Considérons le système linéaire suivant :

\(\left\{ \begin{array}{l} x_1'(t) = 4 x_1(t) + 2 x_2(t) \\ x_2'(t) = x_1(t) + 5 x_2(t) - x_3(t)\\ x_3'(t) = -x_1(t) + x_2(t) + 5 x_3(t) + e^{4t} \end{array} \right.\)

Avec : \(\left\{ \begin{array}{l} x_1(0) = 0;\\x_2(0) = 0;\\x_3(0) = 1. \end{array} \right.\)

On peut l'écrire sous la forme :

\(\left\{ \begin{array}{l} \underbrace{ \begin{pmatrix} x_1(t)\\ x_2(t)\\ x_3(t) \end{pmatrix}'}_{x'(t)} = \underbrace{ \begin{pmatrix} 4 & 2 & 0\\ 1 & 5 & -1\\ -1 & 1 & 5 \end{pmatrix}}_{\bf A} \cdot \underbrace{ \begin{pmatrix} x_1(t)\\ x_2(t)\\ x_3(t) \end{pmatrix}}_{x(t)} + \underbrace{ \begin{pmatrix} 0\\ 0\\ e^{4t} \end{pmatrix}}_{{\bf B}(t)}\\ \\ \underbrace{ \begin{pmatrix} x_1(0)\\ x_2(0)\\ x_3(0) \end{pmatrix}}_{x(0)} = \underbrace{ \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}}_{x_0} \end{array} \right.\)

On reconnaît alors la matrice \({\bf A}\) des exemples précédents.

On avait trouvé :

\({\bf P}^{-1}\cdot{\bf A}\cdot{\bf P} = \begin{pmatrix} 6 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 4 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix}\)

Soit :

\({\bf A} = {\bf P} \cdot \left[ \begin{pmatrix} 6 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 4 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix} \right]\cdot {\bf P}^{-1}\)

D'où :

\({\bf A}t = {\bf P} \cdot \left[ \begin{pmatrix} 6 t & 0 & 0\\ 0 & 4 t & 0\\ 0 & 0 & 4 t \end{pmatrix} + \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & t\\ 0 & 0 & 0 \end{pmatrix} \right]\cdot {\bf P}^{-1}\)

\(\begin{array}{rcl} \exp({\bf A}t) &=& {\bf P} \cdot \left[\exp \begin{pmatrix} 6 t & 0 & 0\\ 0 & 4 t & 0\\ 0 & 0 & 4 t \end{pmatrix} \cdot\exp \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & t\\ 0 & 0 & 0 \end{pmatrix} \right]\cdot {\bf P}^{-1}\\ ~\\ &=& {\bf P} \cdot \begin{pmatrix} e^{6 t} & 0 & 0\\ 0 & e^{4 t} & 0\\ 0 & 0 & e^{4 t} \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & t\\ 0 & 0 & 1 \end{pmatrix} \cdot {\bf P}^{-1} = {\bf P} \cdot \begin{pmatrix} e^{6 t} & 0 & 0\\ 0 & e^{4 t} & t e^{4 t}\\ 0 & 0 & e^{4 t} \end{pmatrix} \cdot {\bf P}^{-1}\\ ~\\ \exp({\bf A}t) &=& \dfrac{1}{2} \begin{pmatrix} e^{6t}+(1-2t)e^{4t} & e^{6t}-(1-2t)e^{4t} & - e^{6t}+(1+2t)e^{4t}\\ e^{6t}-e^{4t} & e^{6t}+e^{4t} & e^{4t}-e^{6t}\\ -2t e^{4t}& 2t e^{4t} & 2(1+t)e^{4t} \end{pmatrix} \end{array}\)

On peut alors appliquer notre théorème :

\(\begin{array}{rcl} X(t) &=& \exp({\bf A}(t-t_0))\cdot X_0 + \displaystyle\int_{t_0}^{t} \exp({\bf A}(t-s))\cdot{\bf B}(s)ds\\ &&\\ &=& \exp({\bf A}(t))\cdot \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} + \displaystyle \int_{0}^{t}{\exp({\bf A}(t-s))\cdot \begin{pmatrix} 0\\ 0\\ e^{4s} \end{pmatrix} ds}\\ &&\\ &=&\displaystyle \frac{1}{2} \begin{pmatrix} -e^{6t}+(1+2t) e^{4t}\\ e^{4t} - e^{6t}\\ 2 (1+t) e^{4t} \end{pmatrix} + \frac{1}{2}\underbrace{ \displaystyle \int_{0}^{t}{ \begin{pmatrix} -e^{6(t-s)} + (1+ 2(t-s))e^{4(t-s)}\\ e^{4(t-s)} - e^{6(t-s)}\\ 2 (1+t-s) e^{4(t-s)} \end{pmatrix} e^{4s} ds}}_{{\bf V}(t)} \end{array}\)

\({\bf V}(t) = \begin{pmatrix} \displaystyle \int_0^t \left(-e^{6t}e^{-2s} + (1+2t) e^{4t} - 2s e^{4t}\right) ds \\ \displaystyle \int_0^t \left(e^{4t} - e^{6t}e^{-2s} \right) ds \\ 2 \displaystyle \int_0^t \left((1+t) e^{4t} - s e^{4t} \right) ds \end{pmatrix} = \begin{pmatrix} e^{6t}\left[\frac{e^{-2s}}{2}\right]_{0}^{t} + (1+2t)e^{4t}\left[s\right]_{0}^{t} - e^{4t}\left[\frac{s^2}{2}\right]_{0}^{t} \\ e^{4t}\left[s\right]_{0}^{t}+ e^{6t}\left[\frac{e^{-2s}}{2}\right]_{0}^{t} \\ 2 (1+t)e^{4t}\left[s\right]_{0}^{t} - e^{4t}\left[s^2\right]_{0}^{t} \end{pmatrix}\)

Finalement :

\(X(t) = \begin{pmatrix} -\frac{3}{2} e^{6t} + \dfrac{e^{4t}}{2}(2t^2+6t+3) \\ -\frac{3}{2} e^{6t} + \dfrac{e^{4t}}{2}(2t+3)\\ e^{4t} (t^2+4t+2) \end{pmatrix},\) soit : \(\left\{ \begin{array}{l} x_1(t) = -\frac{3}{2} e^{6t} + \frac{e^{4t}}{2}(2t^2+6t+3)\\ x_2(t) = -\frac{3}{2} e^{6t} + \frac{e^{4t}}{2}(2t+3)\\ x_3(t) = e^{4t} (t^2+4t+2) \end{array} \right.\)