Non homogeneous systems of linear ODE with constant coefficients

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: $$x(t)=\begin{pmatrix} 1+t & 1 \\ t & 1 \end{pmatrix}\cdot \begin{pmatrix} 3 \\ -4 \end{pmatrix}+ \begin{pmatrix} \ln(t) \\ \ln(t) \end{pmatrix}=\begin{pmatrix} 3+3t-4+\ln(t) \\ 3t-4+\ln(t) \end{pmatrix}=$$ $$=\begin{pmatrix} 3t-1+\ln(t) \\ 3t-4+\ln(t) \end{pmatrix}$$