In this section we begin to study series solutions of a homogeneous linear second order differential equation with a regular singular point at \(x_0=0\), so it can be written as
where \(A\), \(B\), \(C\) are polynomials and \(A(0)\ne0\).
We’ll see that (7.5.1) always has at least one solution of the form
where \(a_0\ne0\) and \(r\) is a suitably chosen number. The method we will use to find solutions of this form and other forms that we’ll encounter in the next two sections is called the method of Frobenius, and we’ll call them Frobenius solutions.
It can be shown that the power series \(\sum_{n=0}^\infty a_nx^n\) in a Frobenius solution of (7.5.1) converges on some open interval \((-\rho,\rho)\), where \(0<\rho\le\infty\). However, since \(x^r\) may be complex for negative \(x\) or undefined if \(x=0\), we’ll consider solutions defined for positive values of \(x\). Easy modifications of our results yield solutions defined for negative values of \(x\). (Exercise 54).
We’ll restrict our attention to the case where \(A\), \(B\), and \(C\) are polynomials of degree not greater than two, so (7.5.1) becomes
where \(\alpha_i\), \(\beta_i\), and \(\gamma_i\) are real constants and \(\alpha_0\ne0\). Most equations that arise in applications can be written this way. Some examples are
where we would multiply the last equation through by \(x\) to put it in the form (7.5.2). However, the method of Frobenius can be extended to the case where \(A\), \(B\), and \(C\) are functions that can be represented by power series in \(x\) on some interval that contains zero, and \(A_0(0)\ne0\) (Exercises 57 and 58).
The next two theorems will enable us to develop systematic methods for finding Frobenius solutions of (7.5.2).
Theorem 7.5.1
Let
and define
Suppose the series
converges on \((0,\rho)\). Then
on \((0,\rho),\) where
Proof We begin by showing that if \(y\) is given by (7.5.3) and \(\alpha\), \(\beta\), and \(\gamma\) are constants, then
where
Differentiating (3) twice yields
and
Multiplying (7.5.7) by \(x\) and (7.5.8) by \(x^2\) yields
and
Therefore
which proves (7.5.6).
Multiplying (7.5.6) by \(x\) yields
Multiplying (7.5.6) by \(x^2\) yields
To use these results, we rewrite
as
From (7.5.6) with \(p=p_0\),
From (7.5.9) with \(p=p_1\),
From (7.5.10) with \(p=p_2\),
Therefore we can rewrite (7.5.11) as
or
which implies (7.5.4) with \(\{b_n\}\) defined as in (7.5.5).
Theorem 7.5.2
Let
where \(\alpha_0\ne0,\) and define
Suppose \(r\) is a real number such that \(p_0(n+r)\) is nonzero for all positive integers \(n.\) Define
Then the Frobenius series
converges and satisfies
on the interval \((0,\rho),\) where \(\rho\) is the distance from the origin to the nearest zero of \(A(x)=\alpha_0+\alpha_1 x+\alpha_2 x^2\) in the complex plane\(.\) \((\)If \(A\) is constant, then \(\rho=\infty\).\()\)
If \(\{a_n(r)\}\) is determined by the recurrence relation (7.5.12) then substituting \(a_n=a_n(r)\) into (7.5.5) yields \(b_0=p_0(r)\) and \(b_n=0\) for \(n\ge1\), so (7.5.4) reduces to (7.5.14). We omit the proof that the series (7.5.13) converges on \((0,\rho)\).∎
If \(\alpha_i=\beta_i=\gamma_i=0\) for \(i=1\), \(2,\) then \(Ly=0\) reduces to the Euler equation
Theorem 7.4.3 shows that the solutions of this equation are determined by the zeros of the indicial polynomial
Since (7.5.14) implies that this is also true for the solutions of \(Ly=0\), we’ll also say that \(p_0\) is the indicial polynomial of (7.5.2), and that \(p_0(r)=0\) is the indicial equation of \(Ly=0\). We’ll consider only cases where the indicial equation has real roots \(r_1\) and \(r_2\), with \(r_1\ge r_2\).
Theorem 7.5.3
Let \(L\) and \(\{a_n(r)\}\) be as in Theorem \(\ref{thmtype:7.5.2},\) and suppose the indicial equation \(p_0(r)=0\) of \(Ly=0\) has real roots \(r_1\) and \(r_2,\) where \(r_1\ge r_2.\) Then
is a Frobenius solution of \(Ly=0\). Moreover\(,\) if \(r_1-r_2\) isn’t an integer then
is also a Frobenius solution of \(Ly=0,\) and \(\{y_1,y_2\}\) is a fundamental set of solutions.
Proof Since \(r_1\) and \(r_2\) are roots of \(p_0(r)=0\), the indicial polynomial can be factored as
Therefore
which is nonzero if \(n>0\), since \(r_1-r_2\ge0\). Therefore the assumptions of Theorem 7.5.2 hold with \(r=r_1\), and (7.5.14) implies that \(Ly_1=p_0(r_1)x^{r_1}=0\).
Now suppose \(r_1-r_2\) isn’t an integer. From (7.5.15),
Hence, the assumptions of Theorem 7.5.2 hold with \(r=r_2\), and (7.5.14) implies that \(Ly_2=p_0(r_2)x^{r_2}=0\). We leave the proof that \(\{y_1,y_2\}\) is a fundamental set of solutions as an exercise (Exercise 52). ∎
It isn’t always possible to obtain explicit formulas for the coefficients in Frobenius solutions. However, we can always set up the recurrence relations and use them to compute as many coefficients as we want. The next example illustrates this.
Example 7.5.1
Find a fundamental set of Frobenius solutions of
Compute just the first six coefficients \(a_0\),…, \(a_5\) in each solution.
