Convert First-order Linear System to Second-order ODE. I'm just getting started on differential equations and am now looking at a task that asks to find the ordinary second-order differential equation that "underlies" a system of differential equations given by: 
$\left(\begin{matrix}\dot u(t)\\\dot v(t)\\ \end{matrix} \right)=\left(\begin{matrix}0&1\\\frac{1}{1-t}&\frac{t}{t-1}\\ \end{matrix} \right)\left(\begin{matrix}v(t)\\u(t)\\ \end{matrix} \right)$
Now, I've read on the reduction of higher-order differential equations to systems, and came up with $x''(t)-\frac{1}{1-t}x'(t)-\frac{t}{t-1}x(t)=0$, which, by taking
$u=x, v=x'$
$u'=x'=v, v'=x''=\frac{1}{1-t}x'(t)+\frac{t}{t-1}x(t)$
reduces exactly to what I was given at the start, which is why I figured my solution was right. However, in the second equation, I am asked to show that $u(t)=\left(\begin{matrix}e^t\\e^t\\ \end{matrix} \right)$, which does not correspond with the second order differential equation that I came up with. Could anybody tell me where I made a mistake or what I should look into?
 A: This line is not correct you inverted $u,v$:
$$\left(\begin{matrix}\dot u(t)\\\dot v(t)\\ \end{matrix} \right)=\left(\begin{matrix}0&1\\\frac{1}{1-t}&\frac{t}{t-1}\\ \end{matrix} \right)\left(\begin{matrix}v(t)\\u(t)\\ \end{matrix} \right)$$
It should be:
$$\left(\begin{matrix}\dot u(t)\\\dot v(t)\\ \end{matrix} \right)=\left(\begin{matrix}0&1\\\frac{1}{1-t}&\frac{t}{t-1}\\ \end{matrix} \right)\left(\begin{matrix}u(t)\\v(t)\\ \end{matrix} \right)$$
So that the second order DE is:
$$x''(t)-\frac{t}{t-1}x'(t)-\frac{1}{1-t}x(t)=0$$

Edit
@psyph  Applying Liouville's formula I got this general solution:
$$|\phi (t)|=c_1 \exp \int \frac {t}{t-1} dt$$
$$|\phi (t)|=c_1e^t(t-1)$$
We also have that:
$$|\phi (t)|=u(t)e^t-v(t)e^t=e^t(u(t)-v(t))$$
$$\implies u(t)e^t-v(t)e^t=c_1e^t(t-1)$$
$$ u(t)-v(t)=c_1(t-1)$$
from the original equation we have:
$$u'(t)=v(t) \implies u'(t)-u(t)=-c_1(t-1)$$
using method of integrating factor
$$ (u(t)e^{-t})'=-c_1(t-1)e^{-t}$$
Integrate:
$$ u(t)e^{-t}=-c_1(-(t-1)e^{-t}-e^{-t})+c_2$$
$$  u(t)=-c_1(-(t-1)-1)+c_2e^{t}$$
$$ \implies u(t)=tc_1+c_2e^{t}$$
And for $v(t)$:
$$v(t)=u'(t)=c_1+c_2e^{t}$$
Finally :
$$\left(\begin{matrix} u(t)\\\ v(t)\\ \end{matrix} \right)=\left(\begin{matrix}c_1t+c_2e^t\\c_1+c_2e^t\\ \end{matrix} \right)$$
$$\left(\begin{matrix} u(t)\\\ v(t)\\ \end{matrix} \right)=c_1\left(\begin{matrix}t\\1\\ \end{matrix} \right)+c_2\left(\begin{matrix}e^t\\e^t\\ \end{matrix} \right)$$
A: Wow, thanks so much for helping me, yet again, really awesome!
