Solve the linear system of differential equations x' = x - 2y, y' = 2x + y.

Step-by-step solution

1. 1

   Write the system as a single matrix equation. With $\mathbf{X} = \begin{pmatrix} x \\ y \end{pmatrix}$ the system is $\mathbf{X}' = A\mathbf{X}$ where

   $A = \begin{pmatrix} 1 & -2 \\ 2 & 1 \end{pmatrix}$

   This matters because the whole solution is then determined by the eigenvalues of $A$ — no substitution or elimination is needed.

2. 2

   Find the eigenvalues. The characteristic equation is

   $\det(A - \lambda I) = (1-\lambda)^2 + 4 = 0 \quad\Longrightarrow\quad (1-\lambda)^2 = -4$

   so $\lambda = 1 \pm 2i$. A complex pair $\alpha \pm \beta i$ always produces $e^{\alpha t}$ times sines and cosines of $\beta t$: growth rate $\alpha = 1$, rotation rate $\beta = 2$.

3. 3

   Find one eigenvector. For $\lambda = 1 + 2i$ the first row of $A - \lambda I$ reads $-2i\,v_1 - 2v_2 = 0$, so $v_2 = -i\,v_1$ and

   $\mathbf{v} = \begin{pmatrix} 1 \\ -i \end{pmatrix}$

   Only one eigenvector is required: the conjugate pair carries no new information.

4. 4

   Split the complex solution into real and imaginary parts. Using $e^{(1+2i)t} = e^{t}(\cos 2t + i\sin 2t)$,

   $e^{(1+2i)t}\mathbf{v} = e^{t}\begin{pmatrix} \cos 2t + i\sin 2t \\ \sin 2t - i\cos 2t \end{pmatrix}$

   Its real part $e^t(\cos 2t,\ \sin 2t)$ and imaginary part $e^t(\sin 2t,\ -\cos 2t)$ are two independent real solutions.

5. 5

   Combine them into the general solution.

   $x(t) = e^{t}\left(C_1\cos 2t + C_2\sin 2t\right), \qquad y(t) = e^{t}\left(C_1\sin 2t - C_2\cos 2t\right)$

6. 6

   Check by substitution. Differentiating $x$ gives $e^t(C_1\cos 2t + C_2\sin 2t) + e^t(-2C_1\sin 2t + 2C_2\cos 2t)$, and $x - 2y = e^t(C_1\cos 2t + C_2\sin 2t) - 2e^t(C_1\sin 2t - C_2\cos 2t)$ — the same expression. The second equation checks the same way. Since $\alpha = 1 > 0$, every trajectory spirals outward from the origin.