Problems from Non homogeneous systems of linear ODE with constant coefficients

Development:

Let's solve firstly the homogeneous part. We have to solve: $$x^{\prime }=\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)\cdot x$$ The corresponding matrix does not diagonalize, let's calculate the reduced Jordan form: $$J=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$$ where $\lambda =0$ is the eigenvalue of multiplicity $2$.

A matrix of eigenvectors and therefore the change of basis matrix $S$ is: $$S=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$$

We know that a fundamental matrix of the homogeneous system is: $${\varphi }_h(t)=S\cdot e^{tJ}=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\cdot \left(\begin{array}{cc}{e}^{0t}& 0\\ t\cdot e^{0t}& e^{0t}\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\cdot \left(\begin{array}{cc}1& 0\\ t& 1\end{array}\right)=\left(\begin{array}{cc}1+t& 1\\ t& 1\end{array}\right)$$

Let's look now for a particular solution to the non homogeneous system of the type: $$x_p(t)={\varphi }_h(t)\cdot u(t)$$ We know that $u(t)$ is such that $u^{\prime }(t)=({\varphi }_h(t){)}^{-1}\cdot b(t)$

And so, as $$({\varphi }_h(t){)}^{-1}={\left(\begin{array}{cc}1+t& 1\\ t& 1\end{array}\right)}^{-1}=\left(\begin{array}{cc}1& -1\\ -t& t+1\end{array}\right)$$ we have that $$u^{\prime }(t)=\left(\begin{array}{cc}1& -1\\ -t& t+1\end{array}\right)=\left(\begin{array}{c}{\displaystyle \frac{1}{t}}\\ {\displaystyle \frac{1}{t}}\end{array}\right)=\left(\begin{array}{c}{\displaystyle \frac{1}{t}}-{\displaystyle \frac{1}{t}}\\ {\displaystyle \frac{-t}{t}}+{\displaystyle \frac{t+1}{t}}\end{array}\right)=\left(\begin{array}{c}0\\ {\displaystyle \frac{1}{t}}\end{array}\right)$$ and solving this separable ODE: $$u(t)=\left(\begin{array}{c}\int 0dt\\ \int {\displaystyle \frac{1}{t}}dt\end{array}\right)=\left(\begin{array}{c}0\\ \ln (t)\end{array}\right)$$ Consequently a particular solution is: $$x_p(t)={\varphi }_h(t)\cdot u(t)=\left(\begin{array}{cc}1+t& 1\\ t& 1\end{array}\right)\cdot \left(\begin{array}{c}0\\ \ln (t)\end{array}\right)=\left(\begin{array}{c}\ln (t)\\ \ln (t)\end{array}\right)$$ The general solution of the system is: $$x(t)=x_h(t)+x_p(t)={\varphi }_h(t)\cdot C+x_p(t)=\left(\begin{array}{cc}1+t& 1\\ t& 1\end{array}\right)\cdot \left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)+\left(\begin{array}{c}\ln (t)\\ \ln (t)\end{array}\right)$$ 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 $(2,-1)$: $$\left(\begin{array}{c}2\\ -1\end{array}\right)=x(1)=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)\cdot \left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)+\left(\begin{array}{c}0\\ 0\end{array}\right)\Rightarrow$$ $$\Rightarrow \left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)\cdot \left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)=\left(\begin{array}{c}2\\ -1\end{array}\right)$$ Solving the linear system, we obtain: $C=\left(\begin{array}{c}3\\ -4\end{array}\right)$.

Therefore the solution is: $$x(t)=\left(\begin{array}{cc}1+t& 1\\ t& 1\end{array}\right)\cdot \left(\begin{array}{c}3\\ -4\end{array}\right)+\left(\begin{array}{c}\ln (t)\\ \ln (t)\end{array}\right)=\left(\begin{array}{c}3+3t-4+\ln (t)\\ 3t-4+\ln (t)\end{array}\right)=$$ $$=\left(\begin{array}{c}3t-1+\ln (t)\\ 3t-4+\ln (t)\end{array}\right)$$

Solution:

$x(t)=\left(\begin{array}{c}3t-1+\ln (t)\\ 3t-4+\ln (t)\end{array}\right)$

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