Solution For the given equation, the polynomials defined in Theorem 7.5.2 are
The zeros of the indicial polynomial \(p_0\) are \(r_1=-3/2\) and \(r_2=-2\), so \(r_1-r_2=1/2\). Therefore Theorem 7.5.3 implies that
form a fundamental set of Frobenius solutions of (7.5.16). To find the coefficients in these series, we use the recurrence relation of Theorem 7.5.2; thus,
Setting \(r=-3/2\) in these equations yields
and setting \(r=-2\) yields
Calculating with (7.5.18) and (7.5.19) and substituting the results into (7.5.17) yields the fundamental set of Frobenius solutions
Special Cases With Two Term Recurrence Relations
For \(n\ge2\), the recurrence relation (7.5.12) of Theorem 7.5.2 involves the three coefficients \(a_n(r)\), \(a_{n-1}(r)\), and \(a_{n-2}(r)\). We’ll now consider some special cases where (7.5.12) reduces to a two term recurrence relation; that is, a relation involving only \(a_n(r)\) and \(a_{n-1}(r)\) or only \(a_n(r)\) and \(a_{n-2}(r)\). This simplification often makes it possible to obtain explicit formulas for the coefficents of Frobenius solutions.
We first consider equations of the form
with \(\alpha_0\ne0\). For this equation, \(\alpha_2=\beta_2=\gamma_2=0\), so \(p_2\equiv0\) and the recurrence relations in Theorem 7.5.2 simplify to
Example 7.5.2
Find a fundamental set of Frobenius solutions of
Give explicit formulas for the coefficients in the solutions.
Solution For this equation, the polynomials defined in Theorem 7.5.2 are
The zeros of the indicial polynomial \(p_0\) are \(r_1=1/3\) and \(r_2=-1\), so \(r_1-r_2=4/3\). Therefore Theorem 7.5.3 implies that
form a fundamental set of Frobenius solutions of (7.5.21). To find the coefficients in these series, we use the recurrence relationss (7.5.20); thus,
Setting \(r=1/3\) in (7.5.22) yields
By using the product notation introduced in Section 7.2 and proceeding as we did in the examples in that section yields
Therefore
is a Frobenius solution of (7.5.21).
Setting \(r=-1\) in (7.5.22) yields
so
Therefore
is a Frobenius solution of (7.5.21), and \(\{y_1,y_2\}\) is a fundamental set of solutions. ∎
We now consider equations of the form
with \(\alpha_0\ne0\). For this equation, \(\alpha_1=\beta_1=\gamma_1=0\), so \(p_1\equiv0\) and the recurrence relations in Theorem 7.5.2 simplify to
Since \(a_1(r)=0\), the last equation implies that \(a_n(r)=0\) if \(n\) is odd, so the Frobenius solutions are of the form
where
Example 7.5.3
Find a fundamental set of Frobenius solutions of
Give explicit formulas for the coefficients in the solutions.
Solution For this equation, the polynomials defined in Theorem 7.5.2 are
The zeros of the indicial polynomial \(p_0\) are \(r_1=2\) and \(r_2=1/2\), so \(r_1-r_2=3/2\). Therefore Theorem 7.5.3 implies that
form a fundamental set of Frobenius solutions of (7.5.25). To find the coefficients in these series, we use the recurrence relation (7.5.24); thus,
Setting \(r=2\) in (7.5.26) yields
so
Therefore
is a Frobenius solution of (7.5.25).
Setting \(r=1/2\) in (7.5.26) yields
so
Therefore
is a Frobenius solution of (7.5.25) and \(\{y_1,y_2\}\) is a fundamental set of solutions.
Remark
Thus far, we considered only the case where the indicial equation has real roots that don’t differ by an integer, which allows us to apply Theorem 7.5.3. However, for equations of the form (7.5.23), the sequence \(\{a_{2m}(r)\}\) in (7.5.24) is defined for \(r=r_2\) if \(r_1-r_2\) isn’t an even integer. It can be shown (Exercise 56) that in this case
form a fundamental set Frobenius solutions of (7.5.23)
.
Using Technology
As we said at the end of Section 7.2, if you’re interested in actually using series to compute numerical approximations to solutions of a differential equation, then whether or not there’s a simple closed form for the coefficents is essentially irrelevant; recursive computation is usually more efficient. Since it’s also laborious, we encourage you to write short programs to implement recurrence relations on a calculator or computer, even in exercises where this is not specifically required.
In practical use of the method of Frobenius when \(x_0=0\) is a regular singular point, we’re interested in how well the functions
approximate solutions to a given equation when \(r_i\) is a zero of the indicial polynomial. In dealing with the corresponding problem for the case where \(x_0=0\) is an ordinary point, we used numerical integration to solve the differential equation subject to initial conditions \(y(0)=a_0,\quad y'(0)=a_1\), and compared the result with values of the Taylor polynomial
We can’t do that here, since in general we can’t prescribe arbitrary initial values for solutions of a differential equation at a singular point. Therefore, motivated by Theorem 7.5.2 (specifically, (7.5.14)), we suggest the following procedure.
Let \(L\) and \(Y_n(x; r_i)\) be defined by
and
where the coefficients \(\{a_n(r_i)\}_{n=0}^N\) are computed as in \(\eqref{eq:7.5.12}\), Theorem \(\ref{thmtype:7.5.2}\). Compute the error
for various values of \(N\) and various values of \(x\) in the interval \((0,\rho)\), with \(\rho\) as defined in Theorem \(\ref{thmtype:7.5.2}\).
The multiplier \(x^{-r_i}/\alpha_0\) on the right of (7.5.27) eliminates the effects of small or large values of \(x^{r_i}\) near \(x=0\), and of multiplication by an arbitrary constant. In some exercises you will be asked to estimate the maximum value of \(E_N(x; r_i)\) on an interval \((0,\delta]\) by computing \(E_N(x_m;r_i)\) at the \(M\) points \(x_m=m\delta/M,\; m=1\), \(2\), …, \(M\), and finding the maximum of the absolute values:
(For simplicity, this notation ignores the dependence of the right side of the equation on \(i\) and \(M\).)
