6.4 Systems of differential equations

1 Find the solution for the system

y_{1}^{'} = 3 y_{1} + y_{2}

y_{2}^{'} = y_{1} + 3 y_{2}

Matrify it:

y^{'} = A y \to [y_{1}^{'} y_{2}^{'}] = [3113] [y_{1}^{'} y_{2}^{'}]

A = [3113]

Find Eigenvalues of A:

3 \pm 3^{2} - 8 \to 3 \pm 1

λ_{1} = 2, λ_{2} = 4

Find Eigenvectors of A:

(A - λ I) v = 0

For $λ = 2$ :

([3113] - 2 [1001]) v = [00]

[1111] v = [00]

From the first row: $v_{1} + v_{2} = 0$ , which can only be true and nonzero if $v_{1} = - v_{2}$

v_{2} = [1 - 1]

For $λ = 2$ :

[- 1 1 1 - 1] v = [00]

From the first row: $- v_{1} + v_{2} = 0$ , which can only be true and nonzero if $v_{1} = v_{2}$

v_{4} = [11]

From Differential Equations, we know the solution for this type of problem is of the form:

y = v e^{λ t}

And so the general solution is:

y = C_{1} e^{2 t} [1 - 1] + C_{2} e^{4 t} v_{4} = [11]

y'_1 = 3y_1 + y_2 $$$$ y'_2 = -2y_1 + y_2 $$

A = \begin{bmatrix} 3 & 1 \ -2 & 1 \end{bmatrix}

E i g e n v a l u es :

2 \pm \sqrt{2 - 5}

\lambda = 2 \pm i

F in d e i g e n v ec t ors (w eo n l y n ee d t o d oo n es in ce * * co m pl e x e i g e n v ec t ors a lw a ysco m e in co nj ug a t e p ai rs * *) :

\begin{bmatrix} 3-(2 + i) & 1 \ -2 & 1-(2 + i) \end{bmatrix} = \begin{bmatrix} 1- i & 1 \ -2 & -1- i \end{bmatrix}

(1-i)v_1 + v_2 = 0 \to v_2 = -1(1-i)v_1

-2v_1 + (-1 - i)v_2 = 0

We can pick $v_1 = 1$, and that is guaranteed to be on an eigen-line:

v_2 = -1(1-i)1

Whi c h f ina ll y g i v ese i g e n v ec t or :

v_+ = \begin{bmatrix} 1 \ -1 + i \end{bmatrix}

an d i t^{'} sco nj ug a t e :

v_- = \begin{bmatrix} 1 \ -1 - i \end{bmatrix}

> Since the general solution for a system with complex eigenvalues $\lambda = \alpha \pm \beta i$ and eigenvector $\mathbf{v}$ involves exponentials with sine and cosine, we write: >

\mathbf{y} = e^{\alpha t} \left( C_1 \operatorname{Re}(\mathbf{v} e^{i\beta t}) + C_2 \operatorname{Im}(\mathbf{v} e^{i\beta t}) \right)

y_1 = e^{2t} (C_1 \cos t + C_2 \sin t)

y_2 = e^{2t} \left( C_1 (-\cos t - \sin t) + C_2 (\cos t - \sin t) \right)