Hey everyone, I am trying to use the Bessel function:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]f(x)= \sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{2^{2k+1}(k!)(k+1)!}[/tex]

To verify the Differential Equation:

[tex]x^2y''+xy'+(x^2-1)y=0[/tex]

So I thought this was gonna be really easy, but I am having some difficulties. First I calculated y,y',y'' to be:

[tex]y= \sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{2^{2k+1}(k!)(k+1)!}[/tex]

[tex]y'= \sum_{k=0}^\infty \frac{(-1)^{k}(2k+1)x^{2k}}{2^{2k+1}(k!)(k+1)!}[/tex]

[tex]y''= \sum_{k=0}^\infty \frac{(-1)^{k}(2k+1)(2k)x^{2k-1}}{2^{2k+1}(k!)(k+1)!}[/tex]

And then I just pluged those into the differential equation to yield:

[tex]x^2 \sum_{k=0}^\infty \frac{(-1)^{k}(2k+1)(2k)x^{2k-1}}{2^{2k+1}(k!)(k+1)!} + x\sum_{k=0}^\infty \frac{(-1)^{k}(2k+1)x^{2k}}{2^{2k+1}(k!)(k+1)!} + (x^2-1)\sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{2^{2k+1}(k!)(k+1)!} = 0 [/tex]

Then I just put the x terms in the summation since they are constant to get:

[tex]\sum_{k=0}^\infty \frac{(-1)^{k}(2k+1)(2k)x^{2k+1}}{2^{2k+1}(k!)(k+1)!} + \sum_{k=0}^\infty \frac{(-1)^{k}(2k+1)x^{2k+1}}{2^{2k+1}(k!)(k+1)!} + \sum_{k=0}^\infty \frac{(-1)^{k}(x^2-1)x^{2k+1}}{2^{2k+1}(k!)(k+1)!} = 0 [/tex]

And since the summations are all indexed the same and all have the same denominator, I combined them to one summation with one fraction:

[tex]\sum_{k=0}^\infty \frac{(-1)^{k}(2k+1)(2k)x^{2k+1} + (-1)^{k}(2k+1)x^{2k+1} + (-1)^{k}(x^2-1)x^{2k+1}}{2^{2k+1}(k!)(k+1)!} = 0[/tex]

But now that I have one huge fraction, after all the simplification I have done, I can't get the numerator to zero out since the DE is homogeneous.

Did I make an error in my calculation? Have I done everything correct, but there is even MORE simplification? And if so, can you help me? Or am I just approaching this problem completely wrong?

Thanks a TON!

Sincerely,

Tony

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# Verifying Bessel's Differential Equation

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