00:01
Okay, we want to solve this differential equation with these initial conditions.
00:09
So the first step is to write out the characteristic equation, r cubed minus four y squared plus 10 y minus 12.
00:27
Okay, so if i substitute in two, i find out that two is a solution so i can factor out r minus two from that.
00:36
I get, so one of my roots is r equals two.
00:49
This one gives me a quadratic, tells me r is two plus or minus the square root of four minus 24 divided by two.
01:12
So that's four minus 24 is negative 20.
01:18
So it's one plus or minus the square root of five.
01:28
So that tells me that y has the following three solutions.
01:33
So arbitrary constant c1 times e to the two x plus we're gonna have, this has a real part of one so it's got e to the x, c2 times the cosine of root five x.
01:52
X plus c3 times the sine of root five x.
02:02
Okay, so then to put in the initial conditions, i gotta take the derivatives of these things.
02:13
So y prime, two c1, e to the two x plus e to the x times all that c2 cosine root five x.
02:28
Plus c3 sine of root five x plus e to the x.
02:52
Now we take that derivative so it's root five and we get minus c2 sine root five x plus c3 cosine root five x.
03:21
There's our y prime, that's an i.
03:33
That's how we get the sines and cosines.
03:43
So y prime, two c1, e to the two x plus e to the x.
04:02
Then the cosine terms gives me c2 minus, excuse me, plus root five times c3 cosine root five x plus and then we get for the sine term, c3 minus c2 root five sine root five x.
04:45
And then we gotta take the second derivative because we still have to set that equal to zero.
04:53
So take another derivative, y double prime of x is gonna give me four c1, e to the two x.
05:08
And we'll get plus e to the x times just what's in there...