A differential equation of the form y+P y+ Q y=0, where P and Q are both functions of x, such that the equation can be represented by a power series, may be solved by the Frobenius method.
The following 4-step procedure may be used in the Frobenius method:
(i) Assume a trial solution of the form y= xc1
a0+a1x+a2x2+a3x3+ ã ã ã +arxr+ ã ã ã2 (ii) differentiate the trial series,
(iii) substitute the results in the given differential equation,
(iv) equate coefficients of corresponding powers of the variable on each side of the equation;
this enables index c and coefficients a1, a2, a3, … from the trial solution, to be determined.
This introductory treatment of the Frobenius method covering the simplest cases is demonstrated, using the above procedure, in the following worked problems.
Problem 7. Determine, using the Frobenius method, the general power series solution of the differential equation: 3xd2y
dx2 +dy
dx−y=0 The differential equation may be rewritten as:
3xy+y−y=0
(i) Let a trial solution be of the form y=xc1
a0+a1x+a2x2+a3x3+ ã ã ã +arxr+ ã ã ã2
(16) where a0=0, i.e. y=a0xc+a1xc+1+a2xc+2+a3xc+3
+ ã ã ã +arxc+r+ ã ã ã (17) (ii) Differentiating equation (17) gives:
y =a0cxc−1+a1(c+1)xc +a2(c+2)xc+1+ ã ã ã +ar(c+r)xc+r−1+ ã ã ã and y =a0c(c−1)xc−2+a1c(c+1)xc−1
+a2(c+1)(c+2)xc+ ã ã ã
+ar(c+r−1)(c+r)xc+r−2 + ã ã ã
(iii) Substituting y, y and y into each term of the given equation 3xy+y−y=0 gives:
3xy =3a0c(c−1)xc−1+3a1c(c+1)xc +3a2(c+1)(c+2)xc+1+ ã ã ã +3ar(c+r−1)(c+r)xc+r−1+ã ã ã(a) y =a0cxc−1+a1(c+1)xc+a2(c+2)xc+1
+ ã ã ã +ar(c+r)xc+r−1+ ã ã ã (b)
−y= −a0xc−a1xc+1−a2xc+2−a3xc+3
− ã ã ã −arxc+r− ã ã ã (c) (iv) The sum of these three terms forms the left- hand side of the equation. Since the right-hand side is zero, the coefficients of each power of x can be equated to zero.
For example, the coefficient of xc−1 is equated to zero giving: 3a0c(c−1)+a0c=0 or a0c[3c−3+1]=a0c(3c−2)=0 (18) The coefficient of xc is equated to zero giving:
3a1c(c+1)+a1(c+1)−a0=0 i.e. a1(3c2+3c+c+1)−a0
=a1(3c2+4c+1)−a0 =0 or a1(3c+1)(c+1)−a0=0 (19)
In each of series (a), (b) and (c) an xc term is involved, after which, a general relationship can be obtained for xc+r, where r≥0.
In series (a) and (b), terms in xc+r−1 are present; replacing r by (r+1) will give the cor- responding terms in xc+r, which occurs in all three equations, i.e.
in series (a), 3ar+1(c+r)(c+r+1)xc+r in series (b), ar+1(c+r+1)xc+r in series (c), −arxc+r
Equating the total coefficients of xc+r to zero gives:
3ar+1(c+r)(c+r+1)+ar+1(c+r+1)
−ar=0 which simplifies to:
ar+1{(c+r+1)(3c+3r+1)}−ar=0 (20) Equation (18), which was formed from the coefficients of the lowest power of x, i.e. xc−1, is called the indicial equation, from which,
Ch52-H8152.tex 23/6/2006 15: 13 Page 499
POWER SERIES METHODS OF SOLVING ORDINARY DIFFERENTIAL EQUATIONS 499
I
the value of c is obtained. From equation (18), since a0=0, then c=0 or c=2
3 (a) When c=0:
From equation (19), if c=0, a1(1×1)−a0 = 0, i.e. a1=a0
From equation (20), if c=0, ar+1(r+1)(3r+1)−ar=0, i.e. ar+1= ar
(r+1)(3r+1) r≥0 Thus, when r=1, a2= a1
(2×4) = a0 (2×4) since a1=a0 when r=2, a3= a2
(3×7) = a0 (2×4)(3×7)
or a0
(2×3)(4×7) when r=3, a4= a3
(4×10)
= a0
(2×3×4)(4×7×10) and so on.
