00:01
So in this problem, we're asked to prove theorem 6 .33b, part b.
00:08
And to do that, we need to let s be the solution space of the differential equation y double prime plus ay prime plus by equals 0.
00:27
And we're going to let gamma 1 and gamma 2 be the roots of the characteristic equation gamma squared plus a y a gamma plus b is equal to zero so then if gamma one and gamma two are equal what we wish to show is that b our basis is e to the gamma 1 t e to the gamma 2 t oh sorry i know i left something out a t i need a t right there in front of that t e to the gamma 2 t okay now, that basis then is a basis for that set, b is a basis for s.
01:49
All right.
01:51
Since s is the solution space of our second order, linear, ordinary, differential equation, because remember by definition, we meant s the solution space of this, which is a second order, linear differential equation, ordinary differential equation, then we see that it has dimension two.
03:07
In other words, that solution space is a two -dimensional space.
03:13
Thus, it suffices for us to show that our basis b is linearly independent and that each element of b, of course, belongs to s.
04:10
Well, since the elements of b are e to the gamma 1t and t, e to the gamma 2t, then these are not proportional.
04:56
All right.
05:07
So since they're not proportional, then we can notice that they're not a linear combination, so they're linearly independent.
05:18
All right...