# Math Help - Solutions of higher order linear equations

1. ## Solutions of higher order linear equations

Hi,

I am going through some solved problems, but I don't understand the answers very well... I would really appreciate it if someone could explain me the answers briefly. I know most of the problems ahead, but I don't have much of a clue about the first problem of this chapter.

Thanks a lot for your help!

The problem says:

"In each of Problems 1 through 6 determine intervals in which solutions are sure to exist.

1) $y^{iv}+4y'''+3y=t$
2) $ty'''+(sin(t))y''+3y=cos(t)$
3) $t(t-1)y^{iv}+e^ty''+4t^2y=0$
4) $y'''+ty''+t^2y'+t^3y=ln(t)$
5) $(x-1)y^{iv}+(x+1)y''+(tan(x))y=0$
6) $(x^2-4)y^{vi}+x^2y'''+9y=0$"

1) $-\infty
2) $t>0$ or t<0
3) $t>1, or 0, or $t<0$
4) $t>0$
5) $..., {-3\pi}/2
6) $-\infty

So well, I understand the first one, and as for the others, I am confused about this:

2) Why can't $t$ be equal to zero?
3) Again, why can't $t$ be equal to 0 or 1?
4) It says that $t$ has to be positive, but why is that the case if $ln(t)$ can take negative numbers? Does that have anything to do with $e^t$?
5) I know tan(x) is not defined when x is equal to ${-3\pi}/2, {-\pi}/2, {\pi}/2, {3\pi}/2$, etc. However, why aren't there any answers when $x=1$?
6) Is this result taken from $(x^2-4)$? If so, why?

2. ## Re: Solutions of higher order linear equations

Linear DEs can only be solved if the leading coefficient is 1. That means if it's not, you will need to divide by it. That should explain why many values of t or x are not possible.