From equation (16), the trial solution was:
y=xc{a0+a1x+a2x2+a3x3+ ã ã ã +arxr+ ã ã ã } Substituting c=0 and the above values of a1, a2, a3, … into the trial solution gives:
y=x0
a0+a0x+ a0
(2×4)
x2 +
a0 (2×3)(4×7)
x3 +
a0
(2×3×4)(4×7×10)
x4+ ã ã ã
i.e. y=a0
1+x+ x2
(2×4)+ x3 (2×3) (4×7)
+ x4
(2ì3ì4) (4ì7ì10)+ ã ã ã
(21) (b) When c=2
3:
From equation (19), if c=2 3, a1(3)
5 3
−a0=0, i.e. a1=a0
5
From equation (20), if c=2 3 ar+1
2
3+r+1
(2+3r+1)−ar=0, i.e. ar+1
r+5
3
(3r+3)−ar
= ar+1(3r2+8r+5)−ar=0, i.e. ar+1= ar
(r+1)(3r+5) r≥0 Thus, when r=1, a2= a1
(2×8) = a0 (2×5×8) since a1= a0
5 when r=2, a3= a2
(3×11)
= a0
(2×3)(5×8×11) when r=3, a4= a3
(4×14)
= a0
(2×3×4)(5×8×11×14) and so on.
From equation (16), the trial solution was:
y=xc{a0+a1x+a2x2+a3x3+ ã ã ã +arxr+ ã ã ã } Substituting c=2
3 and the above values of a1, a2, a3, … into the trial solution gives:
y=x23
a0+a0 5
x+
a0 2×5×8
x2 +
a0
(2×3)(5×8×11)
x3 +
a0
(2×3×4)(5×8×11×14)
x4+ ã ã ã
i.e. y=a0x23
1+x
5+ x2
(2×5×8) + x3
(2×3)(5×8×11)
+ x4
(2ì3ì4)(5ì8ì11ì14)+ ã ã ã
(22) Since a0 is an arbitrary (non-zero) constant in each solution, its value could well be different.
500 DIFFERENTIAL EQUATIONS
Let a0=A in equation (21), and a0=B in equation (22). Also, if the first solution is denoted by u(x) and the second byv(x), then the general solution of the given differential equation is y=u(x)+v(x). Hence, y = A
1+x+ x2
(2×4)+ x3 (2×3) (4×7)
+ x4
(2ì3ì4) (4ì7ì10)+ ã ã ã
+B x23
1+x
5+ x2 (2×5×8)
+ x3
(2×3)(5×8×11)
+ x4
(2ì3ì4)(5ì8ì11ì14)+ ã ã ã
Problem 8. Use the Frobenius method to determine the general power series solution of the differential equation:
2x2d2y dx2 −xdy
dx +(1−x)y=0
The differential equation may be rewritten as:
2x2y−xy+(1−x)y=0
(i) Let a trial solution be of the form y=xc{a0+a1x+a2x2+a3x3+ ã ã ã
+arxr+ ã ã ã } (23) where a0=0,
i.e. y=a0xc+a1xc+1+a2xc+2+a3xc+3 + ã ã ã +arxc+r+ ã ã ã (24) (ii) Differentiating equation (24) gives:
y=a0cxc−1+a1(c+1)xc+a2(c+2)xc+1 + ã ã ã +ar(c+r)xc+r−1+ ã ã ã and y =a0c(c−1)xc−2+a1c(c+1)xc−1
+a2(c+1)(c+2)xc+ ã ã ã
+ar(c+r−1)(c+r)xc+r−2+ ã ã ã (iii) Substituting y, y and y into each term of
