00:01
Okay, so here we want to find the solution for y triple prime plus 3, y double prime minus 4y is equal to e to the negative 2x.
00:11
So we consider the corresponding homogeneous equation of r cubed plus 3r squared minus 4.
00:20
Okay, and so that when we factor it is equal to r minus 1 times r plus 2 squared.
00:31
And so when we set that equal to zero, this is going to have an r equal to 1 and r equal to negative 2 with multiplicity 2.
00:50
So that means our general solution to the homogeneous part, so we'll call that y of h, it's going to be equal to some constant times e to the power of negative 2x plus some other constant times.
01:08
X times e to the power of negative 2x and then plus some third constant times e to the power of x.
01:17
So this x that's up front is here because of the multiplicity of two.
01:29
Okay, so now we want a particular solution and we will look for a particular solution of the form a x squared times e to the negative to x.
01:43
Okay, so from there we get that r.
01:45
Particular solution.
01:47
First derivative is equal to 2ax times e to the negative 2x, minus 2ax squared, e to the negative 2x.
02:04
This gives us a second derivative, which is equal to 2ae to the negative 2x, minus 8 to the negative 2x, minus 8...