To implement this procedure, you’ll have to write a computer program to calculate \(\{a_n(r_i)\}\) from the applicable recurrence relation, and to evaluate \(E_N(x;r_i)\).
The next exercise set contains five exercises specifically identified by L that ask you to implement the verification procedure. These particular exercises were chosen arbitrarily you can just as well formulate such laboratory problems for any of the equations in any of the Exercises 1–10, 14-25, and 28–51
7.5 Exercises
This set contains exercises specifically identified by L that ask you to implement the verification procedure. These particular exercises were chosen arbitrarily you can just as well formulate such laboratory problems for any of the equations in Exercises 1–10, 14-25, and 28–51.
In Exercises 1–10 find a fundamental set of Frobenius solutions. Compute \(a_0\), \(a_{1}\) …, \(a_N\) for \(N\) at least \(7\) in each solution.
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C \(2x^2(1+x+x^2)y''+x(3+3x+5x^2)y'-y=0\)
Show answer
\(y_1=\dst{x^{1/2}\left(1-{1\over5}x-{2\over35}x^2 +{31\over315}x^3+\cdots\right)}\) \(y_2=\dst{x^{-1}\left(1+x+{1\over2}x^2-{1\over6}x^3+\cdots\right)}\);
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C \(3x^2y''+2x(1+x-2x^2)y'+(2x-8x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\left(1-{2\over3}x+{8\over9}x^2- {40\over81}x^3+\cdots\right)}\); \(y_2=\dst{1-x+{6\over5}x^2-{4\over5}x^3+\cdots}\)
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C \(x^2(3+3x+x^2)y''+x(5+8x+7x^2)y'-(1-2x-9x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\left(1-{4\over7}x-{7\over45}x^2+ {970\over2457}x^3+\cdots\right)}\); \(y_2=\dst{x^{-1}\left(1-x^2+{2\over3}x^3+\cdots\right)}\)
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C \(4x^2y''+x(7+2x+4x^2)y'-(1-4x-7x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/4}\left(1-{1\over2}x-{19\over104}x^2+ {1571\over10608}x^3+\cdots\right)}\); \(y_2=\dst{x^{-1}\left(1+2x-{11\over6}x^2-{1\over7}x^3+\cdots\right)}\)
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C \(12x^2(1+x)y''+x(11+35x+3x^2)y'-(1-10x-5x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\left(1-x+{28\over31}x^2-{1111\over1333}x^3 +\cdots\right)}\); \(y_2=\dst{x^{-1/4}\left(1-x+{7\over8}x^2-{19\over24}x^3+ \cdots\right)}\);
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C \(x^2(5+x+10x^2)y''+x(4+3x+48x^2)y'+(x+36x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/5}\left(1-{6\over25}x-{1217\over625}x^2+ {41972\over46875}x^3 +\cdots\right)}\); \(y_2=\dst{x-{1\over4}x^2-{35\over18}x^3+{11\over12}x^4+\cdots}\)
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C \(8x^2y''-2x(3-4x-x^2)y'+(3+6x+x^2)y=0\)
Show answer
\(y_1=\dst{x^{3/2}\left(1-x+{11\over26}x^2-{109\over1326}x^3+ \cdots\right)}\); \(y_2=\dst{x^{1/4}\left(1+4x-{131\over24}x^2+{39\over14}x^3+ \cdots\right)}\)
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C \(18x^2(1+x)y''+3x(5+11x+x^2)y'-(1-2x-5x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\left(1-{1\over3}x+{2\over15}x^2- {5\over63}x^3+\cdots\right)}\); \(y_2=\dst{x^{-1/6}\left(1-{1\over12}x^2+{1\over18}x^3+\cdots\right)}\)
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C \(x(3+x+x^2)y''+(4+x-x^2)y'+xy=0\)
Show answer
\(y_1=\dst{1-{1\over14}x^2+{1\over105}x^3+\cdots}\); \(y_2=\dst{x^{-1/3}\left(1-{1\over18}x-{71\over405}x^2+ {719\over34992}x^3+\cdots\right)}\)
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C \(10x^2(1+x+2x^2)y''+x(13+13x+66x^2)y'-(1+4x+10x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/5}\left(1+{3\over17}x-{7\over153}x^2- {547\over5661}x^3+\cdots\right)}\); \(y_2=\dst{x^{-1/2}\left(1+x+{14\over13}x^2-{556\over897}x^3+\cdots\right)}\)
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L The Frobenius solutions of
\[ 2x^2(1+x+x^2)y''+x(9+11x+11x^2)y'+(6+10x+7x^2)y=0 \]obtained in Example 7.5.1 are defined on \((0,\rho)\), where \(\rho\) is defined in Theorem 7.5.2. Find \(\rho\). Then do the following experiments for each Frobenius solution, with \(M=20\) and \(\delta=.5\rho\), \(.7\rho\), and \(.9\rho\) in the verification procedure described at the end of this section.
Compute \(\sigma_N(\delta)\) (see Eqn. (7.5.28)) for \(N=5\), \(10\), \(15\),…, \(50\).
Find \(N\) such that \(\sigma_N(\delta)<10^{-5}\).
Find \(N\) such that \(\sigma_N(\delta)<10^{-10}\).
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L By Theorem 7.5.2 the Frobenius solutions of the equation in Exercise 4 are defined on \((0,\infty)\). Do experiments (a), (b), and (c) of Exercise 11 for each Frobenius solution, with \(M=20\) and \(\delta=1\), \(2\), and \(3\) in the verification procedure described at the end of this section.
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L The Frobenius solutions of the equation in Exercise 6 are defined on \((0,\rho)\), where \(\rho\) is defined in Theorem 7.5.2. Find \(\rho\) and do experiments (a), (b), and (c) of Exercise 11 for each Frobenius solution, with \(M=20\) and \(\delta=.3\rho\), \(.4\rho\), and \(.5\rho\), in the verification procedure described at the end of this section.
