Solve the Differential Equation: (1 - x^2) y'' - (4x) y' + (6) y = 0

  • Thread starter VinnyCee
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Here is the problem (number 9 in 5.2 of Boyce, DiPrima 8th Edition Book):

[tex](1 - x^2)\,y'' - (4x)\,y' + (6)\,y = 0,\,x_0 = 0[/tex]

Here is what I have so far:

[tex]\sum_{n = 0}^{\infty}\,(n + 2)\,(n + 1)\,a_{n + 2}\,x^n - \sum_{n = 2}^{\infty}\,n(n - 1)\,a_n\,x^n - 4\,\sum_{n = 1}^{\infty}\,n\,a_n\,x^n + 6\,\sum_{n = 0}^{\infty}\,a_n\,x^n = 0[/tex]

Now, I took out the first two terms (n = 0 and n = 1) of the first sum to make it's index go to n = 2 in order to add all the sums together, right?

[tex]2\,a_2 + 6\,a_3\,x[/tex]

Continuing in that fashion to add the sums, i get this:

[tex]2\,a_2\,+\,6\,a_3\,x\,-\,4\,a_1\,x\,+\,6\,a_0\,+\,6\,a_1\,x\,+\,\sum_{n = 2}^{\infty}\,\left[(n\,+\,2)\,(n\,+\,1)\,a_{n\,+\,2}\,-\,n\,(n\,-\,1)\,a_n\,-\,4\,n\,a_n\,+\,6\,a_n\right]\,x^n\,=\,0[/tex]

This is where I am stuck. I am not sure if I pulled out those factors correctly or what, please help. Thank you.
 

Answers and Replies

  • #2
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Whoops!

I copied the problem wrong!

It is supposed to be:

[tex](1 + x^2)\,y'' - (4x)\,y' + (6)\,y = 0,\,x_0 = 0[/tex]

And NOT:

[tex](1 - x^2)\,y'' - (4x)\,y' + (6)\,y = 0,\,x_0 = 0[/tex]

One little minus sign!

The whole solution is worked out as the last problem http://tutorial.math.lamar.edu/AllBrowsers/3401/SeriesSolutions.asp [Broken]!
 
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