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anonymous · Answer

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

anonymous · Answer

So is that:$$y^{(4)}-y'''$$

zepdrix · Answer

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

zepdrix · Answer

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?

zepdrix · Answer

$$\large \lambda^4-\lambda^3=0$$

zepdrix · Answer

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?

zepdrix · Answer

heh