00:01
So we want to solve this differential equation.
00:02
So first of all, we start with the complementary solution.
00:07
And for that, we have to write the characteristic equation, which is r minus 1 equals 0 tells us r equals 1.
00:20
So y sub c is some constant times e to the x.
00:27
I'm also going to work with method of undetermined coefficients coefficients to find my y sub p.
00:43
So if i look at this, there's basically two kinds of terms on the right hand side.
00:49
There's this e to the 2x with an x in front of it, and we also get a constant term.
00:55
Okay, so our y sub p has to have an e to the 2x multiplied by a polynomial in x of order 1, and then we also have a constant term.
01:08
So there's actually three terms in my y sub p.
01:11
It's going to be, and not to be confused with c1, which was on our complementary solution.
01:28
All right, so let's take yp prime.
01:32
So we get this.
01:37
It's e to the 2x.
01:40
Then we get a plus 2ax plus b.
01:48
Okay.
01:48
And we'll substitute all this into the equation.
01:52
So we get, and that has to equal, all right.
02:29
So we collect terms...