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Originally Posted by szpengchao Given the form of the DE, I would assume that the first part requires you to find a solution of the form . Where are you stuck in the second part?
i have problem with the first part
Originally Posted by szpengchao Hint: Use the chain rule Now we need the 2nd derivative using the product and chain rule from here sub into the equation Good luck
no...i mean the 1st question...not the second
Originally Posted by szpengchao no...i mean the 1st question...not the second Are you talking about using the method of frobenius to obtain series solutions to the ODE?
yeah.... i have problem with choosing point for this particular question... would you like to post ur steps?
well, x=0 is a regular singular point, right? then , i set y= sigma( A_n * x^(n+b)) into the equation. then i got something like: sigma( A_n * (n^2+1) = 0) well, this means A_n = 0 for all n...then how can i continue?
Originally Posted by szpengchao well, x=0 is a regular singular point, right? then , i set y= sigma( A_n * x^(n+b)) into the equation. then i got something like: sigma( A_n * (n^2+1) = 0) well, this means A_n = 0 for all n...then how can i continue? I think this is where you are stuck.
yup...then? how can that be zero?
Originally Posted by szpengchao yup...then? how can that be zero? I don't know besides the trivial solution. Maybe someone can find my mistake.
Originally Posted by TheEmptySet I don't know besides the trivial solution. Maybe someone can find my mistake. Using the substitution given in the second part, I get ......
Last edited by mr fantastic; May 19th 2008 at 02:22 AM. Reason: Deleted last sentence
Originally Posted by mr fantastic Given the form of the DE, I would assume that the first part requires you to find a solution of the form . Where are you stuck in the second part? Do this and you get . Therefore by the magic of Euler's equation and the usual identity.
Thanks Mr. Fantastic, I forgot that in this method so so Then the rest follows. Thanks again!
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