the given equation 2x2y−xy+(1−x)y=0
gives:
2x2y =2a0c(c−1)xc+2a1c(c+1)xc+1 +2a2(c+1)(c+2)xc+2+ ã ã ã +2ar(c+r−1)(c+r)xc+r+ ã ã ã
(a)
−xy = −a0cxc−a1(c+1)xc+1
−a2(c+2)xc+2− ã ã ã
−ar(c+r)xc+r− ã ã ã (b) (1−x)y=(1−x)(a0xc+a1xc+1+a2xc+2
+a3xc+3+ ã ã ã +arxc+r+ ã ã ã)
=a0xc+a1xc+1+a2xc+2+a3xc+3 + ã ã ã +arxc+r+ ã ã ã
−a0xc+1−a1xc+2−a2xc+3
−a3xc+4− ã ã ã −arxc+r+1− ã ã ã (c) (iv) The indicial equation, which is obtained by equating the coefficient of the lowest power of x to zero, gives the value(s) of c. Equating the total coefficients of xc (from equations (a) to (c)) to zero gives:
2a0c(c−1)−a0c+a0 =0 i.e. a0[2c(c−1)−c+1] =0 i.e. a0[2c2−2c−c+1] =0 i.e. a0[2c2−3c+1]=0 i.e. a0[(2c−1)(c−1)]=0 from which, c=1 or c=1
2
The coefficient of the general term, i.e. xc+r, gives (from equations (a) to (c)):
2ar(c+r−1)(c+r)−ar(c+r) +ar−ar−1=0 from which,
ar[2(c+r−1)(c+r)−(c+r)+1]=ar−1
and ar= ar−1
2(c+r−1)(c+r)−(c+r)+1 (25) (a) With c=1, ar= ar−1
2(r)(1+r)−(1+r)+1
= ar−1
2r+2r2−1−r+1
= ar−1
2r2+r = ar−1 r(2r+1)
Ch52-H8152.tex 23/6/2006 15: 13 Page 501
POWER SERIES METHODS OF SOLVING ORDINARY DIFFERENTIAL EQUATIONS 501
I
Thus, when r=1, a1= a0
1(2+1) = a0 1×3 when r=2,
a2 = a1
2(4+1) = a1 (2×5)
= a0
(1×3)(2×5) or a0 (1×2)×(3×5) when r=3,
a3 = a2
3(6+1) = a2 3×7
= a0
(1×2×3)×(3×5×7) when r=4,
a4 = a3
4(8+1) = a3 4×9
= a0
(1×2×3×4)×(3×5×7×9) and so on.
From equation (23), the trial solution was:
y=xc1
a0+a1x+a2x2+a3x3+ ã ã ã
+arxr+ ã ã ã2 Substituting c=1 and the above values of a1, a2, a3, … into the trial solution gives:
y=x1
a0+ a0
(1×3)x+ a0
(1×2)×(3×5)x2
+ a0
(1×2×3)×(3×5×7)x3
+ a0
(1×2×3×4)×(3×5×7×9)x4 + ã ã ã
i.e. y=a0x1
1+ x
(1×3) + x2 (1×2)×(3×5)
+ x3
(1×2×3)×(3×5×7)
+ x4
(1×2×3×4)×(3×5×7×9) + ã ã ã
(26)
(b) With c=1 2
ar= ar−1
2(c+r−1)(c+r)−(c+r)+1 from equation (25)
i.e. ar= ar−1
2 1
2+r−1 1
2+r
− 1
2 +r
+1
= ar−1 2
r− 1
2 r+1 2
− 1
2−r+1
= ar−1 2
r2− 1
4
− 1
2−r+1
= ar−1 2r2− 1
2 −1
2−r+1
= ar−1 2r2−r
= ar−1 r(2r−1)
Thus, when r=1, a1= a0
1(2−1) = a0 1×1 when r=2, a2= a1
2(4−1) = a1 (2×3)
= a0 (2×3) when r=3, a3= a2
3(6−1) = a2 3×5
= a0 (2×3)×(3×5) when r=4, a4= a3
4(8−1) = a3 4×7
= a0
(2×3×4)×(3×5×7) and so on.
