 So this is where we left off. This one to to remind you there should be a negative 2 there I'm not sure that we are left off previously which was not today There was a net should be a negative 2. They said we're still trying to solve this set of linear differential equations and We were kind enough to have been given the two answers That's what we're trying to get to the two most important things to remember while trying to solve this of course are these two This matrix a Which is a? Minus lambda I remember that's a scalar It's multiplying it by this identity matrix So it's a matrix minus matrix times this matrix case Equals zero and the way to get to my item values was this determinant of a minus lambda That's how I got to these characteristic values This is a two by two matrix day the two two hundred redeeming us So we're always going to get two item values there So with that with that in mind we now go to this equation here and what that suggests is well Let's have lambda sub one which is going to equal negative two. So let's see what we can get a Minus lambda I So that would be one three five three And I subtract from that this lambda I but lambda sub one is now negative two So there will be negative two zero zero negative two And from then I'm going to get the matrix one minus negative two that's three three minus zero zero five minus zero is five and three minus negative two is What am I doing? three five So that is my a minus lambda sub one I And I've got to multiply that with this two by two matrix K And you remember K sub I that that would just going to be K one and K two in this instance Remember that K was the slowest of of the possible values there. So I've got to multiply these two A minus so this one Multiplied by this two by two matrix times a two by one matrix is going to be three a Two by one matrix. So let's do that in whichever way you want three or three or five or five And we're going to multiply this with case of one in case of two if I multiply that I'm going to get three times case of one That's three times case of two or five times case of one plus five times case of two And it's going to be the matrix multiplication that I did get there I write zero like this, but you remember this is the zero vector the zero Column vector there a zero matrix. So zero zero. So this is going to equal zero zero So either of these I can see three times case of one plus three times case of two is going to equal zero And I can just take out three as a common factor So case of one case of one plus case of two is going to equal zero So I might as well just let case of one case of one equals one That means case of two is going to equal negative one now I can choose anything I could have chosen case of one to be two but try and keep things Can try and keep things simple. So there's my first the values for case of one and case of two so And this makes up this case of one That equals one negative one those are my That is my I can my first I can Vector from this these characteristic values or I can values I get this I can vector one negative one and learn behold. They was one negative one e to the power negative two So what does this mean? It means that x equals? e to the power negative two t and y equals e to the power Or negative one negative e to the power negative two t so that is x of one equals this equals that And if I were to take these and differentiate them I'm going to get back to this. I'm going to get back to this Now let's just use lambda sub two just to see if we can get to the other answer remember that was six So I'm going to have what was three One three five three and subtract from that six six zero Subtract that from that one minus it's negative five three five negative three So that's that and I've got to multiply that by my new Case of one in case of two. Let's call that face of three and face of four So I'm going to be left with negative five times case of three That's three times case of four Equal zero this comes exactly from here In other words, I'm going to have the fact that five times case of three equals three times case of four Or case of three. It's going to equal three fifths of case of four So if I let case of four equal five that means case of three is going to equal three So my second eigenvector is going to be Three five three five Lo and behold Three five exactly what it was there. So these are quite simple to solve