 Good now we saw what to do with the undetermined coefficients, and we know that it doesn't always work We need a better method, and we're going to look at the variation of parameters Remember we still dealing with this non-homogeneous set of differential equations, and now we can use variation of parameters We've done it before where we just had a single differential equation But now we have the system of equations differential equations, and we have to solve them Using parts of those techniques, a few things change though Now we need some preliminary work What we are trying to get at is this particular solution We know how to get the complementary solution set Where we just set f of t equal to 0, we know how to solve x prime equals a x, we know how to solve that So it's before we can get to an equation line for that particular part We need some preliminary work, and we actually need to determine this, what's called the fundamental matrix Now in order to do that, let's just start with a solution set Let's say that x equals c sub 1, x sub 1, c sub 2, x sub 2, and we go on to c sub n, x sub n x sub n Note though that this x, or sub 1, x sub 2, note that you remember that it was just x sub 1, x sub 2 And we go down to x sub n, but that would be 1, for x sub 1, 1, 1 So that was x sub 1 can be written as this column matrix And then we're going to go on to x sub n, we'll have x sub 1, x sub 2, all the way to x sub n But these are n, n, n That's what we have, and I can multiply c sub 1 into that as well So I'll have c sub 1, x sub 1, 1, plus c sub 2, etc, etc until c sub n, x sub 1, n And then all the way down to c sub n, x sub n, 1, plus all the way to c sub n, x sub n, n So I could have written it like that, there are n number of rows and just one column So this is an n by 1 An n by 1 matrix with set of solutions I can rewrite this in another form, all of these will make sense I can rewrite this as x sub, now it's going to be 1, 1, x sub, now we're going to have 1, 2 And eventually we're going to have x sub 1, n, so we're going to have x sub 2, 1 And all the way down to x sub n, 1 And then this will be x sub 2, 2 All the way down to x sub n, 2 And x sub 2, n And all the way down to x sub n, n I can write this like this times this column matrix that I'll call c And c will be c sub 1, c sub 2, all the way down to c sub n If I did this, now remember this is an n by n matrix n by n matrix and this is an n by 1 matrix So if I multiply it in this order from here to there, remember I started off by saying Suppose that this is the set of solutions So this does satisfy The the initial problem, I'm taking that for granted that this is a solution If I do it this way, I can't have c times this, I have to have this times c n times n times n times 1, that's going to give me an n times 1 matrix Which is this matrix that I had here If I call this thing this Matrix c, and that is why we do this, it's the uppercase Phi, phi of t, we call it that matrix, and we call that column matrix column c And this is what we call the fundamental matrix And that is where it is If indeed it's just made up by these individual column matrices That we found that we have And remember that was a k Whatever it was, let's make it i And I have k 1 i k 2 i etc etc Time okay n i and then e to the power whatever that Whatever that Eigenvector was, eigenvalue was And the eigenvector that goes with that So if if I can multiply each out by those With the x sub i with that c sub i was all of those And so that is where these all come from And I can write them like this which you also recognize as the Ronskian So one good thing about this we know that the determinant if These were in constant multiples of each other the Ronskian the determinant of this Was not equal to zero that means Luckily for us this we can get the inverse of that matrix Okay, so this is the fundamental matrix is this all these columns All these column vectors Column matrices written as one large n by n matrix and if I multiply that by These c's I get exactly this all this all this all the same thing But this is what I'm interested in because This fundamental matrix remember they'll all be functions of t where we have an e to the power lambda sub i t All functions of t I can just take now that is a constant so if I were to do that I'll have Up a case phi prime of t times a constant Okay, that constant this is the constant so I can do that but I remember That x prime equals ax x prime equals ax so I'm going to have something like this These two are now related to each other so phi phi prime of t i'm c equals a And what is x well x is to say phi of t c So the c's can go and I have this fact that the prime Of this fundamental matrix is just my original matrix a in the problem that I might have times phi of t And that we're going to use This is just the first part of our preliminary work First part of our preliminary work we're using this fundamental matrix and the fact that it's first derivative equals a times itself And we're going to use that to set up An equation to get that particular solution of a non-originous problem