From equation (23), the trial solution was:
y=xc1
a0+a1x+a2x2+a3x3+ ã ã ã +arxr+ ã ã ã2 Substituting c=1
2 and the above values of a1, a2, a3, … into the trial solution gives:
y=x12
a0+a0x+ a0
(2×3)x2+ a0
(2×3)×(3×5)x3
+ a0
(2ì3ì4)ì(3ì5ì7)x4+ ã ã ã
502 DIFFERENTIAL EQUATIONS
i.e. y=a0x12
1+x+ x2 (2×3) + x3
(2×3)×(3×5)
+ x4
(2×3×4)×(3×5×7) + ã ã ã
(27) Since a0is an arbitrary (non-zero) constant in each solution, its value could well be different.
Let a0=A in equation (26), and a0=B in equa- tion (27). Also, if the first solution is denoted by u(x) and the second byv(x), then the gen- eral solution of the given differential equation is y=u(x)+v(x),
i.e. y= A x
1+ x
(1×3)+ x2 (1×2)×(3×5)
+ x3
(1×2×3)×(3×5×7)
+ x4
(1×2×3×4)×(3×5×7×9) + ã ã ã
+B x12
1+x+ x2 (2×3) + x3
(2×3)×(3×5)
+ x4
(2ì3ì4)ì(3ì5ì7)+ ã ã ã
Problem 9. Use the Frobenius method to deter- mine the general power series solution of the differential equation: d2y
dx2−2y=0
The differential equation may be rewritten as:
y−2y=0
(i) Let a trial solution be of the form y=xc1
a0+a1x+a2x2+a3x3+ ã ã ã +arxr+ ã ã ã2
(28) where a0=0, i.e. y=a0xc+a1xc+1+a2xc+2+a3xc+3
+ ã ã ã +arxc+r+ ã ã ã (29)
(ii) Differentiating equation (29) gives:
y =a0cxc−1+a1(c+1)xc+a2(c+2)xc+1 + ã ã ã +ar(c+r)xc+r−1+ ã ã ã and y=a0c(c−1)xc−2+a1c(c+1)xc−1
+a2(c+1)(c+2)xc+ ã ã ã
+ar(c+r−1)(c+r)xc+r−2+ ã ã ã (iii) Replacing r by (r+2) in
ar(c+r−1)(c+r) xc+r−2gives:
ar+2(c+r+1)(c+r+2)xc+r
Substituting y and yinto each term of the given equation y−2y=0 gives:
y−2y=a0c(c−1)xc−2+a1c(c+1)xc−1 +[a2(c+1)(c+2)−2a0]xc+ã ã ã +[ar+2(c+r+1)(c+r+2)
−2ar] xc+r+ ã ã ã =0 (30) (iv) The indicial equation is obtained by equating the coefficient of the lowest power of x to zero.
Hence, a0c(c−1)=0 from which, c=0 or c=1 since a0=0
For the term in xc−1, i.e. a1c(c+1)=0 With c=1, a1=0; however, when c=0, a1 is indeterminate, since any value of a1 com- bined with the zero value of c would make the product zero.
For the term in xc,
a2(c+1)(c+2)−2a0=0 from which, a2= 2a0
(c+1)(c+2) (31)
For the term in xc+r,
ar+2(c+r+1)(c+r+2)−2ar =0 from which,
ar+2= 2ar
(c+r+1)(c+r+2) (32) (a) When c=0: a1is indeterminate, and from equation (31)
a2= 2a0
(1×2) = 2a0 2! In general, ar+2= 2ar
(r+1)(r+2) and when r=1, a3= 2a1
(2×3)= 2a1
(1×2×3)=2a1 3!
