Question

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Answer 1

Is y^(4) the fourth derivative of y?

Answer 2

So is that:\[y^{(4)}-y'''\]

Answer 3

I don't see the inequality sign, is it suppose to be =0?

Answer 4

Ah :) I see, you bum! lol

Yes, use lambdas. We always called them \(\large r\)'s so it took me a moment to remember what you were talking about :3

You mean breaking it down into the characteristic equation right?

Answer 5

\[\large \lambda^4-\lambda^3=0\]

Answer 6

No, just do some factoring. If we factor \(\large \lambda^3\) out of each term we get,\[\large \lambda^3(\lambda-1)=0\]

And you should be able to find your "roots" pretty easily from here.
Looks like you have a bunch of repeating zeroes.
What do we do with those? I can't quite remember.. something to do with multiplying by x.. right?

Answer 7

heh