System of Differential Equations (Calculus)

Question

I have been stuck at the following differential system problem for so long. Any help would be great.

Question Given the following differentialsystem,

$$ \left\{ \begin{array}{c} x'_1(t)=-x_2(t) \\ x'_2(t)=-x_1(t)\\ \end{array} \right. $$

a) Find all real solutions using the eigenvalue method.

b) Now find the solution for which $x_1(0) = - 1, x_2(0) = 1$.

Any help would be great. Thank you.

Answer 1

Hint. Consider the corresponding system in matrix notation $$x'(t) = Ax(t)$$ where $$A :=\begin{bmatrix}0 & -1\\ -1 & 0\end{bmatrix}$$ and $$x(t) :=\begin{bmatrix} x_1(t)\\ x_2(t)\end{bmatrix}$$ Now calculate the eigenvectors and eigenvalues of $A$.

Solution. We have $$p_A = \det(A - I_2) = X^2 - 1 = (X + 1)(X - 1)$$ so $\lambda_1 = 1$ and $\lambda_2 = -1$ with the corresponing eigenvectors $$v_1 = \begin{bmatrix} -1\\ 1\end{bmatrix} \qquad v_2 = \begin{bmatrix} 1\\ 1\end{bmatrix} $$ Thus the real solution is given by $$x(t) = c_1 \begin{bmatrix} -1\\ 1\end{bmatrix}e^{t} + c_2\begin{bmatrix} 1\\ 1\end{bmatrix}e^{-t}$$ for $c_1,c_2 \in \mathbb{R}$. This solves part a. The initial value problem implies $c_1 = 1$ and $c_2 = 0$. So $$x(t) = \begin{bmatrix} -1\\ 1\end{bmatrix}e^{t}$$ solves the IVP.

Answer 2

Hint:

because the symmetric in two equtions, so the $$x_1(x)=x_2(x)$$