the method of frobenius

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Overview An Example Double Check

The Method of Frobenius

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science


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What is the Method of Frobenius?

1. The method of Frobenius works for differential equationsof the form y′′+P(x)y′+Q(x)y = 0 in which P or Q is notanalytic at the point of expansion x0.

2. But P and Q cannot be arbitrary: (x− x0)P(x) and(x− x0)2Q(x) must be analytic at x0.

3. Instead of a series solution y =∞

∑n=0

cn(x− x0)n, we obtain a

solution of the form y =∞

∑n=0

cn(x− x0)n+r.

4. The method of Frobenius is guaranteed to produce onesolution, but it may not produce two linearly independentsolutions.



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What is the Method of Frobenius?1. The method of Frobenius works for differential equations

of the form y′′+P(x)y′+Q(x)y = 0 in which P or Q is notanalytic at the point of expansion x0.

2. But P and Q cannot be arbitrary: (x− x0)P(x) and(x− x0)2Q(x) must be analytic at x0.

3. Instead of a series solution y =∞

∑n=0

cn(x− x0)n, we obtain a

solution of the form y =∞

∑n=0

cn(x− x0)n+r.

4. The method of Frobenius is guaranteed to produce onesolution, but it may not produce two linearly independentsolutions.



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What is the Method of Frobenius?

5. As for series solutions, we substitute the series and itsderivatives into the equation to obtain an equation for r anda set of equations for the cn.

6. These equations will allow us to compute r and the cn.7. For each value of r (typically there are two), we can

compute the solution just like for series.That’s it.



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What is the Method of Frobenius?5. As for series solutions, we substitute the series and its

derivatives into the equation to obtain an equation for r anda set of equations for the cn.





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6. These equations will allow us to compute r and the cn.

7. For each value of r (typically there are two), we cancompute the solution just like for series.

That’s it.



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compute the solution just like for series.

That’s it.



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Solve the Differential Equation9x2y′′+3x2y′+2y = 0

9x2y′′+3x2y′+2y = 0

9x2

(∞

∑n=0

cnxn+r

)′′

+3x2

(∞

∑n=0

cnxn+r

)′

+2

(∞

∑n=0

cnxn+r

)= 0



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9x2y′′+3x2y′+2y = 0

9x2

(∞

∑n=0

cnxn+r

)′′

+3x2

(∞

∑n=0

cnxn+r

)′

+2∞

∑n=0

cnxn+r = 0



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9x2y′′+3x2y′+2y = 0

9x2

(∞

∑n=0

cnxn+r

)′′

+3x2∞

∑n=0

cn(n+ r)xn+r−1 +2∞

∑n=0

cnxn+r = 0



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9x2y′′+3x2y′+2y = 0

9x2∞

∑n=0

cn(n+ r)(n+ r−1)xn+r−2 +3x2∞

∑n=0

cn(n+ r)xn+r−1 +2∞

∑n=0

cnxn+r = 0

∞

∑n=0

9(n+ r)(n+ r−1)cnxn+r +∞

∑n=0

3(n+ r)cnxn+r+1 +∞

∑n=0

2cnxn+r = 0

∞

∑k=0

9(k + r)(k + r−1)ckxk+r +∞

∑k=1

3(k + r−1)ck−1xk+r +∞

∑k=0

2ckxk+r = 0

(9r(r−1)+2

)c0xr+

∞

∑k=1

[9(k+r)(k+r−1)ck +3(k+r−1)ck−1+2ck

]xk+r = 0



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Indicial equation:(9r(r−1)+2

)c0 = 0

9r(r−1)+2 = 0

9r2−9r +2 = 0

r1,2 =9±

√81−7218

=9±3

18

=23,

13



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)c0 = 0

9r(r−1)+2 = 0

9r2−9r +2 = 0

r1,2 =9±

√81−7218

=9±3

18

=23

,13



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)c0 = 0

9r(r−1)+2 = 0

9r2−9r +2 = 0

r1,2 =9±

√81−7218

=9±3

18

=23,

13



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Recurrence Relation.

