solve $x_1'(t) =x_1 (t) + 2x_2(t) , x_2'(t)=2x_1(t)+4x_2(t)$

Question

I have found eigenvalues and eigenvectors, how do I find x with these values?

Answer 1

Hint: Note that both $$ {\rm e}^{\lambda_it} \mathbf{v}_i $$ are (linearly independent) solutions of

$$ \mathbf{x}'(t) =A \mathbf{x}(t) $$

Use the definition of eigenvalues/eigenvectors: $A\mathbf{v}_i = \lambda_i \mathbf{v}_i$.

Edit: Taking linear combinations, we get the general solution:

$$ \mathbf{x}(t) = k_1 {\rm e}^{\lambda_1 t} \mathbf{v}_1 + k_2 {\rm e}^{\lambda_2 t} \mathbf{v}_2 $$

($k_1$, $k_2$ arbitrary constants).

Answer 2

Based on your work, we have

$\pmatrix{x_1(t)\\x_2(t)}'=\pmatrix{1&2\\2&4}\pmatrix{x_1(t)\\x_2(t)}=\pmatrix{1&-2\\2&1}\pmatrix{5&0\\0&0}\pmatrix{1&-2\\2&1}^{-1}\pmatrix{x_1(t)\\x_2(t)}$

with the solution

$\pmatrix{x_1(t)\\x_2(t)}=\pmatrix{1&-2\\2&1}\pmatrix{e^{5t}&0\\0&0} \pmatrix{\frac15&\frac25\\-\frac25&\frac45}\pmatrix{c_1\\c_2}=\pmatrix{\frac15e^{5t}+\frac45&\frac25e^{5t}-\frac25\\\frac25e^{5t}-\frac25&\frac45e^{5t}+\frac15}\pmatrix{c_1\\c_2}.$