6.4 Systems of differential equations

1 Find the solution for the system

$$y_1^{\prime }=3y_1+y_2$$ $$y_2^{\prime }=y_1+3y_2$$

Matrify it:

$$y^{\prime }=Ay\to \left[\begin{array}{c}{y}_1^{\prime }\\ y_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}3& 1\\ 1& 3\end{array}\right]\left[\begin{array}{c}{y}_1^{\prime }\\ y_2^{\prime }\end{array}\right]$$ $$A=\left[\begin{array}{cc}3& 1\\ 1& 3\end{array}\right]$$

Find Eigenvalues of A:

$$3\pm \sqrt{3^2-8}\to 3\pm 1$$ $${\lambda }_1=2,{\lambda }_2=4$$

Find Eigenvectors of A:

$$(A-\lambda I)v=0$$

For $\lambda =2$:

$$\left(\left[\begin{array}{cc}3& 1\\ 1& 3\end{array}\right]-2\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right)\vec{v}=\left[\begin{array}{c}0\\ 0\end{array}\right]$$ $$\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\vec{v}=\left[\begin{array}{c}0\\ 0\end{array}\right]$$

From the first row: $v_1+v_2=0$, which can only be true and nonzero if $v_1=-v_2$

$$\overrightarrow{v_2}=\left[\begin{array}{c}1\\ -1\end{array}\right]$$

For $\lambda =2$:

$$\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]\vec{v}=\left[\begin{array}{c}0\\ 0\end{array}\right]$$

From the first row: $-v_1+v_2=0$, which can only be true and nonzero if $v_1=v_2$

$$\overrightarrow{v_4}=\left[\begin{array}{c}1\\ 1\end{array}\right]$$

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

$$y=v{e}^{\lambda t}$$

And so the general solution is:

$$y=C_1{e}^{2t}\left[\begin{array}{c}1\\ -1\end{array}\right]+C_2{e}^{4t}\overrightarrow{v_4}=\left[\begin{array}{c}1\\ 1\end{array}\right]$$

2 Find the general solution for the system

y'_1 = 3y_1 + y_2 $$$$ y'_2 = -2y_1 + y_2 $$A=\left[\begin{array}{cc}3& 1\\ -2& 1\end{array}\right]$$

Eigenvalues:

$$2\pm \sqrt{2-5}$$

Eigenvalues are complex:

$$\lambda =2\pm i$$

Find eigenvectors (we only need to do one since complex eigenvectors always come in conjugate pairs):

$$\left[\begin{array}{cc}3-(2+i)& 1\\ -2& 1-(2+i)\end{array}\right]=\left[\begin{array}{cc}1-i& 1\\ -2& -1-i\end{array}\right]$$ $$(1-i)v_1+v_2=0\to v_2=-1(1-i)v_1$$ $$-2v_1+(-1-i)v_2=0$$

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

$$v_2=-1(1-i)1$$

Which finally gives eigenvector:

$$v_{+}=\left[\begin{array}{c}1\\ -1+i\end{array}\right]$$

and it’s conjugate:

$$v_{-}=\left[\begin{array}{c}1\\ -1-i\end{array}\right]$$

Again, from Differential Equations, we know that:

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:

$$>\mathbf{y}=e^{\alpha t}\left(C_1\mathrm{Re}(\mathbf{v}{e}^{i\beta t})+C_2\mathrm{Im}(\mathbf{v}{e}^{i\beta t})\right)>$$ $$y_1=e^{2t}(C_1\cos t+C_2\sin t)$$ $$y_2=e^{2t}\left(C_1(-\cos t-\sin t)+C_2(\cos t-\sin t)\right)$$