$\displaystyle x''=\frac{t+1}{t}x'-\frac{1}{t}x$

with $\displaystyle x_1(t)=e^t$

The goal is to find the other solution.

my attempt

$\displaystyle x_2=ux_1$

$\displaystyle x_2'=e^{t}u'+e^{t}u$

$\displaystyle x_2''=e^tu''+2e^tu'+e^tu$

Substituting into the original equation I get...

$\displaystyle u''=-\frac{t-1}{t}u'$

substituting with u'' with v'...

$\displaystyle v'=-\frac{t-1}{t}v$

I integrate to

$\displaystyle ln v=ln t-t+c1$

$\displaystyle v=t\cdot e^{-t}=u'$

integrating again...

$\displaystyle u=(-t-1)e^{-t}$

$\displaystyle x_2=(-t-1)e^{-t}\cdot e^t$

$\displaystyle x_2=-t-1$

but the answers supposed to be $\displaystyle t-1$ where did I go wrong? thanks