00:02
All right, so here again, we are using the auxiliary method to find solutions to some of these homogeneous equations.
00:10
And here, the equation is the second derivative of y plus the first derivative of y plus y.
00:17
And so our auxiliary equation is r squared plus r plus r plus one is equal to zero.
00:25
Again, no roots are jumping out at me as being obvious.
00:28
So i'm going to use the quadratic formula.
00:31
Quadratic formula states that for an equation, a x squared plus bx plus c equals zero.
00:39
We can find the roots by solving for negative b plus or minus the square root of b squared minus 4 times a times c all over 2 times a.
00:55
Okay, so applying that to our problem now, we get negative 1, plus or minus the square root b squared is going to be one minus four times a times one.
01:09
C is also one, so this will be four times one minus four, all over two times one...