Question

Please help me with this question?

(my book only has answers for odd numbered exercises) :

If y = sin (ax +b) is a general solution to a second order differential equation, find it.

So far I got:

y' = a cos (ax +b)
y'' = - a^2 sin (ax + b)


y'' = - a^2 sin (ax + b)
+ a^2 y = a^2 sin (ax +b)

y'' + a^2y = 0

a^2 y = - y''
a^2 = - y'' / y

Answer 1

That looks fine, you can probably leave it as
y'' = -a^2 y, or y'' + a^2y = 0.

Answer 2

No, I can't. I need to find a order 2 DE for which y = sin (ax + b) is the general solution, so far none of the exercises I'm doing have the constants "a" or "b" in them.

Answer 3

I see. Weird. You solved the other problems in the section in a similar way?

Answer 4

Yes, and in none of them one of the constants "a" or "b" remained in the DE. I could always get something just in terms of

f(x) (nought) y'' + f(x) (sub - one) y' + f(x) (sub - two) y = 0

Answer 5

I don't know, this question is frustrating. It does say that "If y = sin (ax +b) is a general solution to a second order differential equation, find it." so maybe it's a trick question and it really isn't a general solution to anything? I'm not sure how to prove it, aside from taking the DE we found and finding what the general solution really is.

Answer 6

I guess it's ok if the DE has one of the constants in it, it's just that since it didn't happen with any of the others, I thought there might be some way to fin "a" and "b" solely in terms of y, y' or y'', however hard.

Well, thanks for your help. :)

Answer 7

Yeah. By the way, I found this:
http://www.physicsforums.com/showthread.php?t=227578
They work the same problem and even check that it works for y''+a^2y=0. It's weird, since the constants c_1 and c_2 end up being cos(b) and sin(b) thanks to a trig identity.

Answer 8

I'll check it.

Answer 9

Cool! : )