In Exercises 14–25 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients in each solution.
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\(2x^2y''+x(3+2x)y'-(1-x)y=0\)
Show answer
\(y_1=\dst{x^{1/2}\sum_{n=0}^\infty{(-2)^n\over\prod_{j=1}^n(2j+3)}x^n}\); \(y_2=\dst{x^{-1}\sum_{n=0}^\infty{(-1)^n\over n!}x^n}\)
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\(x^2(3+x)y''+x(5+4x)y'-(1-2x)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\sum_{n=0}^\infty{(-1)^n\prod_{j=1}^n(3j+1)\over9^nn!}x^n}\); \(x^{-1}\)
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\(2x^2y''+x(5+x)y'-(2-3x)y=0\)
Show answer
\(y_1=\dst{x^{1/2}\sum_{n=0}^\infty {(-1)^n\over2^nn!}x^n}\); \(y_2=\dst{{1\over x^2}\sum_{n=0}^\infty{(-1)^n\over\prod_{j=1}^n(2j-5)} x^n}\)
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\(3x^2y''+x(1+x)y'-y=0\)
Show answer
\(y_1=\dst{x\sum_{n=0}^\infty{(-1)^n\over\prod_{j=1}^n(3j+4)}x^n}\); \(y_2=\dst{x^{-1/3}\sum_{n=0}^\infty{(-1)^n\over3^nn!}x^n}\)
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\(2x^2y''-xy'+(1-2x)y=0\)
Show answer
\(y_1=\dst{x\sum_{n=0}^\infty{2^n\over n!\prod_{j=1}^n(2j+1)}x^n}\); \(y_2=\dst{x^{1/2}\sum_{n=0}^\infty{2^n\over n!\prod_{j=1}^n(2j-1)}x^n}\)
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\(9x^2y''+9xy'-(1+3x)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\sum_{n=0}^\infty{1\over n!\prod_{j=1}^n(3j+2)} x^n}\); \(y_2=\dst{x^{-1/3}\sum_{n=0}^\infty{1\over n!\prod_{j=1}^n(3j-2)} x^n}\)
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\(3x^2y''+x(1+x)y'-(1+3x)y=0\)
Show answer
\(y_1=\dst{x\left(1+{2\over7}x+{1\over70}x^2\right)}\); \(y_2=\dst{x^{-1/3}\sum_{n=0}^\infty{(-1)^n\over3^nn!}\left(\prod_{j=1}^n {3j-13\over3j-4}\right) x^n}\)
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\(2x^2(3+x)y''+x(1+5x)y'+(1+x)y=0\)
Show answer
\(y_1=\dst{x^{1/2}\sum_{n=0}^\infty(-1)^n\left(\prod_{j=1}^n{2j+1 \over6j+1}\right); x^n}\) \(y_2=\dst{x^{1/3}\sum_{n=0}^\infty{(-1)^n\over9^nn!} \left(\prod_{j=1}^n(3j+1)\right)x^n}\)
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\(x^2(4+x)y''-x(1-3x)y'+y=0\)
Show answer
\(y_1=\dst{x\sum_{n=0}^\infty{(-1)^n(n+2)!\over2\prod_{j=1}^n(4j+3)}; x^n}\) \(y_2=\dst{x^{1/4}\sum_{n=0}^\infty{(-1)^n\over16^nn!}\prod_{j=1}^n(4j+5) x^n}\)
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\(2x^2y''+5xy'+(1+x)y=0\)
Show answer
\(y_1=\dst{x^{-1/2}\sum_{n=0}^\infty{(-1)^n\over n!\prod_{j=1}^n(2j+1)} x^n}\); \(y_2=\dst{x^{-1}\sum_{n=0}^\infty{(-1)^n\over n!\prod_{j=1}^n(2j-1)} x^n}\)
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\(x^2(3+4x)y''+x(5+18x)y'-(1-12x)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\sum_{n=0}^\infty{(-1)^n\over n!}\left(2\over9\right)^n\left(\prod_{j=1}^n(6j+5)\right) x^n}\); \(y_2=\dst{x^{-1}\sum_{n=0}^\infty(-1)^n2^n\left(\prod_{j=1}^n {2j-1\over3j-4}\right) x^n}\)
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\(6x^2y''+x(10-x)y'-(2+x)y=0\)
Show answer
\(y_1=4\dst{x^{1/3}\sum_{n=0}^\infty{1\over6^nn!(3n+4)} x^n}\); \(x^{-1}\)
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L By Theorem 7.5.2 the Frobenius solutions of the equation in Exercise 17 are defined on \((0,\infty)\). Do experiments (a), (b), and (c) of Exercise 11 for each Frobenius solution, with \(M=20\) and \(\delta=3\), \(6\), \(9\), and \(12\) in the verification procedure described at the end of this section.
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L The Frobenius solutions of the equation in Exercise 22 are defined on \((0,\rho)\), where \(\rho\) is defined in Theorem 7.5.2. Find \(\rho\) and do experiments (a), (b), and (c) of Exercise 11 for each Frobenius solution, with \(M=20\) and \(\delta=.25\rho\), \(.5\rho\), and \(.75\rho\) in the verification procedure described at the end of this section.
In Exercises 28–32 find a fundamental set of Frobenius solutions. Compute coefficients \(a_0\), …, \(a_N\) for \(N\) at least \(7\) in each solution.
