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Rct33
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Homework Statement
I am trying to find the recursion relation for the coefficients of the series around x=0 for the ODE: [itex]y'''+x^2y'+xy=0[/itex]
The Attempt at a Solution
Therefore letting:
[itex]y=\sum_{m=0}^\infty y_mx^m[/itex]
[itex]\therefore y'=\sum_{m=1}^\infty my_mx^{m-1}[/itex]
[itex]\therefore y''=\sum_{m=2}^\infty m(m-1)y_mx^{m-2}[/itex]
[itex]\therefore y'''=\sum_{m=3}^\infty m(m-1)(m-2)y_mx^{m-3}[/itex]
Subbing this back in gives:
[itex]\sum_{m=3}^\infty m(m-1)(m-2)y_mx^{m-3}+x^2\sum_{m=1}^\infty my_mx^{m-1}+x\sum_{m=0}^\infty y_mx^m=0[/itex]
Fixing [itex]y'''[/itex]:
[itex]\sum_{m=3}^\infty m(m-1)(m-2)y_mx^{m-3}=\sum_{m=2}^\infty (m+1)(m+2)(m+3)y_{m+3}x^{m}+6y_3+24xy_4[/itex]
Fixing [itex]y'[/itex]:
[itex]x^2\sum_{m=1}^\infty my_mx^{m-1}=\sum_{m=2}^\infty (m-1)y_{m-1}x^m[/itex]
Fixing [itex]y[/itex]:
[itex]x\sum_{m=0}^\infty y_mx^m=\sum_{m=2}^\infty y_{m-1}x^m+xy_0[/itex]
Therefore combining these terms gives:
[itex]\sum_{m=2}^\infty\left[(m+1)(m+2)(m+3)y_{m+3}+(m-1)y_{m-1}+y_{m-1}\right]x^m+6y_3+xy_0+24xy_4=0[/itex]
Therefore I have two equations which sum to 0, with one of them being this:
[itex]6y_3+xy_0+24xy_4=0[/itex]
[itex]\therefore y_3=0[/itex] and [itex]y_4=-\frac{1}{24}y_0[/itex]
Using the other equation:
[itex](m+1)(m+2)(m+3)y_{m+3}+(m-1)y_{m-1}+y_{m-1}=0[/itex]
[itex]\therefore (m+1)(m+2)(m+3)y_{m+3}+y_{m-1}((m-1)+1)=0[/itex]
[itex]\therefore y_{m+3}=-\frac{y_{m-1}m}{ (m+1)(m+2)(m+3)}[/itex]
This gives me:
[itex]y_3=0[/itex]
[itex]y_4=-\frac{1}{24}y_0[/itex]
[itex]y_5=-\frac{1}{30}y_1[/itex]
[itex]y_6=-\frac{1}{40}y_2[/itex]
The problem is I am not sure how to relate these together to solve the recursion! Any help is much appreciated.
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