ODE system - finding the general solution given 2 solutions Given the following ode system  
$\begin{cases}tx_1'=2x_2+2x_3+t^3e^t\\tx_2'=-x_1+3x_2+x_3+t^4\\tx_3'=-x_1+x_2+3x_3+t^3e^t\end{cases}$
and $2$ solutions of the system $$u_{(1)}=\begin{pmatrix}(a+b)t^2\\at^2\\bt^2\end{pmatrix},\quad u_{(2)}=\begin{pmatrix}2\ln t+1\\ \ln t+1\\\ln t+1\end{pmatrix}$$

What is the system's general solution ?

Attempt
I think it is quite clear that some sort of parameters variation is required due to given solutions. The only problem is that I am not sure how to approach this solution. 
I tried to develop some sort of a matrix  $$\psi(t)=\begin{pmatrix}(a+b)t^2&2\ln t+1\\at^2&\ln t+1\\bt^2&\ln t+1\end{pmatrix}$$ such that for some $v=\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$ fulfills $\psi(t)v'(t)=\begin{pmatrix}t^2e^t\\t^3\\t^2e^t\end{pmatrix}$;
the same method we solve linear ode systems, but clearly this matrix multipication is undefined. 
How should I approach this? Thank you very much!
 A: The system matrix $$A=\pmatrix{0&2&2\\-1&3&1\\-1&1&3}$$ has the characteristic polynomial
$$
\det(A-\lambda I)
%=\det\pmatrix{-λ&2&2\\-1&3-λ&1\\-1&1&3-λ}
=-λ(3-λ)^2-4+λ+4(3-λ)
\\
=-λ^3+6λ^2-12λ+8=(-λ+2)^3
$$
so that substituting $u_k=x_k/t^2$, $u_k'=(tx_k'-2x_k)/t^3$, results in the system
$$
\left\{\begin{aligned}
tu_1'&=-2u_1+2u_2+2u_3+te^t\\
tu_2'&=-u_1+u_2+u_3+t^2\\
tu_3'&=-u_1+u_2+u_3+te^t
\end{aligned}\right.
$$
As one can see, the homogeneous part is always a multiple of $v=-u_1+u_2+u_3$ with
$$
tv'=t^2\implies v=\frac12t^2+c
$$
and consequently
\begin{align}
tu_1'&=2v+te^t&\implies u_1 &= \frac12t^2+2c\ln t+e^t+d_1\\
tu_2'&=v+t^2&\implies u_2&=\frac34t^2+c\ln t+d_2\\
tu_3'&=v+te^t&\implies u_3&=\frac14t^2+c\ln t+e^t+d_3
\end{align}
with $c=-d_1+d_2+d_3$. With setting $d_2=a+c$, $d_3=b+c$, $d_1=d_2+d_3-c=a+b+c$ one gets expressions close to those you cited, but with extra terms.
A: *Another approach *
$$\cases{tx_1'=2x_2+2x_3+t^3e^t\\tx_2'=-x_1+3x_2+x_3+t^4\\tx_3'=-x_1+x_2+3x_3+t^3e^t}$$
We can observe that $r_2-r_3$ gives :
$$t(x'_2-x'_3)=2(x_2-x_3)+t^4-t^3e^t$$
Thats easy to integrate  (substitute $z=x_2-x_3)$
$$ \implies tz'=2z+t^4-t^3e^t$$ 
$$ \implies t^2z'-2tz=t^5-t^4e^t \implies (\frac z {t^2})'=t-e^t $$
$$ \boxed{\implies x_2=x_3+\frac {t^4}2-e^tt^2+Kt^2}$$
You have $x_2$ as a function of $x_3$ and t
We also have $r_1-2r_3$ which gives another easy integrable equation
$$t(x'_1-2x'_3)=2(x_1-2x_3)-t^3e^t$$
$$\implies tv'=2v-t^3e^t$$
Solve that equation with the same method as for the first one
$$ \boxed{\implies x_1=2x_3-e^tt^2+Ct^2}$$
You get $x_1$ in function of $x_3$ and t
Then take any equation and substitute these values ..and integrate
$$tx_3'=-x_1+x_2+3x_3+t^3e^t$$
$$tx_3'-2x_3=t^3e^t+\frac{t^4}2+Kt^2-Ct^2$$
$$(\frac {x_3}{t^2})'=e^t+\frac t 2+\frac 1t(K-C)$$
$$\boxed{\implies x_3=t^2e^t+\frac {t^4}4+(K-C)t^2\ln|t|+At^2 }$$
A: A quite general solution to these kinds questions is tried here,
The system can be written as
$$
x'(t) = A(t)x(t)+b(t).
$$, with $A(t)$ commuting for all $t$. This has a general solution written as
$$
x(t) = \exp(A(t))\left (x(0) + \int_0^t\exp(-A\tau)b(\tau)\,d\tau \right )
$$
We can solve for $x(0)$ and get
$$
x(0) = \exp(-A(t))x(t) - \int_0^t\exp(-A\tau)b(\tau)\,d\tau 
$$
Let $c(t) = \int_0^t\exp(-A\tau)b(\tau)\,d\tau$, hence
$$
x(0) = \exp(-A(t)) x(t) - c(t)
$$
If you have two solutions $u_1,u_2$, then you get two solutions $x_1(0),x_2(0)$. Assume $x_1(0),x_2(0)$ linearly independent, then you can take $x_3(0) = x_1(0)\times x_2(0)$ and write the general solution as
$$
a_1 x_1(0)+a_2 x_2(0) + a_3 x_3(0) = \exp(-A(t)) x_{a}(t) - c(t)
$$
L.H.S is the same as
$$
a_1 (\exp(-A(t))x_1(t)-c(t)) + a_2 (\exp(-A(t))x_2(t)-c(t)) +a_3(\exp(-A(t))x_1(t)-c(t))\times (\exp(-A(t))x_2(t)-c(t))=L_a(t)
$$
And we get,
$$
x_a(t) = \exp(A(t))(L_a(t) + c(t))
$$ And you can express all solutions with the solutions $x_1(t)$ and $x_2(t)$
