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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    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















  • 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












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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











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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  • 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




    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










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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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    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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      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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        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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        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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