I'm trying to solve a given system:
$x_1'=x_1+2x_2+5e^{4t}$
$x_2'=2x_1+x_2$
However, depending on if I apply (D-1) to the first or the second equation, then substitute in and use the annihilator method, I get two different answers.
Answer 1:
$x_1=c_1e^{3t}e^{-t}+3e^{4t}$
$x_2=c_1e^{3t}+c2e^{-t}+2e^{4t}$
Answer 2:
$x_1=c_1e^{3t}+c_2e^{-t}+3e^{4t}$
$x_2=-c_1e^{3t}+c_2e^{-t}+2e^{4t}$
Is this possible? I've tried playing around with negative signs but they don't equate, but I've checked and both solutions solve the system. I've also encountered other issues like this before when I'm solving systems (both homogeneous and nonhomogeneous)
Apologies for the shoddy notation; I tried to use Latex but apparently I don't have enough reputation to insert it.
Careful!
$$x_1=c_1e^{3t}+c_2e^{-t}+3e^{4t}$$ $$x_2=-c_1e^{3t}+c_2e^{-t}+2e^{4t}$$ does not satisfy your system.
Note that $$x_1'=3c_1e^{3t}-c_2e^{-t}+12e^{4t}$$ while $$ x_1+2x_2+5e^{4t} = -c_1e^{3t}+3c_2e^{-t}+12e^{4t}$$ Which are not the same.