find the interval: third order differential equation

(sin x)y''' − 3xy'' + 2y = tan x

You are interested in listing the interval of validities of the third-order differential equation $(\sin x)y'''-3xy''+2y = \tan x$.That explains why you didn't specify initial condition like $y(0) = 4$.

Here's how to get it...

Modifying the differential equation to look like this:

$$y'''+p(x)y''+q(x)y'+r(x)y = g(x)$$

It becomes...

$$(\sin x)y'''-3xy''+2y = \tan x$$ $$\frac{(\sin x)y'''-3xy''+2y = \tan x}{\sin x}$$ $$y'''-\frac{3x}{\sin x} y'' + \frac{2}{\sin x}y = \frac{\tan x}{\sin x}$$ $$y'''-\frac{3x}{\sin x} y'' + \frac{2}{\sin x}y = \frac{1}{\cos x}$$

To determine the intervals, we need to see where the $p(x)$ , $r(x)$, and $g(x)$ to be continuous. We must avoid those points that render $p(x)$ , $r(x)$, and $g(x)$ to be discontinuous. Those points we must avoid are:

Setting $\sin x$ to be zero makes $p(x)$ and $r(x)$ to be discontinuous. Sine function goes to zero every $n\pi $ interval.

$$\sin x = 0 \implies x = \color{red}{n}\pi\,, n \in \mathbb Z$$

Setting $\cos x$ to be zero makes $g(x)$ to be discontinuous. Cosine function goes to zero every $(2n+1)\frac{\pi}{2}$ intervals.

$$\cos x =0 \implies x =(2n+1)\color{blue}{\frac{\pi}{2}}\,, n\in \mathbb Z$$

Therefore, the intervals of the differential equation $(\sin x)y'''-3xy''+2y = \tan x$ would be...

$$\mathbb R \setminus \{n\pi \mid \, n \in \mathbb Z\} \setminus \{(2n+1)\frac{\pi}{2} \mid \, n \in \mathbb Z\}$$

It refers to the fact that the solution set is $\mathbb R$ minus than the values $2nπ$ and $(2n+1)\frac{π}{2}$. It is the difference of sets. It an has interval ranging from $-\infty$ to $\infty$, excluding $2nπ$ and $(2n+1)\frac{π}{2}.$