00:02
Okay, to solve this differential equation, we first think about it as a polynomial equation, and we try and solve that.
00:10
So, oh, i know i can't factor 29, except for enda 1 in 29, so we have to use the quadratic formula on it.
00:16
And remember, it goes like this.
00:18
If a x squared plus bx plus c equals 0, then x equals minus b plus minus the square root of b squared minus 4ac .a.
00:33
All of it over to a.
00:36
So in our problem, a is 1, b is 4, c is 29, m is x.
00:45
So m equals minus 4 plus or minus 4 squared, minus 4 times 1 times 29, all over 2, minus minus 4 plus or minus 16, you know what, instead of that, i'm going to factor the 4 out.
01:12
So i have four times four minus 29.
01:17
I made the 16 into four times four.
01:19
Factored out this four.
01:24
So i get minus four plus or minus the square root of four times minus 25 all over two.
01:37
So that's minus four plus or minus.
01:40
I'm going to write that as minus one times 25.
01:42
Square root of four, two, square root of minus one, i, square root of 25, five.
01:48
Oh, that's a weird way to write it.
01:50
Let me write it in a better order.
01:53
Square to minus 1.
01:54
I, square root of 4, 2, square to 25, 5.
02:02
Now i'm going to factor 2 out of everything.
02:05
So i have a minus 2 plus or minus 5i all over 2.
02:10
So i get minus 2 plus or minus 5i.
02:16
So then the solution is going to be c1, e to the real part, minus 2x, cosine imaginary part, 5x, plus c2, e to the minus 2x, sign 5x.
02:36
Okay, now we use the initial conditions to find c1 and c2...