Answer
The solutions of the given equation are sure to exist on $\mathbb{R}- \{1,\pi (k+\frac{1}{2})\}\;,\;k\in \mathbb{Z}$
Work Step by Step
The given higher order linear equation is written in the form :
$\frac{d^ny}{dt^n}\;+\;p_{1}(t)\frac{d^{n-1}y}{dt^{n-1}}\;+\;.........\;+\;p_{n-1}\frac{dy}{dt}\;+\;p_{n}(t)y\;=\;g(t)\\\\ $
$n=4\;\;\;\;\;\;\;\;\;\;p_{1}(t)=0\;\;\;\;\;\;\;\;\;\;p_{2}=\frac{x+1}{x-1}\;\;\;\;\;\;\;\;p_{3}(t)=0\;\;\;\;\;\;\;\;\;\;\;\;\;p_{4}(t)=\frac{tan(x)}{x-1}\;\;\;\;\;\;\;\;\;\;g(t)=0\\\\$
$p_{1},p_{3},g$: continuous everywhere on $\mathbb{R}$.
$p_{2}$ is continuous everywhere on its domain $(-\infty,1) \cup (1,+\infty)$
$p_{4}$ is continuous everywhere except $x=\pi (k+\frac{1}{2})$ where $k$ is an integer.
The solutions of the given equation are sure to exist on $\mathbb{R}- \{1,\pi (k+\frac{1}{2})\}\;,\;k\in \mathbb{Z}$
