How to find the general solution of a system of ODEs?

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}

• I just noticed: it should be a $-3$ in $A$, not a $3$. Although, this does not affect the eigenvalues, but it will affect the eigenvectors. – Dave Oct 14 '17 at 15:16
• I calculated the eigenvectors using -3 on my notebook so those are correct. It's just a typo here. – mm-crj Oct 14 '17 at 15:22
• @Harry49, so can you suggest what am I doing wrong? – mm-crj Oct 14 '17 at 16:06

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.
You can use this fact. Given $\dot x_1(t)=-x_1-3x_2$ and $\dot x_2=2x_2$. Adding this two equation gives $\dot x_1+\dot x_2=-(x_1+x_2)$, so that $\frac{d}{dt}(\dot x_1+\dot x_2)=-(x_1+x_2)$ which gives $(x_1+x_2)(t)=c_1e^{-t}$ and this very well agrees with the answer you found out.
• This fact is true, even if it's nothing more than writing the time evolution of $y=P^{-1} x$. However, if $c=x (0)$, it must be $(x_1 + x_2)(t) = (c_1 + c_2)e^{-t}$. – Harry49 Oct 14 '17 at 19:44
The solution $x_1=y_1−y_2$, $x_2=y_2$ is also correct as mentioned by @Nobagoto. To get to the answer in the book you need to use equation $(1)$. The matrices $P$ and $P^{-1}$ are known so $x(t)$ can be easily calculated.