A community for students.
Here's the question you clicked on:
 0 viewing
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, \((n2)(n3)y^{(n)} = n(n1)y^{(n2)}\).
(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)?
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, \((n2)(n3)y^{(n)} = n(n1)y^{(n2)}\). (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)?

This Question is Closed

Callisto
 2 years ago
Best 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...

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

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

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

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

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.2Not 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...

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.2But 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.