Solving inhomogenous continuous-time system with non-diagonalisable system matrix

Question

I have an exercise where i have to find the general solution to this problem:

$$ X'=\left( \begin{matrix} 2&-1\\ 4&-2 \end{matrix} \right)X + \begin{pmatrix} 2\\1 \end {pmatrix}. $$ Obviously the system matrix has linearly dependent vectors, which means that it can't be diagonalsed. Another method I tried is to use an expansion and using it I got a homogeneous solution:

$$ X_h = e^{tA} \begin{pmatrix}C_1\\C_2\end{pmatrix} $$ Where $$ e^{tA} = I + tA + \frac{t^2A^2}{2!} + ... , where A^2 = 0 $$ The homogeneous solution is then $$ \begin{pmatrix} 1+2t&-t\\ 4t&1-2 \end{pmatrix} * \begin{pmatrix} C_1\\C_2 \end{pmatrix} $$ The problem is I dont know how to get the particular solution. I tried using $$ X_p = (sI - A)^{-1}\pmatrix{2\\1}e^{st} $$ Where s in this case is zero, but the matrix -A is not invertible. Can anyone help me?

Answer 1

If $A^2=0$ you can find easily $e^{At}=I+tA$ by the Taylor expansion. Now the solution is given by the formula $$ X(t)=e^{At}X(0)+e^{At}\int_0^te^{-As}\left[\matrix{2\\1}\right]\,ds.\tag1 $$ P.S. To get $(1)$ use the integrating factor $e^{-At}$ for the system $$ \underbrace{e^{-At}(X'-AX)}_{(e^{-At}X(t))'}=e^{-At}\left[\matrix{2\\1}\right]\quad\Leftrightarrow\quad e^{-At}X(t)-X(0)=\int_0^t e^{-As}\left[\matrix{2\\1}\right]\,ds. $$

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P.P.S. How the integrating factor works

1. We collect all terms that depend on $X$ on the left hand side and check if the derivative has the unit coefficient $$ X'(t)-AX(t)=f(t). $$
2. Now we would like to get a complete derivative in the left hand side after multiplying both sides of the equation by a proper chosen function, let call it $\Phi(t)$ $$ \color{red}{\Phi(t)}(X'-AX)=\color{red}{\Phi(t)}f(t)\qquad\Leftrightarrow\qquad \Phi(t)\cdot X'(t)+\underbrace{\Phi(t)(-A)}_{=\Phi'(t)}X(t)=\Phi(t)f(t) $$
3. We want the left hand side to be the complete derivative of $\Phi\cdot X$, which leaves us almost no other choice for $\Phi$ than $\Phi(t)=e^{-At}$. With this choice we have $$ (e^{-At}X(t))'=e^{-At}f(t). $$
4. Now integrate both sides from $0$ (or any other initial value for $t$) to $t$. Because the variable $t$ is occupied in the integral limit, we change the integration variable name to your favorite name (I like $s$) to avoid confusions $$ \int_0^t(e^{-As}X(s))'\,ds=\int_0^t e^{-As}f(s)\,ds\qquad\Leftrightarrow\qquad \underbrace{\left[e^{-As}X(s)\right]_0^t}_{=e^{-At}X(t)-I\cdot X(0)}=\int_0^t e^{-As}f(s)\,ds. $$
5. Move $X(0)$ to the right and multiply both sides by $e^{At}$ to get $(1)$.