Question

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)?

Answer 1

\[\bar y^{(n)} = -(\frac{n(n-1)}{(n-2)(n-3)})\bar y^{(n-2)}\]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...

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

So far so good? Or...?

Answer 6

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...

Answer 7

*

Answer 8

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.