00:01
For this problem, we first rewrite the ode system with an operator d equals to d over d.
00:08
So this is our system of ode.
00:14
For equation one, we use an operator d plus 3 acting on it.
00:25
So d plus 3 acting on minus 2, it gives us minus 6.
00:30
And for the second equation, we just multiply everything by 3.
00:38
So this new system of ode is.
00:42
Have as the same term here.
00:45
So you can do the elimination.
00:47
We use the first one, subtract the second one.
00:49
So we have d squared minus 2d minus 3, x equals to minus 3.
00:56
So this is a homogeneous, a non -homogeneous ode with, so the homogeneous part has a characteristic equation r squared minus 2r minus 3 equals to 0.
01:09
That gives us r equals to 3 and the minus 1.
01:13
So for the homogeneous part, the general solution should be c1e to 3t plus c2e to the minus t...