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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)?
 2 years ago
 2 years 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)?
 2 years ago
 2 years 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...
 2 years ago

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

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

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