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C \(x^2(8+x)y''+x(2+3x)y'+(1+x)y=0\)
Show answer
\(y_1=\dst{x^{1/2}\left(1-{9\over40}x+{5\over128}x^2-{245\over39936}x^3 +\cdots\right)}\); \(y_2=\dst{x^{1/4}\left(1-{25\over96}x+{675\over14336}x^2- {38025\over5046272}x^3 +\cdots\right)}\)
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C \(x^2(3+4x)y''+x(11+4x)y'-(3+4x)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\left(1+{32\over117}x-{28\over1053}x^2+ {4480\over540189}x^3+\cdots\right)}\); \(y_2=\dst{x^{-3}\left(1+{32\over7}x+{48\over7}x^2\right)}\)
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C \(2x^2(2+3x)y''+x(4+11x)y'-(1-x)y=0\)
Show answer
\(y_1=\dst{x^{1/2}\left(1-{5\over8}x+{55\over96}x^2 -{935\over1536}x^3+\cdots\right)}\); \(y_2=\dst{x^{-1/2}\left(1+{1\over4}x-{5\over32}x^2 -{55\over384}x^3+\cdots\right)}\).
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C \(x^2(2+x)y''+5x(1-x)y'-(2-8x)y\)
Show answer
\(y_1=\dst{x^{1/2}\left(1-{3\over4}x+{5\over96}x^2+{5\over4224}x^3 +\cdots\right)}\); \(y_2=\dst{x^{-2}\left(1+8x+60x^2-160x^3+\cdots\right)}\)
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C \(x^2(6+x)y''+x(11+4x)y'+(1+2x)y=0\)
Show answer
\(y_1=\dst{x^{-1/3}\left(1-{10\over63}x+{200\over7371}x^2- {17600\over3781323}x^3; +\cdots\right)}\); \(y_2=\dst{x^{-1/2}\left(1-{3\over20}x+{9\over352}x^2 -{105\over23936}x^3 +\cdots\right)}\)
In Exercises 33–46 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients in each solution.
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\(8x^2y''+x(2+x^2)y'+y=0\)
Show answer
\(y_1=\dst{x^{1/2}\sum_{m=0}^\infty{(-1)^m\over8^mm!}\left(\prod_{j=1}^m{ 4j-3\over8j+1}\right)x^{2m}}\); \(y_2=\dst{x^{1/4}\sum_{m=0}^\infty{(-1)^m\over16^mm!}\left(\prod_{j=1}^m{ 8j-7\over8j-1}\right)x^{2m}}\)
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\(8x^2(1-x^2)y''+2x(1-13x^2)y'+(1-9x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/2}\sum_{m=0}^\infty\left(\prod_{j=1}^m{8j-3\over8j+ 1}\right) x^{2m}}\); \(y_2=\dst{x^{1/4}\sum_{m=0}^\infty{1\over2^mm!}\left(\prod_{j=1}^m(2j-1) \right)x^{2m}}\)
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\(x^2(1+x^2)y''-2x(2-x^2)y'+4y=0\)
Show answer
\(y_1=\dst{x^4\sum_{m=0}^\infty(-1)^m(m+1)x^{2m}}\); \(y_2=-\dst{x\sum_{m=0}^\infty(-1)^m(2m-1)x^{2m}}\)
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\(x(3+x^2)y''+(2-x^2)y'-8xy=0\)
Show answer
\(y_1=\dst{x^{1/3}\sum_{m=0}^\infty{(-1)^m\over18^mm!}\left(\prod_{j=1}^m (6j-17)\right) x^{2m}}\); \(y_2=\dst{1+{4\over5}x^2+{8\over55}x^4}\)
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\(4x^2(1-x^2)y''+x(7-19x^2)y'-(1+14x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/4}\sum_{m=0}^\infty\left(\prod_{j=1}^m{8j+1\over8j+5} \right)x^{2m}}\); \(y_2=\dst{x^{-1}\sum_{m=0}^\infty{\prod_{j=1}^m(2j-1)\over2^mm!} x^{2m}}\)
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\(3x^2(2-x^2)y''+x(1-11x^2)y'+(1-5x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/2}\sum_{m=0}^\infty{1\over8^mm!}\left(\prod_{j=1}^m(4j-1) \right)x^{2m}}\); \(y_2=\dst{x^{1/3}\sum_{m=0}^\infty2^m\left(\prod_{j=1}^m{3j-1\over12j-1} \right) x^{2m}}\)
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\(2x^2(2+x^2)y''-x(12-7x^2)y'+(7+3x^2)y=0\)
Show answer
\(y_1=\dst{x^{7/2}\sum_{m=0}^\infty(-1)^m{\prod_{j=1}^m(4j+5)\over8^mm!}x^{2m}}\); \(y_2=\dst{x^{1/2}\sum_{m=0}^\infty{(-1)^m\over4^m}\left(\prod_{j=1}^m {4j-1\over2j-3} \right)x^{2m}}\)
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\(2x^2(2+x^2)y''+x(4+7x^2)y'-(1-3x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/2}\sum_{m=0}^\infty{(-1)^m\over4^m}\left(\prod_{j=1}^m {4j-1\over2j+1}\right)x^{2m}}\); \(y_2=\dst{x^{-1/2}\sum_{m=0}^\infty{(-1)^m\over8^mm!}\left(\prod_{j=1}^m (4j-3)\right)x^{2m}}\)
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\(2x^2(1+2x^2)y''+5x(1+6x^2)y'-(2-40x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/2}\sum_{m=0}^\infty{(-1)^m\over m!}\left(\prod_{j=1}^m(2j+1)\right) x^{2m}}\); \(y_2=\dst{{1\over x^2}\sum_{m=0}^\infty(-2)^m\left(\prod_{j=1}^m{4j-3\over 4j-5}\right) x^{2m}}\)
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\(3x^2(1+x^2)y''+5x(1+x^2)y'-(1+5x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\sum_{m=0}^\infty(-1)^m\left(\prod_{j=1}^m {3j-4\over3j+2}\right) x^{2m}}\); \(y_2=\dst{x^{-1}(1+x^2)}\)
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\(x(1+x^2)y''+(4+7x^2)y'+8xy=0\)
