 So back to our preliminary work. We know this now. We also know that x prime Let's say we have a equals phi b times c That's what we that's what we had on the board before we have this now Let's carry on Let's carry on with us we have this now now we're going to have x prime Instead of the complementary set Or having it like this. It's got an equal It's got an equal x c plus x p X c the complementary set of solutions you know how to do we need to get p Now x x x p instead of this c we're going to do something else We're going to say phi of t and we're going to multiply it by this column vector u of t Such that the u of t is going to equal this u1, u2, u3 for u n this column vector We're going to suggest that through the variation of parameters that instead of c for the constants for the complementary set we're going to have this other Convector u of t that if we multiply those two with each other we're going to get x sub p This is just what we suppose now this is a Matrix and this is a matrix. So it's a product of two matrices. So if I get the first derivative of this I better use the product rule as far as that's concerned So if I get x prime x prime of p I've got to get and I bring it in a certain way you'll see why so I'm going to have phi of t u prime of t Plus I'm going to have phi prime of t u of t I'm going to have that I'm going to have that Those are the symbols I wanted to use but I also remember that x prime equals a times x plus the f of t and then equals a times now the x prime is this Phi of t u of t So I have that plus the f of t In other words, I have two statements For this might as well be the p there and the p there So I have two statements for x of p two equations So I can equate these two to each other But I also know something else. I have x prime of t here Phi prime of t there and I have phi prime of t here So just going to rewrite that phi prime of t as such. So I'm going to have a Phi of t u prime of t Plus instead of that, I'm going to have this I'm going to have this a I'm going to have a phi of t and a phi of t for that one and then u of t and it is going to be equal a Phi of t u of t plus f of t Lo and behold both sides of the equation I have these two expressions So they can go and I have the fact that I have Phi prime of t u of t Is that right? Is that what I wanted to do? It's my prime Where that little prime Apologies for that. It's equal to f of t equals the f of t Now fortunately for us this fundamental matrix remember I started off by saying let's suppose we do have a solution set And the round skin or the determinant of that could not be zero therefore. It is a non-singular matrix So I have more than just the whole zero solution set so That means I can't get the inverse of this and if I multiply the inverse out of both sides The inverse of this matrix times itself. Well, nothing is nothing remains. So I have the u prime of t Stoicius says I but anyway and then the inverse inverse negative 1 of t f of t on both sides and very Elegantly if I take the derivative the integral of this side I'm going to be left with the u of t equals the integral of the inverse of the fundamental matrix times the f of t d t And there I have u of t, but I remember what my x of p was was this my x of p was that so x p There we have it is going to be Phi of t times the integral of the inverse Phi f of t d t And there we have an equation beautiful for the particular solution And all I have to remember is that x equals x Complementary set plus let's say in other words x is going to be X of c and that is actually there Phi of t times c and then I'm going to have plus plus this Phi t integral of the inverse of t f of t d t I'm sure you can't see that anymore So I have this whole solution just for x and that is the variation of parameters just developing this very beautiful Equation for the particular solution set is the fundamental matrix times the integral of the inverse of the fundamental Matrix times the f of t which in an original problem would be x of prime equals I have a bx plus f of t which makes it equal to zero matrix color matrix that means it's non-homogeneous So just times that f of t d t and all you have to know is how to get the How to multiply these matrices with each other and there was a specific reason why we did the product rule that way around So that we could develop this equation for the u of t and if we do multiply these out That it is it is allowed is a matrix multiplication that is allowed by the size of the matrices You multiply by the fundamental set and it takes a speed to still remember them To add it to the complementary set which we know now how to do and that is a method that works for us Many more instances are much more useful. I should say and determine coefficients