Recent content by NoMoreExams

I'm also not sure why you think this is a nonlinear DE? I googled for "solving second order linear DEs", this is one of the links you might want to read: http://silmaril.math.sci.qut.edu.au/~gustafso/mab112/topic12/

Find the roots of q^2 + a*q = 0

Don't you just have x'' + ax' = -bx^n So first solve x'' + ax' = 0, do you know how to do that? Then assume a solution to solve the = -bx^n part?

Don't you solve the associated quadratic?

Why don't you differentiate your answer and see if you got what you started with...

Well in this case, which numbers squared, divided by 3 give a remainder of 1?

Nope. The remainder of when you divide 1, 4, 16, 25, etc. by 3 is 1, hence it's in your equivalence class of [1]

That's not what remainder is. If you take x and divide it by n then you have a*q + r so r is the remainder, for example 4 divided by 3 is 4 = 3*1 + 1, so the remainder is 1.

No, there will never be a remainder of 3, you are dividing by 3, how can you have a remainder of 3?

n^2 mod 3 means take a number n, square it, divide it by 3 and tell me the remainder.

Which numbers squared are 1 mod 3? let's see... 1^2 = 1 mod 3, 2^2 = 4 = 1 mod 3, 4^2 = 16 = 1 mod 3... see a pattern?

I was responding to the OP saying that he thought they were the same function if the domains were the same, etc. and pointing out that the domains are not the same.

In post 7, the OP says y'(0) = 0 which seems to be in addition to y(0) = 0?

Didn't the OP already say they solved it?

You should probably write down the question, show some attempt at it, say where you are stuck and people will give you hints.