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Callisto

  • 2 years ago

Given that \(y=x^2cosx\) satisfies \(x^2y^{(2)} - 4xy^{(1)} + (x^2 +6)y =0\) , where \( y^{(n)} = \frac{d^ny}{dx^n}\). (a) Show that when x=0, \((n-2)(n-3)y^{(n)} = -n(n-1)y^{(n-2)}\). (b) If we use the notation \(\bar y^{(n)} = \frac{d^ny}{dx^n}|_{x=0}\), find \(\bar y^{(2m)}\) and \(\bar y^{(2m+1)}\). How to do part (b)?

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  1. Callisto
    • 2 years ago
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    \[\bar y^{(n)} = -(\frac{n(n-1)}{(n-2)(n-3)})\bar y^{(n-2)}\]When n=2m \[\bar y^{(2m)} = -(\frac{2m(2m-1)}{(2m-2)(2m-3)})\bar y^{(2m-2)}\]\[\bar y^{(2m)} = -\frac{m(2m-1)}{(m-1)(2m-3)}\bar y^{(2m-2)}\]Then... I don't know how to continue.. When m=0, \[\bar y^{(0)} = 0\] When m=1, \[\bar y^{(2)} = \frac{d^2y}{dx^2}(x^2cosx)|_{x=0} = 2\] When m=2, \[\bar y^{(4)} = \frac{2(4-1)}{(2-1)(4-3)}\bar y^{(4-2)} =2\times 3 \times (2) = 12\]Hmm...

  2. Callisto
    • 2 years ago
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    Oh wait. It should be -12. My bad!

  3. Callisto
    • 2 years ago
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    Well, I guess 0 is not in the domain of m, so just ignore it...

  4. Callisto
    • 2 years ago
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    Hmm.... By observing the pattern, I got \[\bar y^{(2m)} = (-1)^{m-1}(2m-1)(2m)\]

  5. Callisto
    • 2 years ago
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    So far so good? Or...?

  6. Callisto
    • 2 years ago
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    Not that good. 0 is not a domain when n=2m. Since in part a, domain of n is positive integers. So, for m, it can't be 0. Hmm...

  7. TuringTest
    • 2 years ago
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    *

  8. Callisto
    • 2 years ago
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    But for n=2m+1, m can be 0 Since the domain of n is positive integers, When m=0, 2m+1 = 1 >0 => it's in the domain of n, so for n=2m+1, domain of m is integers greater than or equal to 0.

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