Solve the following linear system with initial conditions: $$$\displaystyle \left\{\begin{array}{rcl} x' & = & \begin{pmatrix} 1 & -1 \\ 1 & -1\end{pmatrix}\cdot x+ \begin {pmatrix} \frac{1}{t} \\ \frac{1}{t}\end{pmatrix} \\ x(1) & = & \begin{pmatrix} 2 \\ -1 \end{pmatrix} \end{array}\right.$$$
Development:
Let's solve firstly the homogeneous part. We have to solve: $$$x'=\begin{pmatrix} 1 & -1 \\ 1 & -1\end{pmatrix}\cdot x$$$ The corresponding matrix does not diagonalize, let's calculate the reduced Jordan form: $$$J=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$$$ where $$\lambda=0$$ is the eigenvalue of multiplicity $$2$$.
A matrix of eigenvectors and therefore the change of basis matrix $$S$$ is: $$$S=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$$
We know that a fundamental matrix of the homogeneous system is: $$$\phi_h(t)=S\cdot e^{tJ}=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} e^{0t} & 0 \\ t\cdot e^{0t} & e^{0t} \end{pmatrix}= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\cdot \begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix}=\begin{pmatrix} 1+t & 1 \\ t & 1 \end{pmatrix} $$$
Let's look now for a particular solution to the non homogeneous system of the type: $$$x_p(t)=\phi_h(t)\cdot u(t)$$$ We know that $$u(t)$$ is such that $$u'(t)=(\phi_h(t))^{-1}\cdot b(t)$$
And so, as $$$(\phi_h(t))^{-1}=\begin{pmatrix} 1+t & 1 \\ t & 1 \end{pmatrix}^{-1}=\begin{pmatrix} 1 & -1 \\ -t & t+1 \end{pmatrix}$$$ we have that $$$u'(t)=\begin{pmatrix} 1 & -1 \\ -t & t+1 \end{pmatrix}=\begin{pmatrix} \dfrac{1}{t} \\ \dfrac{1}{t} \end{pmatrix}=\begin{pmatrix} \dfrac{1}{t}-\dfrac{1}{t} \\ \dfrac{-t}{t}+\dfrac{t+1}{t} \end{pmatrix} = \begin{pmatrix} 0 \\ \dfrac{1}{t} \end{pmatrix}$$$ and solving this separable ODE: $$$u(t)=\begin{pmatrix} \int 0 dt \\ \int\dfrac{1}{t}dt \end{pmatrix} = \begin{pmatrix} 0 \\ \ln(t) \end{pmatrix}$$$ Consequently a particular solution is: $$$x_p(t)=\phi_h(t)\cdot u(t)=\begin{pmatrix} 1+t & 1 \\ t & 1 \end{pmatrix}\cdot \begin{pmatrix} 0 \\ \ln(t) \end{pmatrix}=\begin{pmatrix} \ln(t) \\ \ln(t) \end{pmatrix}$$$ The general solution of the system is: $$$x(t)=x_h(t)+x_p(t)=\phi_h(t)\cdot C+x_p(t)=\begin{pmatrix} 1+t & 1 \\ t & 1 \end{pmatrix}\cdot \begin{pmatrix} C_1 \\ C_2 \end{pmatrix}+\begin{pmatrix} \ln(t) \\ \ln(t) \end{pmatrix}$$$ Now it is only necessary to impose the initial conditions, so we will find the vector $$C$$.
Let's evaluate the solution at $$t=1$$ and let's impose that the solution is $$(2,-1)$$: $$$\begin{pmatrix} 2 \\ -1 \end{pmatrix}=x(1)= \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\cdot \begin{pmatrix} C_1 \\ C_2 \end{pmatrix}+\begin{pmatrix} 0 \\ 0 \end{pmatrix} \Rightarrow $$$ $$$\Rightarrow \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\cdot \begin{pmatrix} C_1 \\ C_2 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$$$ Solving the linear system, we obtain: $$C=\begin{pmatrix} 3 \\ -4 \end{pmatrix} $$.
Therefore the solution is: $$$x(t)=\begin{pmatrix} 1+t & 1 \\ t & 1 \end{pmatrix}\cdot \begin{pmatrix} 3 \\ -4 \end{pmatrix}+ \begin{pmatrix} \ln(t) \\ \ln(t) \end{pmatrix}=\begin{pmatrix} 3+3t-4+\ln(t) \\ 3t-4+\ln(t) \end{pmatrix}=$$$ $$$=\begin{pmatrix} 3t-1+\ln(t) \\ 3t-4+\ln(t) \end{pmatrix}$$$
Solution:
$$x(t)=\begin{pmatrix} 3t-1+\ln(t) \\ 3t-4+\ln(t) \end{pmatrix}$$