_{n} + (n + 1/2 - 3)a_{n-1 } = 0 n
1

which gives the recurrence relation

Step 5. Set up column and use multiplication devise to obtain a_{n}. Substituting into 6) we get,
for different values of n

a_{0} arbitrary

-------------

Multiplying these equations together and simplifying we get

which can be simplified to

Step 6. Write solution. Using a_{0} = 1, the a_{n} from 7) and c = 1/2, the solution is given by

Computation of solution corresponding to the second root.

In computing the solution for the second root it is obvious that the a’s will be different from those of the first root. To avoid confusion let us therefore change notation and use b’s in place of a’s.

Step 7. Derive recurrence relation for second root. Substituting c = 0 into the second equation of 5) we obtain

(n)(2n- 1)b_{n} + (n - 3)b_{n-1 } = 0 n
1

which gives the recurrence relation

Step 8. Set up column and use multiplication devise to obtain b_{n}. Substituting into 9) we get,
for different values of n

b_{0} arbitrary

-------------

We note that b_{3} and all subsequent b’s are zero. Using b_{0} = 1, we obtain for b_{1} and b_{2}

b_{1} = 2

b_{2} = 1/3

The second solution is then

The solutions we have derived, y_{1} and y_{2}, are linearly independent particular solutions to the
equation and the general solution is given by

y = Ay_{1} + By_{2}

Summary of procedure. Difference of indicial roots a non-integer.

1. Assume a solution of the form

y = x^{c}[a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ....... ] = a_{0}x^{c} + a_{1}x^{c+1 }+ a_{2}x^{c+2 } + a_{3}x^{c+3 } + .......

2. Substitute assumed solution into differential equation to get basic equation L(y) = 0.

3. Collect in like powers of x

4. Shift index

5. Compute roots of indicial equation.

6. Derive recurrence relation using first root.

7. Set up column and use multiplication devise to obtain a_{n}(c). Choose a_{0} = 1.

8. Write solution.

9. Repeat steps 6 - 8 using second root to obtain second solution y_{2}.

10. General solution is

y = Ay_{1} + By_{2}

where A and B are arbitrary constants

********************************************************************

Case 2. Roots of indicial equation equal.

Problem. Solve the equation

1) x^{2}y" + 3xy' + (1 - 2x)y = 0

about the point x = 0.

Solution. This equation has a regular singular point at x = 0.

Step 1. Assume solution. Assume a solution of the form

2) y = x^{c}[a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ....... ] = a_{0}x^{c} + a_{1}x^{c+1 }+ a_{2}x^{c+2 } + a_{3}x^{c+3 } + .......

where a_{0}
0.

Step 2. Substitute assumed solution into differential equation. Substitute the assumed series 2) into the differential equation 1), collect like terms, and do an index shift in such a way as to bring all the exponents of x down to the smallest one present.

Substitution of 2) into 1) gives

or

or

We now shift the index to bring all exponents down to the smallest one present by replacing all n’s in the second summation by (n -1)

Step 3. Compute roots of indicial equation. The coefficient of the lowest power of x, A_{0}(c, a_{0
}), n = 0, is

A_{0}(c, a_{0 }) = (c+1)^{2}a_{0}

and so the indicial equation is

(c+1)^{2 } = 0 .

The roots of the indicial equation are c_{1} = -1, c_{2} = -1. We thus have the case of equal indicial
roots. Now we know that the general solution of our differential equation consists of a linear
combination of two linearly independent particular solutions. Using the method of the previous
case we can get from two equal roots only one particular solution. By what method might we
obtain another linearly independent solution? Luckily someone has discovered a method. We
proceed as follows.

1. We refrain from choosing a value for c for the present and remove, for the present, the requirement that the function L(y) equal zero.

2. We set equal to zero all the coefficients A_{i}(c, a_{0, }a_{1, }... ) of the various powers of x in 4)
except for the A_{0}(c, a_{0 }) coefficient of the n = 0 term.

With what we have done the function L(y) of 4) reduces to a single term, the A_{0}(c, a_{0 }), n = 0,
term i.e.

