00:01
So for this problem, we are asked to find the general solution and solve the initial value problem for the given differential equation, 9y double prime minus 6y prime plus 10 y equals 0 to start.
00:11
So the way that we go about doing this is since this is a second order differential equation with constant, pardon me, constant coefficients, we make the substitution y equals e to the power of rx.
00:26
When we do that, we'll have, as we've seen previously, we'd get the characteristic equation, 9r squared minus 6r plus 10 must be equal to 0.
00:41
Now we then want to solve this equation for its roots.
00:46
Particularly, we have that we can factor out, or actually not factor out, what we'd want to do is apply the quadratic formula.
00:56
We'd have that r must be equal to negative b so that's positive 6 over 2a so that's positive 6 over 18 plus or minus over 2a so over 18 the square root of b squared 36 minus 4 times 9 times 10 in this case so we'll have that we can simplify this down express it as r equals 1 over 3 plus or minus i and we'll note that in this complex roots case, we have that the general solution would be y of x equals e to the power, or actually i'll say, if we have r equals alpha plus beta i, then we'd expect that y of x is going to be equal to e to the power of alpha t times c1 cos of beta t plus, or excuse me i mean i swapped from x to t partway through there y of t would be each power of alpha t times c1 c1 ccccc sine of beta t so i'm going to stick with having said that it was actually no i'll go with treating this as being a t so we have then that our general solution will be y of t equals e to the power of t over three times c1 t c1 c1 c1 c1 cc of t plus c2 sine of t then we apply sign of t then we apply our initial conditions to solve the differential equation or the initial value problem we have y of zero equals negative 3 so that e to the power of t over 3 goes to 1 out front cos of 0 is 1 and sine of 0 is 0 so we have y of 0 would be equal to c1 which then equals negative 3.
03:05
And we have y prime of t would be equal to 1 over 3, e to power of t over 3 times.
03:17
Actually, it's a little bit more complicated than that one second here.
03:22
Sorry about that...