Determine the general solution of the system
$$ \begin{split} x'&=(3t-1)x-(1-t)y+te^{t^{2}}\\ y'&=-(t+2)x+(t-2)y-e^{t^{2}} \end{split} $$
Hint: The homogeneous system has a solution of the form $(x(t),y(t))= (\phi(t), -\phi(t))$
For this problem I attempted using the Method of Variation of Constants.
First I have the matrix $Y(t) = \begin{bmatrix} (3t-1)& -(1-t) \\ -(t+2) & (t-2) \end{bmatrix}$ and $b(t) = \begin{bmatrix} te^{t^{2}} \\ -e^{{t^2}} \end{bmatrix}$
Then $$Y^{-1}(t) = \frac{1}{4t^{2}-6t} \begin{bmatrix} t-2& 1-t \\ t+2 & 3t-1 \end{bmatrix}$$
I am not sure how to continue.
Inspired by the solution of the homogeneous equation, make a transformation of variables to $$ v=\frac{x+y}2\\u=\frac{x-y}2 $$ or $$ x = v+u\\y=v-u $$ The straightforward algebra on the two equations, substituting for $x$ and $y$, gives $$ v' = \frac12 (t-1)e^{t^2}+(2t-3)v \\ u' = \frac12 (t+1)e^{t^2}+2tu (2t+1)v $$ The second of these is not immediately useful but the first is an ordinary differential equation for $v(t)$ and its solution is $$ v(t) = \left( C e^{-3t} + \frac{t}6-\frac29\right)e^{t^2} $$ Now we can make use of the $u'$ equation, buy substituting this expression for $v(t)$. Again we get an easily solved ordinatry differential equation; the solution is: $$ u(t) = \left(+\frac{2t^3+2t^2+5t}{18} -\frac{6t+5}9 C e^{-3t} +K \right)e^{t^2} $$ Adding $v+u$ gives $x$; subtracting gives $y$.