9(k + r)(k + r−1)ck +3(k + r−1)ck−1 +2ck = 0[9(k + r)(k + r−1)+2

]ck +3(k + r−1)ck−1 = 0[

9(k + r)(k + r−1)+2]ck = −3(k + r−1)ck−1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1



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9(k + r)(k + r−1)ck +3(k + r−1)ck−1 +2ck = 0

[9(k + r)(k + r−1)+2

]ck +3(k + r−1)ck−1 = 0[

9(k + r)(k + r−1)+2]ck = −3(k + r−1)ck−1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1



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]ck +3(k + r−1)ck−1 = 0

[9(k + r)(k + r−1)+2

]ck = −3(k + r−1)ck−1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1



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]ck +3(k + r−1)ck−1 = 0[

9(k + r)(k + r−1)+2]ck = −3(k + r−1)ck−1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1 =−1

6

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252=− 1

567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1

=3−3k−3 · 2

3

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1 =−1

6

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252=− 1

567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1 =−1

6

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252=− 1

567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1

=1−3k

9k2 +3kck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1 =−1

6

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252=− 1

567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1 =−1

6

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252=− 1

567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1

=− 212

·1 =−16

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252=− 1

567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1

=−16

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252=− 1

567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1 =−1

6

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252=− 1

567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1 =−1

6

c2 =1−3 ·2

9 ·22 +3 ·2c2−1

=− 542

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252=− 1

567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1 =−1

6

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)

=5

252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252=− 1

567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1 =−1

6

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252=− 1

567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1 =−1

6

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1

=− 890

5252

=− 1567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1 =−1

6

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252

=− 1567



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r =23

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 23

3(k + 2

3

)3(k + 2

3 −1)+2

ck−1

=1−3k

(3k +2)(3k−1)+2ck−1 =

1−3k9k2 +3k

ck−1

c1 =1−3 ·1

9 ·12 +3 ·1c1−1 =− 2

12·1 =−1

6

c2 =1−3 ·2

9 ·22 +3 ·2c2−1 =− 5

42

(−1

6

)=

5252

c3 =1−3 ·3

9 ·32 +3 ·3c3−1 =− 8

905

252=− 1

567



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r =23

c4 =1−3 ·4

9 ·42 +3 ·4c4−1

=− 11156

(− 1

567

)=

1188452

y1 = x23 − 1

6x

23 +1 +

5252

x23 +2− 1

567x

23 +3 +

1188452

x23 +4

= x23 − 1

6x

53 +

5252

x83 − 1

567x

113 +

1188452

x143



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r =23

c4 =1−3 ·4

9 ·42 +3 ·4c4−1 =− 11

156

(− 1

567

)

=11

88452

y1 = x23 − 1

6x

23 +1 +

5252

x23 +2− 1

567x

23 +3 +

1188452

x23 +4

= x23 − 1

6x

53 +

5252

x83 − 1

567x

113 +

1188452

x143



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r =23

c4 =1−3 ·4

9 ·42 +3 ·4c4−1 =− 11

156

(− 1

567

)=

1188452

y1 = x23 − 1

6x

23 +1 +

5252

x23 +2− 1

567x

23 +3 +

1188452

x23 +4

= x23 − 1

6x

53 +

5252

x83 − 1

567x

113 +

1188452

x143



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r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 13

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1 =

2−3k9k2−3k

ck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1 =−1

6

c2 =2−3 ·2

9 ·22−3 ·2c2−1 =− 4

30

(−1

6

)=

145

c3 =2−3 ·3

9 ·32−3 ·3c3−1 =− 7

721

45=− 7

3240



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r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1

=3−3k−3 · 1

3

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1 =

2−3k9k2−3k

ck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1 =−1

6

c2 =2−3 ·2

9 ·22−3 ·2c2−1 =− 4

30

(−1

6

)=

145

c3 =2−3 ·3

9 ·32−3 ·3c3−1 =− 7

721

45=− 7

3240



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r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 13

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1 =

2−3k9k2−3k

ck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1 =−1

6

c2 =2−3 ·2

9 ·22−3 ·2c2−1 =− 4

30

(−1

6

)=

145

c3 =2−3 ·3

9 ·32−3 ·3c3−1 =− 7

721

45=− 7

3240



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r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 13

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1

=2−3k

9k2−3kck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1 =−1

6

c2 =2−3 ·2

9 ·22−3 ·2c2−1 =− 4

30

(−1

6

)=

145

c3 =2−3 ·3

9 ·32−3 ·3c3−1 =− 7

721

45=− 7

3240



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r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 13

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1 =

2−3k9k2−3k

ck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1 =−1

6

c2 =2−3 ·2

9 ·22−3 ·2c2−1 =− 4

30

(−1

6

)=

145

c3 =2−3 ·3

9 ·32−3 ·3c3−1 =− 7

721

45=− 7

3240



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r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 13

