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Solve Using Frobenius Method
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[QUOTE="cjc0117, post: 4527911, member: 305265"] Thanks for the replies. I had found the two s values to be [itex]s_{1}=3[/itex] and [itex]s_{2}=0[/itex]. [itex]s_{1}-s_{2}[/itex] is equal to a positive integer. I thought in this case, there are two independent solutions: [itex]y_{1}=\sum^{∞}_{m=0}a_{m}x^{m+s_{1}}[/itex][RIGHT][B](Eqn. 1)[/B][/RIGHT] [itex]y_{2}=ky_{1}ln(x)+\sum^{∞}_{m=0}b_{m}x^{m+s_{2}}[/itex] [RIGHT][B](Eqn. 2)[/B][/RIGHT] I thought this was because when [itex]s_{1}-s_{2}[/itex] [itex](s_{1}>s_{2})[/itex] is equal to a positive integer, you end up getting a trivial solution when trying to find an independent solution in the form of Eqn. 1 for [itex]s_{2}[/itex]. However, the recurrence relation [itex]a_{m}=a_{m+1}[/itex] does not result in a trivial solution for either [itex]s_{1}[/itex] or [itex]s_{2}[/itex] when using independent solutions in the form of Eqn. 1. The general solution for this particular problem seems to be: [itex]y=a_{0}\sum^{∞}_{m=0}x^{m+3}+b_{0}\sum^{∞}_{m=0}x^{m}[/itex] This turns out to be true if you just let both independent solutions take the form of Eqn. 1 right from the get go, or if you go the long way and only let [itex]s_{1}[/itex] take the form of Eqn. 1, and then let [itex]s_{2}[/itex] take the form of Eqn. 2 (because it turns out that k ends up being equal to zero and the recurrence relation ends up still being [itex]b_{m+1}=b_{m}[/itex]) I just don't get how you can know whether you'll have to let one of your independent solutions take the form of Eqn. 2. Do you just have to see through trial and error whether you'll get a trivial solution or not? [/QUOTE]
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