## Callisto 3 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)?

1. Callisto

$\bar y^{(n)} = -(\frac{n(n-1)}{(n-2)(n-3)})\bar y^{(n-2)}$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

Oh wait. It should be -12. My bad!

3. Callisto

Well, I guess 0 is not in the domain of m, so just ignore it...

4. Callisto

Hmm.... By observing the pattern, I got $\bar y^{(2m)} = (-1)^{m-1}(2m-1)(2m)$

5. Callisto

So far so good? Or...?

6. Callisto

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

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8. Callisto

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.