Finding general solution when given only particular solution

The differential equation is given by $$ty''+ (1-2t)y'+ (t-1)y = 0.$$
The particular solution give is $e^t.$

I haven't seen an equation like this so far in my class and so far haven't found anything of use by searching on google.

I've tried finding the complimentary equation but it doesn't make sense where
where $tr^2 + (1-2t)r + (t-1) = 0.$ This isn't a constant coefficients or var of parameters is it?

Then I've tried building another solution by using the formula:
$$y_2= y_1 \int \frac{e^{-\int P}}{y_1^2}$$ But so far haven't had any success.
What should I try doing? Any help would be appreciated.

Hint: Let $y_2(t)=e^tv(t)$ and take first an second derivatives and substitute into the equation to find $v(t)$(This metthod is called reduction of order). Your equation for $v$ will be $tv''+v'=0$, from which $v=\ln t$ and so $y_2(t)=e^t\ln t$.
$$W = C e^{-\int dt \frac{1-2 t}{t}} = C \frac{e^{2 t}}{t}$$
$$y_2 = C e^t \int dt \frac{e^{2 t}/t}{(e^t)^2} = C e^{t} \log{t}$$