For each of the following linear systems of differential equations, (i) find the general real solution (ii) show that the solutions are linearly independent (iii) draw the phase portrait
a. $$\dot x=\begin{bmatrix}6&-3\\2&1\end{bmatrix}x$$
find the characteristic equation: $$(6-λ)(1-λ)-(-3*2)=0$$ $$(λ-3)(λ-4)=0$$ with roots $λ_1=3$ and $λ_2=4$
for the eigenvalue, $λ_1=3$: $$(A-λ_1I)v_1=(A-3I)v_1=0$$ $$=\begin{bmatrix}6&-3\\2&1\end{bmatrix}-3\begin{bmatrix}1&0\\0&1\end{bmatrix}v_1$$ $$=\begin{bmatrix}3&-3\\2&-2\end{bmatrix}v_1$$
row reduced form: $$=\begin{bmatrix}1&-1\\0&0\end{bmatrix}v_1=0$$
find the eigenvector $v_1$: $$\begin{bmatrix}1&-1\\0&0\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=0$$ $$a-b=0$$ $$a=b$$ $$v_1=\begin{bmatrix}1\\1\end{bmatrix}$$
first solution: $x_1(t)=e^{3t}\begin{bmatrix}1\\1\end{bmatrix}$
for the eigenvalue, $λ_2=4$: $$(A-λ_2I)v_2=(A-4I)v_2=0$$ $$=\begin{bmatrix}6&-3\\2&1\end{bmatrix}-4\begin{bmatrix}1&0\\0&1\end{bmatrix}v_2$$ $$=\begin{bmatrix}2&-3\\2&-3\end{bmatrix}v_2$$
row reduced form: $$=\begin{bmatrix}1&-1\\0&0\end{bmatrix}v_2=0$$
find the eigenvector $v_2$: $$\begin{bmatrix}1&-1\\0&0\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=0$$ $$a-b=0$$ $$a=b$$ $$v_2=\begin{bmatrix}2\\3\end{bmatrix}$$
second solution: $x_2(t)=e^{4t}\begin{bmatrix}1\\1\end{bmatrix}$
matrix solution: $\begin{bmatrix}e^{3t}2e^{4t}\\e^{3t}&3e^{4t}\end{bmatrix}$
wronskian: $$W(t)=det\begin{bmatrix}e^{3t}&2e^{4t}\\e^{3t}&3e^{4t}\end{bmatrix}$$ $$=e^{7t}$$
t=0:
$$W(O)=det(v_1,v_2)=det\begin{bmatrix}1&2\\1&3\end{bmatrix}=1$$
So it's linearly independent.
I am really unsure about my work for the eigenvector,does this look correcT? If not, could you explain how I would find the eigenvector?
(I know I am not done answering the problem, still working through it)