Why does the solution correspond to larger root if we solve 2nd ODE with Frobenius method?

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In my book (Boyce Diprima 2000. 7th ed) page 271 first paragraph it is written that: In all cases it is possible to find at least one solution whether $r_1$ and $r_2$ ( two roots of the indicial equation) differ by integer or the same or complex. If they are differ by an integer the solution corresponds to larger root. Why not smaller root ? I couldnot make sufficient reasoning after playing with indices and sums...

differential-equations power-series

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asked Jul 22 at 14:16

user2312512851

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  If I remember right, if you start with the smaller root, $r_1<r_2=r_1+k$ then the recursion equation forces $0=a_0=a_1=...=a_k-1$ so that effectively you get the same solution as if you had started with the larger root.$
  – LutzL
  Jul 22 at 14:21
  

  
  
  
  

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In my book (Boyce Diprima 2000. 7th ed) page 271 first paragraph it is written that: In all cases it is possible to find at least one solution whether $r_1$ and $r_2$ ( two roots of the indicial equation) differ by integer or the same or complex. If they are differ by an integer the solution corresponds to larger root. Why not smaller root ? I couldnot make sufficient reasoning after playing with indices and sums...

differential-equations power-series

share|cite|improve this question

asked Jul 22 at 14:16

user2312512851

1,139521

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If I remember right, if you start with the smaller root, $r_1<r_2=r_1+k$ then the recursion equation forces $0=a_0=a_1=...=a_k-1$ so that effectively you get the same solution as if you had started with the larger root.$
  – LutzL
  Jul 22 at 14:21
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

In my book (Boyce Diprima 2000. 7th ed) page 271 first paragraph it is written that: In all cases it is possible to find at least one solution whether $r_1$ and $r_2$ ( two roots of the indicial equation) differ by integer or the same or complex. If they are differ by an integer the solution corresponds to larger root. Why not smaller root ? I couldnot make sufficient reasoning after playing with indices and sums...

differential-equations power-series

share|cite|improve this question

asked Jul 22 at 14:16

user2312512851

1,139521

In my book (Boyce Diprima 2000. 7th ed) page 271 first paragraph it is written that: In all cases it is possible to find at least one solution whether $r_1$ and $r_2$ ( two roots of the indicial equation) differ by integer or the same or complex. If they are differ by an integer the solution corresponds to larger root. Why not smaller root ? I couldnot make sufficient reasoning after playing with indices and sums...

differential-equations power-series

share|cite|improve this question

asked Jul 22 at 14:16

user2312512851

1,139521

share|cite|improve this question

share|cite|improve this question

asked Jul 22 at 14:16

user2312512851

1,139521

asked Jul 22 at 14:16

user2312512851

1,139521

asked Jul 22 at 14:16

user2312512851

1,139521

1,139521

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If I remember right, if you start with the smaller root, $r_1<r_2=r_1+k$ then the recursion equation forces $0=a_0=a_1=...=a_k-1$ so that effectively you get the same solution as if you had started with the larger root.$
  – LutzL
  Jul 22 at 14:21
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If I remember right, if you start with the smaller root, $r_1<r_2=r_1+k$ then the recursion equation forces $0=a_0=a_1=...=a_k-1$ so that effectively you get the same solution as if you had started with the larger root.$
  – LutzL
  Jul 22 at 14:21
  

  
  
  
  

2

2

If I remember right, if you start with the smaller root, $r_1<r_2=r_1+k$ then the recursion equation forces $0=a_0=a_1=...=a_k-1$ so that effectively you get the same solution as if you had started with the larger root.$
– LutzL
Jul 22 at 14:21

If I remember right, if you start with the smaller root, $r_1<r_2=r_1+k$ then the recursion equation forces $0=a_0=a_1=...=a_k-1$ so that effectively you get the same solution as if you had started with the larger root.$
– LutzL
Jul 22 at 14:21

add a comment | 

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accepted

You can solve for smaller root, but the issue is that sometimes we can find more than one group of the linearly independent solutions at the same time. For example in Find the form of a second linear independent solution when the two roots of indicial equation are different by a integer.

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answered Jul 25 at 13:02

doraemonpaul

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1 Answer
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1 Answer
1

active

oldest

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oldest

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up vote
1
down vote



accepted

You can solve for smaller root, but the issue is that sometimes we can find more than one group of the linearly independent solutions at the same time. For example in Find the form of a second linear independent solution when the two roots of indicial equation are different by a integer.

share|cite|improve this answer

answered Jul 25 at 13:02

doraemonpaul

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up vote
1
down vote



accepted

You can solve for smaller root, but the issue is that sometimes we can find more than one group of the linearly independent solutions at the same time. For example in Find the form of a second linear independent solution when the two roots of indicial equation are different by a integer.

share|cite|improve this answer

answered Jul 25 at 13:02

doraemonpaul

12k31660

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up vote
1
down vote



accepted

up vote
1
down vote



accepted

You can solve for smaller root, but the issue is that sometimes we can find more than one group of the linearly independent solutions at the same time. For example in Find the form of a second linear independent solution when the two roots of indicial equation are different by a integer.

share|cite|improve this answer

answered Jul 25 at 13:02

doraemonpaul

12k31660

You can solve for smaller root, but the issue is that sometimes we can find more than one group of the linearly independent solutions at the same time. For example in Find the form of a second linear independent solution when the two roots of indicial equation are different by a integer.

share|cite|improve this answer

answered Jul 25 at 13:02

doraemonpaul

12k31660

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 13:02

doraemonpaul

12k31660

answered Jul 25 at 13:02

doraemonpaul

12k31660

answered Jul 25 at 13:02

doraemonpaul

12k31660

12k31660

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