How to solve this ordinary differential equation $tx'''+3x''-tx'-x=0$?

For equation $tx''' +3x''-tx'-x=0$, we know a special solution $x_1=\frac{1}{t}$, how to general solution?

I firstly attempted $d(tx''+2x'-tx)=0$,then $tx'' +2x' -tx = C$,$C$ is a constant.But in next step , I found that my solution is wrong.Since $x_1=\frac{1}{t}$ is a special solution of $tx''' +3x''-tx'-x=0$,we found $x_2 = -\frac{1}{t}$ is a solution of equation.Then $x = x_1-x_2 =\frac{2}{t}$ is a solution of $tx''+2x'-tx=0$.As you can see ,the step is wrong.

Then I attemped other way to solve this equation ,but all failed.Could help me solve this equation?Thans!

We can use Reduction of Order, we have

$$tx''' +3x''-tx'-x=0, ~~~~x_1 = \dfrac{1}{t}$$

Let

$$x_2 = v~ x_1 = \dfrac{v}{t}$$

Taking derivatives, substituting and simplifying into $tx_2''' +3x_2''-tx_2'-x_2=0$

$$v''' - v' = 0$$

Let $w = v'$, so we have

$$w'' - w = 0 \implies w = a e^t + b e^{-t}$$

This gives

$$v = a e^t + b e^{-t} + c$$

From the initial substitution

$$x_2 = \dfrac{v}{t} = \dfrac{a e^t + b e^{-t} + c}{t}$$