I am trying to understand an example of Lawrance Perko, the problem being
\begin{align}
\dot{x}_1 &=-x_1-3x_2\\
\dot{x}_2 &=2x_2
\end{align}
Which can be written in the form $\dot{x}=Ax$ where :
\begin{align}
A=
\begin{bmatrix}
-1 & -3 \\
0 & 2
\end{bmatrix}
\end{align}
The eigen values of A are $\lambda_{1}=-1$,$\lambda_{1}=2$ and the eigenvectors are $v_1=[1,0]$ and $v_2=[-1,1]$ respectively. The matrix P and the decoupling matrix $P^{-1}$ are given by,
\begin{align}
P=\begin{bmatrix}
1 &-1 \\
0 & 1
\end{bmatrix},P^{-1}=
\begin{bmatrix}
1 & 1 \\
0 & 1
\end{bmatrix}
\end{align}
Under the co-ordinate transform $y=P^{-1}x$ the linearly uncoupled system is given by:
\begin{align}
\dot{y_1}=-y_1\\
\dot{y_2}=2y_2
\end{align}
Which has the general solution $y_1=c_1\exp(-t)$, $y_2=c_2\exp(2t)$. Now the general solution of the original system is iven by:
\begin{align}\tag{1}
x(t)=P\begin{bmatrix}
e^{-t} & 0 \\
0 & e^{2t}
\end{bmatrix}P^{-1}c
\end{align}
Where $c=x(0)$, or equivalently by
What I don't understand is the last step. According to my calculation $y=P^{-1}x$ so $x_1=y_1-y_2$, $x_2 = y_2$. So $x_1$ should be $c_1 e^{-t}-c_2e^{2t}$ and not $c_1 e^{-t}+c_2(e^{-t}-e^{2t})$ \ref{1}
Best Answer
Setting $t=0$ in the solution proposed by the author of the OP gives $x_1 (0) = c_1 - c_2$, which is in contradiction with the definition of $c = x (0)$. Equation $(1)$ is correct. If $y = P^{-1} x$, then we have $$y_1 (t) = e^{-t} y_1 (0) ,\quad y_2 (t) = e^{2t} y_2 (0),$$ and in particular $$x_1 (t) = y_1 (t) - y_2 (t) . $$ Now, it remains to express $y (0)$ in terms of $x (0) = c$, i.e. $$y_1 (0) = c_1 + c_2 ,\quad y_2 (0) = c_2,$$ which gives the result of the book.