5) L(y) = (c+1)^{2}a_{0}x^{c}

Step 4. Derive recurrence relation. We now derive the recurrence relation retaining the variable c in the relation. For n 1 we have

(n + c + 1)^{2}a_{n} - 2a_{n -1 } = 0 n
1

or

Step 5. Set up column and use multiplication devise to obtain a_{n}. Substituting into 6) we get,
for different values of n

a_{0} arbitrary

-------------

Multiplying these equations together and simplifying we get

Step 6. Write function y(x, c). To obtain a particular solution we choose a_{0} = 1 and, from these
a’s, write down the function y(x, c):

where

Now by 5) above

9) L[y(x, c)] = (c+1)^{2}x^{c}

A solution of the original differential equation is a function y for which L(y) = 0. We note that a choice of c = -1 yields a solution since it makes the right member of 9) zero.

Let us now take the partial derivative of both sides of 9) with respect to c. We get

We make the observation that the value c = -1 will also reduce the right side of this equation to zero. Might ∂ L[y(x, c)] / ∂c provide us with that second linearly independent solution that we desire? Indeed it will. We now make the following observation. The factor (c + 1) occurs squared in 9) and the fact that it is squared is an automatic consequence of the equality of the roots of the indicial equation. Thus whenever we have two equal indicial roots we can expect this squared factor. Moreover we know from calculus that when a function contains a factor to a certain power the derivative of that function contains the same factor to a power one lower than the original. Thus we can expect ∂ L[y(x, c)] / ∂c to provide us with a second solution whenever the indicial roots are equal. It can be shown that

As a consequence of all this it can be seen that two solutions of the equation L(y) = 0 are

We have y(x, c) from 7) and we need ∂y(x, c) /∂c. Taking the partial derivative with respect to c of 7) we get

which simplifies to

where

The solutions y_{1} and y_{2} will be obtained by putting c = -1 in equations 7) and 13) i.e. they are
given by

Thus we need to evaluate a_{n}(c) and a_{n}'(c) at c = -1. We know that

The following theorems are often useful in evaluating a_{n}'(c).

Theorem 7. Let

_{
}

where each of the u’s is a function of x. Let differentiation with respect to x be denoted by primes. Then

Theorem 8. Let

Then

Writing 16) as

a_{n}(c) = 2^{n} [(c + 2)^{-1} (c + 3)^{-1} ....... (c + n + 1)^{-1}] [ (c + 2)^{-1} (c + 3)^{-1} ....... (c + n + 1)^{-1}]

and applying Theorem 8 we get

Substituting c = -1 into 16) gives

and

A frequently used notation for the partial sum of a harmonic series is

Using this notation 18) can be written more simply as

The desired solutions are then

The general solution, valid for all finite x 0, is given by

y = Ay_{1} + By_{2}

where A and B are arbitrary constants.

Summary of procedure for equal roots.

1. Assume a solution of the form

y = x^{c}[a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ....... ] = a_{0}x^{c} + a_{1}x^{c+1 }+ a_{2}x^{c+2 } + a_{3}x^{c+3 } + .......

2. Substitute assumed solution into differential equation to get basic equation L(y) = 0.

3. Collect in like powers of x

4. Shift index

5. Compute roots of indicial equation.

6. Derive recurrence relation

7. Set up column and use multiplication devise to obtain a_{n}(c). Choose a_{0} = 1.

8. Derive expression for a_{n}'(c) from a_{n}(c)

9. Solutions are given by

********************************************************************

Case 3. Difference of indicial roots an integer. Non-logarithmic case.

Problem. Solve the equation

1) xy" - (4 + x)y' + 2y = 0

about the point x = 0.

Solution. This equation has a regular singular point at x = 0.

Step 1. Assume solution. Assume a solution of the form

^{c}[a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ....... ] = a_{0}x^{c} + a_{1}x^{c+1 }+ a_{2}x^{c+2 } + a_{3}x^{c+3 } + .......

where a_{0}
0.

Step 2. Substitute assumed solution into differential equation. Substitute the assumed series 2) into the differential equation 1), collect like terms, and do an index shift in such a way as to bring all the exponents of x down to the smallest one present.

