# ODE system, find initial conditions

I am trying to solve this problem:

Given the system $$x_1'=-x_2$$$$x_2'=2x_1+3x_2$$

Find the general solution and the set of initial conditions such that the solution tends to $0$ when $t$ tends to $+\infty$ and when $t$ tends to $-\infty$.

I could find a general solution by first trying to find a solution of the form $X(t)=\xi e^{\alpha}t$ with $\xi$ in $\mathbb R^2$ and $\alpha$ a real number. By assuming there is a solution of this form one gets to linear independent solutions $X_1(t)=c_1e^t\left(\begin{array}{r} 1 \\ -1 \\ \end{array}\right),X_2(t)=c_2e^{2t}\left(\begin{array}{r} 1 \\ -2 \\ \end{array}\right)$

Since the vector space of solutions is two dimensional, then any solution is of the form $$X(t)=c_1e^t\left(\begin{array}{r} 1 \\ -1 \\ \end{array}\right)+c_2e^{2t}\left(\begin{array}{r}1 \\ -2 \\ \end{array}\right)$$

I don't know what to do in the second part, any help would be appreciated.

• Can you derive the asymptotics of solution when $t \rightarrow + \infty$ or $t \rightarrow - \infty$? You have to choose coefficients $c_1$ and $c_2$ in such a way that solution tends to zero. – Evgeny Aug 12 '15 at 5:40
• Something's wrong with your solution. The eigenvalues are clearly not 1 and 2, since the determinant of the system's matrix is 5... – Hans Lundmark Aug 12 '15 at 6:58
• I've corrected the first equation, it wasn't $x_1'=x_1-x_2$, sorry for my mistake. – user16924 Aug 12 '15 at 15:07
• @user16924 The advice still applies. You know the exact solution. You can figure out what happens with its magnitude when $t$ goes to $+ \infty$ or $- \infty$. That definitely depends on constants $c_1$ and $c_2$. – Evgeny Aug 12 '15 at 16:03