Show answer
\(y_1=\dst{\sum_{m=0}^\infty(-1)^m{2^m(m+1)!\over\prod_{j=1}^m(2j+3)}x^{2m}}\); \(y_2=\dst{{1\over x^3 }\sum_{m=0}^\infty(-1)^m{\prod_{j=1}^m(2j-1)\over2^mm!}x^{2m}}\)
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\(x^2(2+x^2)y''+x(3+x^2)y'-y=0\)
Show answer
\(y_1=\dst{x^{1/2}\sum_{m=0}^\infty{(-1)^m\over8^mm!}\left(\prod_{j=1}^m {(4j-3)^2\over4j+3}\right)x^{2m}}\); \(y_2=\dst{x^{-1}\sum_{m=0}^\infty{(-1)^m\over2^mm!}\left(\prod_{j=1}^m {(2j-3)^2\over4j-3}\right) x^{2m}}\)
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\(2x^2(1+x^2)y''+x(3+8x^2)y'-(3-4x^2)y=0\)
Show answer
\(y_1=\dst{x\sum_{m=0}^\infty(-2)^m\left(\prod_{j=1}^m{2j+1\over 4j+5}\right) x^{2m}}\); \(y_2=\dst{x^{-3/2}\sum_{m=0}^\infty{(-1)^m\over4^mm!}\left(\prod_{j=1}^m(4j-3) \right)x^{2m}}\)
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\(9x^2y''+3x(3+x^2)y'-(1-5x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\sum_{m=0}^\infty{(-1)^m\over2^m\prod_{j=1}^m(3j+1)} x^{2m}}\); \(y_2=\dst{x^{-1/3}\sum_{m=0}^\infty{(-1)^m\over6^mm!} x^{2m}}\)
In Exercises 47–51 find a fundamental set of Frobenius solutions. Compute the coefficients \(a_0\), …, \(a_{2M}\) for \(M\) at least \(7\) in each solution.
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C \(6x^2y''+x(1+6x^2)y'+(1+9x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/2}\left(1-{6\over13}x^2+{36\over325}x^4- {216\over12025}x^6 +\cdots\right)}\); \(y_2=\dst{x^{1/3}\left(1-{1\over2}x^2+{1\over8}x^4-{1\over48}x^6 +\cdots\right)}\)
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C \(x^2(8+x^2)y''+7x(2+x^2)y'-(2-9x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/4}\left(1-{13\over64}x^2+{273\over8192}x^4- {2639\over524288}x^6 +\cdots\right)}\); \(y_2=\dst{x^{-1}\left(1-{1\over3}x^2+{2\over33}x^4-{2\over209}x^6 +\cdots\right)}\)
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C \(9x^2(1+x^2)y''+3x(3+13x^2)y'-(1-25x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\left(1-{3\over4}x^2+{9\over14}x^4-{81\over140}x^6 +\cdots\right)}\); \(y_2=\dst{x^{-1/3}\left(1-{2\over3}x^2+{5\over9}x^4- {40\over81}x^6+\cdots\right)}\)
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C \(4x^2(1+x^2)y''+4x(1+6x^2)y'-(1-25x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/2}\left(1-{3\over2}x^2+{15\over8}x^4-{35\over16}x^6 +\cdots\right)}\); \(y_2=\dst{x^{-1/2}\left(1-2x^2+{8\over3}x^4-{16\over5}x^6 +\cdots\right)}\)
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C \(8x^2(1+2x^2)y''+2x(5+34x^2)y'-(1-30x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/4}\left(1-x^2+{3\over2}x^4-{5\over2}x^6 +\cdots\right)}\); \(y_2=\dst{x^{-1/2}\left(1-{2\over5}x^2+{36\over65}x^4- {408\over455}x^6+\cdots\right)}\)
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Suppose \(r_1>r_2\), \(a_0=b_0=1\), and the Frobenius series
\[ y_1=x^{r_1}\sum_{n=0}^\infty a_nx^n\quad\mbox{ and } \quad y_2=x^{r_2}\sum_{n=0}^\infty b_nx^n \]both converge on an interval \((0,\rho)\).
Show that \(y_1\) and \(y_2\) are linearly independent on \((0,\rho)\).
Hint
Show that if \(c_1\) and \(c_2\) are constants such that \(c_1y_1+c_2y_2\equiv0\) on \((0,\rho)\), then
\[ c_1x^{r_1-r_2}\sum_{n=0}^\infty a_nx^n+ c_2\sum_{n=0}^\infty b_nx^n=0,\quad 0<x<\rho. \]Then let \(x\to0+\) to conclude that \(c_2=0\).
Use the result of (b) to complete the proof of Theorem 7.5.3.
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The equation
\begin{equation} x^2y''+xy'+(x^2-\nu^2)y=0 \tag{7.5.1}\end{equation}is Bessel’s equation of order \(\nu\). (Here \(\nu\) is a parameter, and this use of “order” should not be confused with its usual use as in “the order of the equation.”) The solutions of (7.5.1) are Bessel functions of order \(\nu\).
Assuming that \(\nu\) isn’t an integer, find a fundamental set of Frobenius solutions of (7.5.1).
If \(\nu=1/2\), the solutions of (7.5.1) reduce to familiar elementary functions. Identify these functions.
Show answer
(a) \(y_1=\dst{x^\nu\sum_{m=0}^\infty{(-1)^m\over4^mm!\prod_{j=1}^m(j+\nu)}x^{2m}}\); \(y_2=\dst{x^{-\nu}\sum_{m=0}^\infty{(-1)^m\over4^mm!\prod_{j=1}^m(j-\nu)}x^{2m}}\)
\(y_1=\dst{\sin x\over\sqrt x}\); \(y_2=\dst{\cos x\over\sqrt x}\)
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Verify that
\[ {d\over dx}\left(|x|^rx^n\right)=(n+r)|x|^rx^{n-1}\mbox{\quad and \quad} {d^2\over dx^2}\left(|x|^rx^n\right)=(n+r)(n+r-1)|x|^rx^{n-2} \]if \(x\ne0\).