Substitution of 2) into 1) gives

or

Step 3. Compute roots of indicial equation. The coefficient of the lowest power of x, A_{0}(c, a_{0
}), n = 0, is

A_{0}(c, a_{0 }) = c(c - 5)a_{0}

and so the indicial equation is

c(c - 5) = 0 .

The roots of the indicial equation are c_{1} = 0, c_{2} = 5. Thus we are dealing with the case where the
difference between indicial roots is an integer. In this case the solution may include a logarithm
or it may not. We don’t yet know. The two type solutions are handled differently. We don’t
currently know which we are dealing with. To find out whether we need to use the non-logarithmic or logarithmic procedure we proceed as follows: Let s = c_{2} - c_{1}, where c_{2} is the
larger of the roots, making s positive. It may be possible to obtain two power series solutions,
one starting with an x^{0} term and the other starting with an x^{s} term. If it happens that both a_{0} and
a_{s} turn out to be arbitrary we can obtain the general solution by this method. Otherwise we must
use the logarithmic procedure. If both a_{0} and a_{s} are arbitrary we assume a solution of the form 2)
above and use the smaller root i.e.c_{1}. The series we get will then include both solutions.

Step 4. Choose smallest root, c_{1}. Write out the relation for the coefficients up through n = ,
c_{2} - c_{1} substituting in the smallest root, c_{1}, for the variable c. Write recurrence relation.
Substituting c = c_{1} = 0 into 3) above gives

The relationships for the coefficients are

n = 0; 0·a0 = 0 (a_{0} arbitrary)

n
1: n(n - 5)a_{n} - (n - 3)a_{n -1 } = 0

Because division by (n - 5) cannot be accomplished until n > 5, we write the relationships out
explicitly through the critical one for a_{5}.

n = 1: -4a_{1} + 2a_{0} = 0

n = 2: -6a_{2} + a_{1} = 0

n = 3: -6a_{3} + 0·a_{2} = 0

n = 4: -4a_{4} - a_{3} = 0

n = 5: 0·a_{5} - 2a_{4} = 0

Recurrence relation:

From the above relationships we derive

Thus we see that a_{5} is arbitrary and we can use the non-logarithmic method. The a_{n} for n > 5 will
be obtained in the usual way.

Step 5. Set up column and use multiplication devise to obtain a_{n}. Substituting into 5) we get,
for different values of n

a_{5} arbitrary

-------------

Multiplying these equations together and simplifying we get

The general solution is then

where a_{0} and a_{5} are arbitrary constants.

********************************************************************

Case 4. Difference of indicial roots an integer. Logarithmic case.

Problem. Solve the equation

1) x^{2}y" + x(1 - x)y' - (1 + 3x)y = 0

about the point x = 0.

Solution. This equation has a regular singular point at x = 0.

Step 1. Assume solution. Assume a solution of the form

^{c}[a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ....... ] = a_{0}x^{c} + a_{1}x^{c+1 }+ a_{2}x^{c+2 } + a_{3}x^{c+3 } + .......

where a_{0}
0.

Substitution of 2) into 1) gives

Doing an index shift gives

Step 3. Compute roots of indicial equation. The coefficient of the lowest power of x, A_{0}(c, a_{0
}), n = 0, is

A_{0}(c, a_{0 }) = (c + 1)(c - 1)a_{0}

and so the indicial equation is

(c + 1)(c - 1) = 0.

The roots of the indicial equation are c_{1} = -1, c_{2} = 1. This represents a case where the difference
between indicial roots is an integer. The solution may involve a logarithm or it may not. We
don’t yet know. However, we note that there is no power series solution that starts with
and
so we suspect the presence of a logarithmic term.

