Recent content by Omzyma

So that would mean A=\tfrac{2}{3} and B = \tfrac{3}{5}! Thanks for your help. Just needed a little nudge I guess.

Thanks for your reply! Picking x_p = A \left ( e^{4t} \right )+ B \left ( e^{6t} \right ) as my particular solution I ended up with 3A \left ( e^{4t} \right )+ 5B \left ( e^{6t} \right ) = 2K_2 \left ( e^{4t} \right ) + 3K_1 \left ( e^{6t} \right ) At this point do I solve for both A and B...

Hello! I have the following initial value problem: \[ x' = x + 2y + 3z \] \[ y' = 4y + 5z \] \[ z' = 6z \] All I'm looking to do is find the general solution to this system, and as long as I'm doing this correctly I have these answers: \[ y(t) = K_2e^{4t} + \tfrac{5K_1}{2}e^{6t} \] \[...