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1 =

2−3k9k2−3k

ck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1

=−16

c2 =2−3 ·2

9 ·22−3 ·2c2−1 =− 4

30

(−1

6

)=

145

c3 =2−3 ·3

9 ·32−3 ·3c3−1 =− 7

721

45=− 7

3240



logo1



r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 13

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1 =

2−3k9k2−3k

ck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1 =−1

6

c2 =2−3 ·2

9 ·22−3 ·2c2−1 =− 4

30

(−1

6

)=

145

c3 =2−3 ·3

9 ·32−3 ·3c3−1 =− 7

721

45=− 7

3240



logo1



r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 13

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1 =

2−3k9k2−3k

ck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1 =−1

6

c2 =2−3 ·2

9 ·22−3 ·2c2−1

=− 430

(−1

6

)=

145

c3 =2−3 ·3

9 ·32−3 ·3c3−1 =− 7

721

45=− 7

3240



logo1



r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 13

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1 =

2−3k9k2−3k

ck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1 =−1

6

c2 =2−3 ·2

9 ·22−3 ·2c2−1 =− 4

30

(−1

6

)

=145

c3 =2−3 ·3

9 ·32−3 ·3c3−1 =− 7

721

45=− 7

3240



logo1



r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 13

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1 =

2−3k9k2−3k

ck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1 =−1

6

c2 =2−3 ·2

9 ·22−3 ·2c2−1 =− 4

30

(−1

6

)=

145

c3 =2−3 ·3

9 ·32−3 ·3c3−1 =− 7

721

45=− 7

3240



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r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 13

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1 =

2−3k9k2−3k

ck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1 =−1

6

c2 =2−3 ·2

9 ·22−3 ·2c2−1 =− 4

30

(−1

6

)=

145

c3 =2−3 ·3

9 ·32−3 ·3c3−1

=− 772

145

=− 73240



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r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 13

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1 =

2−3k9k2−3k

ck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1 =−1

6

c2 =2−3 ·2

9 ·22−3 ·2c2−1 =− 4

30

(−1

6

)=

145

c3 =2−3 ·3

9 ·32−3 ·3c3−1 =− 7

721

45

=− 73240



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r =13

, c0 = 1

ck =3−3k−3r

9(k + r)(k + r−1)+2ck−1 =

3−3k−3 · 13

3(k + 1

3

)3(k + 1

3 −1)+2

ck−1

=2−3k

(3k +1)(3k−2)+2ck−1 =

2−3k9k2−3k

ck−1

c1 =2−3 ·1

9 ·12−3 ·1c1−1 =−1

6

c2 =2−3 ·2

9 ·22−3 ·2c2−1 =− 4

30

(−1

6

)=

145

c3 =2−3 ·3

9 ·32−3 ·3c3−1 =− 7

721

45=− 7

3240



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r =13

c4 =2−3 ·4

9 ·42−3 ·4c4−1

=− 10132

(− 7

3240

)=

742768

y2 = x13 − 1

6x

13 +1 +

145

x13 +2− 7

3240x

13 +3 +

742768

x13 +4

y2 = x13 − 1

6x

43 +

145

x73 − 7

3240x

103 +

742768

x133



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r =13

c4 =2−3 ·4

9 ·42−3 ·4c4−1 =− 10

132

(− 7

3240

)

=7

42768

y2 = x13 − 1

6x

13 +1 +

145

x13 +2− 7

3240x

13 +3 +

742768

x13 +4

y2 = x13 − 1

6x

43 +

145

x73 − 7

3240x

103 +

742768

x133



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r =13

c4 =2−3 ·4

9 ·42−3 ·4c4−1 =− 10

132

(− 7

3240

)=

742768

y2 = x13 − 1

6x

13 +1 +

145

x13 +2− 7

3240x

13 +3 +

742768

x13 +4

y2 = x13 − 1

6x

43 +

145

x73 − 7

3240x

103 +

742768

x133



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Checking Solutions When Only The First FewTerms are Available

As for series solutions, because the “solution” is a truncatedseries,we cannot expect that the differential equation is exactlysatisfied.



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As before, for a correct approximation, low order terms shouldcancel.

We went to order143

for the first solution and to order133

forthe second solution.And, of course, we use a computer algebra system.



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As before, for a correct approximation, low order terms shouldcancel.We went to order

143


forthe second solution.

And, of course, we use a computer algebra system.



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As before, for a correct approximation, low order terms shouldcancel.We went to order

143


forthe second solution.And, of course, we use a computer algebra system.



the method of frobenius

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