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Callisto
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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, \((n2)(n3)y^{(n)} = n(n1)y^{(n2)}\).
(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)?
 one year ago
 one year ago
Callisto Group Title
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, \((n2)(n3)y^{(n)} = n(n1)y^{(n2)}\). (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)?
 one year ago
 one year ago

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Callisto Group TitleBest ResponseYou've already chosen the best response.2
\[\bar y^{(n)} = (\frac{n(n1)}{(n2)(n3)})\bar y^{(n2)}\]When n=2m \[\bar y^{(2m)} = (\frac{2m(2m1)}{(2m2)(2m3)})\bar y^{(2m2)}\]\[\bar y^{(2m)} = \frac{m(2m1)}{(m1)(2m3)}\bar y^{(2m2)}\]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, \[\bar y^{(4)} = \frac{2(41)}{(21)(43)}\bar y^{(42)} =2\times 3 \times (2) = 12\]Hmm...
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
Oh wait. It should be 12. My bad!
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
Well, I guess 0 is not in the domain of m, so just ignore it...
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
Hmm.... By observing the pattern, I got \[\bar y^{(2m)} = (1)^{m1}(2m1)(2m)\]
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
So far so good? Or...?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
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...
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago
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