Let
\[ Ly= x^2(\alpha_0+\alpha_1x+\alpha_2x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y' +(\gamma_0+\gamma_1x+\gamma_2x^2)y=0. \]Show that if \(x^r\sum_{n=0}^\infty a_nx^n\) is a solution of \(Ly=0\) on \((0,\rho)\) then \(|x|^r\sum_{n=0}^\infty a_nx^n\) is a solution on \((-\rho,0)\) and \((0,\rho)\).
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Deduce from Eqn. (7.5.20) that
\[ a_n(r)=(-1)^n\prod_{j=1}^n{p_1(j+r-1)\over p_0(j+r)}. \]Conclude that if \(p_0(r)=\alpha_0(r-r_1)(r-r_2)\) where \(r_1-r_2\) is not an integer, then
\[ y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n\quad\mbox{ and }\quad y_2=x^{r_2}\sum_{n=0}^\infty a_n(r_2)x^n \]form a fundamental set of Frobenius solutions of
\[ x^2(\alpha_0+\alpha_1x)y''+x(\beta_0+\beta_1x)y'+(\gamma_0+\gamma_1x)y=0. \]Show that if \(p_0\) satisfies the hypotheses of (b) then
\[ y_1=x^{r_1}\sum_{n=0}^\infty {(-1)^n\over n!\prod_{j=1}^n(j+r_1-r_2)} \left(\gamma_1\over\alpha_0\right)^nx^n \]and
\[ y_2=x^{r_2}\sum_{n=0}^\infty {(-1)^n\over n!\prod_{j=1}^n(j+r_2-r_1)} \left(\gamma_1\over\alpha_0\right)^nx^n \]form a fundamental set of Frobenius solutions of
\[ \alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_1x)y=0. \]
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Let
\[ Ly=x^2(\alpha_0+\alpha_2x^2)y''+x(\beta_0+\beta_2x^2)y'+ (\gamma_0+\gamma_2x^2)y=0 \]and define
\[ p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0\quad\mbox{ and }\quad p_2(r)=\alpha_2r(r-1)+\beta_2r+\gamma_2. \]Use Theorem 7.5.2 to show that if
\begin{equation} \begin{array}{rcl} a_0(r)&=&1,\\ p_0(2m+r)a_{2m}(r)+p_2(2m+r-2)a_{2m-2}(r)&=&0,\quad m\ge1, \end{array} \tag{7.5.1}\end{equation}then the Frobenius series \(y(x,r)=x^r\sum_{m=0}^\infty a_{2m}x^{2m}\) satisfies \(Ly(x,r)=p_0(r)x^r\).
Deduce from (7.5.1) that if \(p_0(2m+r)\) is nonzero for every positive integer \(m\) then
\[ a_{2m}(r)=(-1)^m\prod_{j=1}^m{p_2(2j+r-2)\over p_0(2j+r)}. \]Conclude that if \(p_0(r)=\alpha_0(r-r_1)(r-r_2)\) where \(r_1-r_2\) is not an even integer, then
\[ y_1=x^{r_1}\sum_{m=0}^\infty a_{2m}(r_1)x^{2m}\quad\mbox{ and }\quad y_2=x^{r_2}\sum_{m=0}^\infty a_{2m}(r_2)x^{2m} \]form a fundamental set of Frobenius solutions of \(Ly=0\).
Show that if \(p_0\) satisfies the hypotheses of (c) then
\[ y_1=x^{r_1}\sum_{m=0}^\infty {(-1)^m\over 2^mm!\prod_{j=1}^m(2j+r_1-r_2)} \left(\gamma_2\over\alpha_0\right)^mx^{2m} \]and
\[ y_2=x^{r_2}\sum_{m=0}^\infty {(-1)^m\over 2^mm!\prod_{j=1}^m(2j+r_2-r_1)} \left(\gamma_2\over\alpha_0\right)^mx^{2m} \]form a fundamental set of Frobenius solutions of
\[ \alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_2x^2)y=0. \]
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Let
\[ Ly=x^2q_0(x)y''+xq_1(x)y'+q_2(x)y, \]where
\[ q_0(x)=\sum_{j=0}^\infty \alpha_jx^j,\quad q_1(x)=\sum_{j=0}^\infty \beta_jx^j,\quad q_2(x)=\sum_{j=0}^\infty \gamma_jx^j, \]and define
\[ p_j(r)=\alpha_jr(r-1)+\beta_jr+\gamma_j,\quad j=0,1,\dots. \]Let \(y=x^r\sum_{n=0}^\infty a_nx^n\). Show that
\[ Ly=x^r\sum_{n=0}^\infty b_nx^n, \]where
\[ b_n=\sum_{j=0}^np_j(n+r-j)a_{n-j}. \] -
Let \(L\) be as in Exercise 57. Show that if
\[ y(x,r)=x^r\sum_{n=0}^\infty a_n(r)x^n \]where
\begin{eqnarray*} a_0(r)&=&1, \\ a_n(r)&=&-\dst{1\over p_0(n+r)}\sum_{j=1}^n p_j(n+r-j)a_{n-j}(r),\quad n\ge1, \end{eqnarray*}then
\[ Ly(x,r)=p_0(r)x^r. \]Conclude that if
\[ p_0(r)=\alpha_0(r-r_1)(r-r_2) \]where \(r_1-r_2\) isn’t an integer then \(y_1=y(x,r_1)\) and \(y_2=y(x,r_2)\) are solutions of \(Ly=0\).