Step 4. Choose smallest root, c_{1}. Write out the relation for the coefficients up through n =
c_{2} - c_{1} , substituting in the smallest root, c_{1}, for the variable c. Write recurrence relation.
Substituting c = c_{1} = -1 into 3) above gives

The relationships for the coefficients are

n = 0; 0·a_{0} - a_{-1} = 0

n
1: n(n - 2)a_{n} - (n + 1)a_{n -1 } = 0

n = 1: -a_{1} - 2a_{0} = 0

n = 2: 0 · a_{2} - 3a_{1} = 0

From the above we derive

a_{1} = -2a_{0}

0 · a_{2} = 3a_{1} = -6a_{0}

The only way these relationships could be satisfied is by choosing a_{0} = 0. And if we did that a_{2}
would be the only arbitrary constant. We would only get a single solution, that corresponding to
the large value of c, c_{2} = 1. A logarithmic solution is indicated.

We now proceed in the same manner that we did in obtaining a logarithmic solution for the case of equal roots:

1. We refrain from choosing a value for c for the present and remove, for the present, the requirement that the function L(y) equal zero.

2. We set equal to zero all the coefficients A_{i}(c, a_{0, }a_{1, }... ) of the various powers of x in 3)
except for the A_{0}(c, a_{0 }) coefficient of the n = 0 term.

With what we have done the function L(y) of 3) reduces to a single term, the A_{0}(c, a_{0 }), n = 0,
term.

We now write from 3) the recurrence relation, retaining the variable c.

Step 5. Set up column and use multiplication devise to obtain a_{n}. Substituting into 5) we get,
for different values of n

a_{0} arbitrary

-------------

Multiplying these equations together and simplifying we get

or

We now write down the function y(x, c):

or

Since all terms in the power series have been made to vanish this function y(x, c) must satisfy

8) L(y) = (c+1)(c-1)a_{0}x^{c}

Only one solution can be obtained from the larger root, c_{2} = 1. Two solutions could be obtained
from the smaller root, c_{1} = -1, if the right member of 8) contained the factor (c + 1)^{2}, instead of
just (c + 1) to the first power. However, a_{0} is still arbitrary so if we were to take a_{0} = (c + 1) we
would get the desired square in the right member of 8). This is what we do.

With a_{0} = (c + 1), 7) becomes

for which

L(y) = (c+1)^{2}(c-1)x^{c}

By the same argument that we used for the case of equal indicial roots we now assert that the following constitute two linearly independent solutions of our equation:

These solutions will have the form

where c_{1} is the smaller indicial root and a_{n}(c) is given by

One should perform the cancellation of the factor (c + 1) from the numerator and denominator in the terms of the series in 9). However, the factor (c + 1) does not enter the denominator until the term n = 2. Thus it is best to write out the terms that far separately. We rewrite 9) as

Step 6. Derive a_{n}'(c). We now derive a_{n}'(c), n
1 utilizing Theorem 8 from the “Case 2,
Equal roots” section. Before obtaining a_{n}'(c), however, we must replace the factor a_{0} that a_{n}(c)
contains with the factor (c - c_{1}) and perform any cancellations of this factor that may be possible
in some of the terms as we did in 13) above.

We are now in a position to evaluate y_{1} and y_{2} using 11) above at c = -1. Substituting into 11)
and using the expressions for a_{n}'(c) that we have derived we obtain

These results can be written more concisely as

Summary of procedure for difference of indicial roots an integer. Logarithmic case.

1. Assume a solution of the form

y = x^{c}[a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ....... ] = a_{0}x^{c} + a_{1}x^{c+1 }+ a_{2}x^{c+2 } + a_{3}x^{c+3 } + .......

2. Substitute assumed solution into differential equation to get basic equation L(y) = 0.

3. Collect in like powers of x

4. Shift index

5. Compute roots of indicial equation.

6. Derive recurrence relation

7. Set up column and use multiplication devise to obtain a_{n}(c). Choose a_{0} = c - c_{1} where c_{1} is the
smaller of the roots.

8. Replace the factor a_{0} that a_{n}(c) contains with the factor (c - c_{1}) and perform any cancellations of
this factor that may be possible in the various terms before deriving a_{n}'(c).

9. Derive expression for a_{n}'(c) from a_{n}(c)

10. Solutions are given by

References

1. Murray R. Spiegel. Applied Differential Equations.

2. James B. Scarborough. Differential Equations and Applications.

3. Frank Ayres. Differential Equations (Schaum).

4. Earl Rainville. Elementary Differential Equations.

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