Ch52-H8152.tex 23/6/2006 15: 13 Page 503
POWER SERIES METHODS OF SOLVING ORDINARY DIFFERENTIAL EQUATIONS 503
I
when r=2, a4= 2a2
3×4 = 4a0 4! Hence, y=x0
a0+a1x+ 2a0
2! x2+2a1 3! x3 +4a0
4! x4+ ã ã ã
from equation (28)
=a0
1+2x2 2! +4x4
4! + ã ã ã
+a1 3
x+2x3 3! + 4x5
5! + ã ã ã 4
Since a0 and a1 are arbitrary constants depending on boundary conditions, let a0=P and a1=Q, then:
y=P
1+2x2 2! +4x4
4! + ã ã ã
+Q
x+2x3 3! +4x5
5! + ã ã ã
(33) (b) When c=1: a1=0, and from equa- tion (31),
a2= 2a0
(2×3) = 2a0 3!
Since c=1, ar+2= 2ar
(c+r+1)(c+r+2)
= 2ar (r+2)(r+3) from equation (32) and when r=1,
a3= 2a1
(3×4) =0 since a1=0 when r=2,
a4= 2a2
(4×5) = 2
(4×5) ×2a0 3! = 4a0
5! when r=3,
a5= 2a3 (5×6) =0 Hence, when c=1,
y=x1
a0+2a0
3! x2+ 4a0
5! x4+ ã ã ã
from equation (28) i.e. y=a0
3 x+2x3
3! +4x5 5! +. . .
4
Again, a0is an arbitrary constant; let a0=K,
then y=K
x+2x3
3! +4x5 5! + ã ã ã
However, this latter solution is not a separate solu- tion, for it is the same form as the second series in equation (33). Hence, equation (33) with its two arbitrary constants P and Q gives the general solu- tion. This is always the case when the two values of c differ by an integer (i.e. whole number). From the above three worked problems, the following can be deduced, and in future assumed:
(i) if two solutions of the indicial equation differ by a quantity not an integer, then two inde- pendent solutions y=u(x)+v(x) results, the general solution of which is y=Au+Bv(note:
Problem 7 had c=0 and 2
3 and Problem 8 had c=1 and 1
2; in neither case did c differ by an integer)
(ii) if two solutions of the indicial equation do differ by an integer, as in Problem 9 where c=0 and 1, and if one coefficient is indeterminate, as with when c=0, then the complete solution is always given by using this value of c. Using the second value of c, i.e. c=1 in Problem 9, always gives a series which is one of the series in the first solution.
Now try the following exercise.
Exercise 197 Further problems on power series solution by the Frobenius method 1. Produce, using Frobenius’ method, a power
series solution for the differential equation:
2xd2y dx2 +dy
dx −y=0
⎡
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎣ y=A
1+x+ x2 (2×3) + x3
(2ì3)(3ì5)+ ã ã ã
+B x12
1+ x
(1×3)+ x2 (1×2)(3×5)
+ x3
(1ì2ì3)(3ì5ì7)+ ã ã ã
⎤
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎦
504 DIFFERENTIAL EQUATIONS
2. Use the Frobenius method to determine the general power series solution of the differen- tial equation: d2y
dx2+y=0
⎡
⎢⎢
⎢⎢
⎢⎢
⎢⎣ y=A
1−x2
2!+x4 4! − ã ã ã
+B
x−x3 3!+x5
5! − ã ã ã
=P cos x+Q sin x
⎤
⎥⎥
⎥⎥
⎥⎥
⎥⎦
3. Determine the power series solution of the differential equation: 3xd2y
dx2 +4dy
dx −y=0 using the Frobenius method.
⎡
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣ y=A
1+ x
(1×4) + x2 (1×2)(4×7)
+ x3
(1ì2ì3)(4ì7ì10)+ ã ã ã
+Bx−13
1+ x
(1×2) + x2 (1×2)(2×5)
+ x3
(1ì2ì3)(2ì5ì8)+ ã ã ã
⎤
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
4. Show, using the Frobenius method, that the power series solution of the differential equation: d2y
dx2−y=0 may be expressed as y=P cosh x+Q sinh x, where P and Q are constants. [Hint: check the series expansions for cosh x and sinh x on page 48]