by Guest » Fri Jun 05, 2020 12:04 pm
Have you NO idea how to even start? Can you not show any thing you have done or even tried on this? One problem with not showing any attempt of your own is that there are many different ways to do a problem like this (or any math problem) and we have no idea which you would be expected to use!
This is what I would do: I would change form the matrix form to two simultaneous equations-
[tex]\frac{dx}{dt}= -2x+ y- e^{2t}[/tex]
[tex]\frac{dy}{dt}= -3x+ 2y+ 6e^{2t}[/tex].
Now change from two first order equations to one second order equation.
Differentiate the first equation again
[tex]\frac{d^2x}{dt^2}= -2\frac{dx}{dt}+ \frac{dy}{dt}- 2e^{2t}[/tex].
Replace [tex]\frac{dy}{dt}[/tex] using the second equation
[tex]\frac{d^2x}{dt^2}= -2\frac{dx}{dt}+ (-3x+ 2y+ 6e^{2t})- 2e^{2t}[/tex]
[tex]\frac{d^2x}{dt^2}= -2\frac{dx}{dt}- 3x+ 2y+ 4e^{2t}[/tex].
From the first equation, [tex]y= \frac{dx}{dt}+ 2x+ e^{2t}[/tex] so we can write
[tex]\frac{d^2x}{dt^2}= -2\frac{dx}{dt}- 3x+ 2(\frac{dx}{dt}+ 2x+ e^{2t})+ 4e^{2t}[/tex]
[tex]\frac{d^2x}{dt^2}= x+ 6e^{2t}[/tex].
That has "associated homogeneous equation" [tex]\frac{d^2x}{dt^2}= x[/tex] which has "characteristic equation" [tex]r^2= 1[/tex] so r= 1 or -1. The general solution to that is [tex]x(t)= Ae^{t}+ Be^{-t}[/tex].
Now, the "undetermined coefficients method" is to look for a solution to the entire equation of the form [tex]x(t)= Ae^{2t}[/tex] where "A" is the "undetermined coefficient" that we have to determine. Then [tex]\frac{dx}{dt}= 2Ae^{2t}[/tex] and [tex]\frac{d^2x}{dt^2}= 4Ae^{2t}[/tex].
Putting those into the equation [tex]\frac{d^2x}{dt^2}= x+ 6e^{2t}[/tex], [tex]4Ae^{2t}= Ae^{2t}+ 6e^{2t}[/tex]. [tex]3Ae^{2t}= 6e^{2t}[/tex] so A= 2.
The general solution to the entire equation is [tex]x(t)= Ae^{t}+ Be^{-t}+ 2e^{2t}[/tex].
Use [tex]y= \frac{dx}{dt}+ 2x+ e^{2t}[/tex] to get y(t).