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Let
\[ Ly=x^2(\alpha_0+\alpha_qx^q)y''+x(\beta_0+\beta_qx^q)y'+ (\gamma_0+\gamma_qx^q)y \]where \(q\) is a positive integer, and define
\[ p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0\quad\mbox{ and }\quad p_q(r)=\alpha_qr(r-1)+\beta_qr+\gamma_q. \]Show that if
\[ y(x,r)=x^{r}\sum_{m=0}^\infty a_{qm}(r)x^{qm} \]where
\begin{equation} \begin{array}{rcl} a_0(r)&=&1,\\ a_{qm}(r)&=&-\dst{p_q\left(q(m-1)+r\right)\over p_0(qm+r)}a_{q(m-1)}(r),\quad m\ge1, \end{array} \tag{7.5.1}\end{equation}then
\[ Ly(x,r)=p_0(r)x^r. \]Deduce from (7.5.1) that
\[ a_{qm}(r)=(-1)^m\prod_{j=1}^m{p_q\left(q(j-1)+r\right)\over p_0(qj+r)}. \]Conclude that if \(p_0(r)=\alpha_0(r-r_1)(r-r_2)\) where \(r_1-r_2\) is not an integer multiple of \(q\), then
\[ y_1=x^{r_1}\sum_{m=0}^\infty a_{qm}(r_1)x^{qm}\quad\mbox{ and }\quad y_2=x^{r_2}\sum_{m=0}^\infty a_{qm}(r_2)x^{qm} \]form a fundamental set of Frobenius solutions of \(Ly=0\).
Show that if \(p_0\) satisfies the hypotheses of (c) then
\[ y_1=x^{r_1}\sum_{m=0}^\infty {(-1)^m\over q^mm!\prod_{j=1}^m(qj+r_1-r_2)} \left(\gamma_q\over\alpha_0\right)^mx^{qm} \]and
\[ y_2=x^{r_2}\sum_{m=0}^\infty {(-1)^m\over q^mm!\prod_{j=1}^m(qj+r_2-r_1)} \left(\gamma_q\over\alpha_0\right)^mx^{qm} \]form a fundamental set of Frobenius solutions of
\[ \alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_qx^q)y=0. \]
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Suppose \(\alpha_0,\alpha_1\), and \(\alpha_2\) are real numbers with \(\alpha_0\ne0\), and \(\{a_n\}_{n=0}^\infty\) is defined by
\[ \alpha_0a_1+\alpha_1a_0=0 \]and
\[ \alpha_0a_n+\alpha_1a_{n-1}+\alpha_2a_{n-2}=0,\quad n\ge2. \]Show that
\[ (\alpha_0+\alpha_1x+\alpha_2x^2)\sum_{n=0}^\infty a_nx^n=\alpha_0a_0, \]and infer that
\[ \sum_{n=0}^\infty a_nx^n={\alpha_0a_0\over\alpha_0+\alpha_1x+\alpha_2x^2}. \]With \(\alpha_0,\alpha_1\), and \(\alpha_2\) as in (a), consider the equation
\begin{equation} x^2(\alpha_0+\alpha_1x+\alpha_2 x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y'+ (\gamma_0+\gamma_1x+\gamma_2x^2)y=0, \tag{7.5.1}\end{equation}and define
\[ p_j(r)=\alpha_jr(r-1)+\beta_jr+\gamma_j,\quad j=0,1,2. \]Suppose
\[ {p_1(r-1)\over p_0(r)}= {\alpha_1\over\alpha_0},\qquad {p_2(r-2)\over p_0(r)}= {\alpha_2\over\alpha_0}, \]and
\[ p_0(r)=\alpha_0(r-r_1)(r-r_2), \]where \(r_1>r_2\). Show that
\[ y_1={x^{r_1}\over\alpha_0+\alpha_1x+\alpha_2x^2}\quad\mbox{ and }\quad y_2={x^{r_2}\over\alpha_0+\alpha_1x+\alpha_2x^2} \]form a fundamental set of Frobenius solutions of (7.5.1) on any interval \((0,\rho)\) on which \(\alpha_0+\alpha_1x+\alpha_2x^2\) has no zeros.
In Exercises 61–68 use the method suggested by Exercise 60 to find the general solution on some interval \((0,\rho)\).
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\(2x^2(1+x)y''-x(1-3x)y'+y=0\)
Show answer
\(y_1=\dst{x^{1/2}\over1+x}\); \(y_2=\dst{x\over1+x}\)
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\(6x^2(1+2x^2)y''+x(1+50x^2)y'+(1+30x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\over1+2x^2}\); \(y_2=\dst{x^{1/2}\over1+2x^2}\)
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\(28x^2(1-3x)y''-7x(5+9x)y'+7(2+9x)y=0\)
Show answer
\(y_1=\dst{x^{1/4}\over1-3x}\); \(y_2=\dst{x^2\over1-3x}\)
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\(9x^2(5+x)y''+9x(5+3x)y'-(5-8x)y=0\)
Show answer
\(y_1=\dst{x^{1/3}\over5+x}\); \(y_2=\dst{x^{-1/3}\over5+x}\)
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\(8x^2(2-x^2)y''+2x(10-21x^2)y'-(2+35x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/4}\over2-x^2}\); \(y_2=\dst{x^{-1/2}\over2-x^2}\)
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\(4x^2(1+3x+x^2)y''-4x(1-3x-3x^2)y'+3(1-x+x^2)y=0\)
Show answer
\(y_1=\dst{x^{1/2}\over1+3x+x^2}\); \(y_2=\dst{x^{3/2}\over1+3x+x^2}\)
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\(3x^2(1+x)^2y''-x(1-10x-11x^2)y'+(1+5x^2)y=0\)
Show answer
\(y_1=\dst{x\over(1+x)^2}\); \(y_2=\dst{x^{1/3}\over(1+x)^2}\)
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\(4x^2(3+2x+x^2)y''-x(3-14x-15x^2)y'+(3+7x^2)y=0\)
Show answer
\(y_1=\dst{x\over3+2x+x^2}\); \(y_2=\dst{x^{1/4}\over3+2x+x^2}\)