CHAPTER 08.01: PRIMER ON ORDINARY DIFFERENTIAL EQUATIONS: Exact Solution of 1st Order ODE: Another Example

In this segment, we're going to find an exact solution to a first-order ordinary differential equation.  So first-order ordinary differential equation, we're going to find the exact solution to it by using our classical solution techniques of finding the homogenous part and the particular part.  So we're taking a first-order ordinary differential equation with fixed coefficients, and we're going to find out what the exact solution is. So let's suppose somebody gives me a differential equation like this, 3 dy by dx, plus 6 y is equal to 12 e to the power -2 x, with the initial condition y of 0 equal to 5. So somebody is giving me a differential equation like this, and wants me to find out y as a function of x. So again, the coefficients are constant, 3 and 6 here, and I'm going to use the classical solution techniques of finding the homogenous and the particular part.  I'm going to write down this equation, this differential equation in the form of the . . . form of the differential operator, I'm going to get 3 D, plus 6, operating on y is equal to 12 e to the power -2 x, y of 0 equal to 5.  And the reason why I wrote it like this is because by it writing in the operator form, I directly get what the characteristic equation is, so I've got to find the homogenous part of the solution.  So the characteristic equation is 3 m, plus 6 equal to 0.  3 times m, plus 6 is equal to 0 is the characteristic equation which is going to give me the homogenous part, help me to find the homogenous part. This gives me m is equal to -2.  So what that means is the homogenous part of the solution is k e to the power -2 x.  That is the homogenous part of the solution, which means the homogenous part of the solution, meaning that if I plug this homogenous part of the solution into my differential equation right here, I'll get 0 as the solution.  Now, let's go ahead and see what the particular part should be.  Now, you've got to realize that the particular part is going to be of the form of the right-hand side, in this case, it is some constant multiplied by e to the power -2 x. So the particular part is going to be some form of e to the power -2 x and all its possible derivatives, and we know that derivatives of e to the power -2 x are just e to the power -2 x again, e to the power -2 x again, so the form itself and derivatives itself are of the form e to the power -2 x, but one of the things which you are seeing here is that the homogenous part of the solution is k e to the power -2 x, so I cannot choose my particular part, I cannot choose my particular part to be of the form A e to the power -2 x.  So I have the form of the right-hand side and its derivatives is of the form of e to the power -2 x, so somebody might say, hey, then the particular part should be of this particular form.  But I cannot choose this to be the form of the particular part of the solution, because the homogenous part looks like that, and since the homogenous part, by substituting it into the differential equation gives me 0, I cannot have 0 equal to 12 e to the power -2 x.  So that's why what I'll have to do is I'll have to choose the next possible independent solution, it will be A x e to the power -2 x.  I'm going to get yP is equal to A times x times e to the power -2 x, that will be my particular part of the solution which I should be choosing there.  So let's go ahead and see that what the value of A is.  So we have yP, I have just established that this should be the form of my particular part of the solution.  So now I am going to plug it back into my differential equation, which is 3 d by dx of y, which is A x e to the power -2 x here, plus 6 times y, which is A x e to the power -2 x is equal to 12 e to the power -2 x, that's the right-hand side of the differential equations.  That's how we're going to find k, so . . . A.  So I'm going to take the derivative of this, this is by using derivative of multiple of two variables here, I'm going to get 3, I take the derivative of x, I'll get 1, so I get A e to the power -2 x, plus A times x, and I take the derivative of this, and that's -2 e to the power -2 x, So that's what I get as the derivative of this quantity here, plus 6 times A x e to the power -2 x is equal to 12 e to the power -2 x.  I again do this, I get 3 A e to the power -2 x, minus 6 A x e to the power -2 x, plus 6 A x e power -2 x is equal to 12 e to the power -2 x. So when I expand this, this is what I get, and of course, this has to cancel with this, otherwise you won't be able to solve the equation, and you can get from here, you're going to get 3 A e to the power -2 x is equal to 12 e to the power -2 x, and that gives me A is equal to 4.  So that is the particular part of the solution, it is nothing but 4 x e to the power - 2 x, it's A x e to the power -2 x, I just found out that A is equal to 4, so I found out what the particular part of the solution looks like.  So that's how you find it, find the particular part of the solution.  So we know that the complete solution is the homogenous part plus the particular part, that the complete solution is the homogenous part plus the particular part.  What is the homogenous part?  Homogenous part is, which we just found out was k e to the power -2 x, particular part we just found out to be 4 x e to the power -2 x. Now, what I have to do is, this is the complete solution now for the differential equation, but there is still one unknown, which is k, which I am going to find out from my initial condition. I'm given that y of 0 is equal to 5, so that makes it k e to the power -2 times 0, plus 4 times 0 e to the power -2 times 0, and that gives me 5 is equal to k, because e to the power 0 is one, this gives me just 0, so k is equal to 5.  So I get y is equal to k, I'm going to substitute 5, e to the power -2 x, plus 4 x e to the power -2 x.  So what I was just showing you in this segment is just another example of solving a first-order ordinary differential equation with fixed coefficients, constants coefficients, but in this case, we are using something, the example which I took purposefully had the homogenous part matching with the form of the particular part which you would have taken, but you have to take the next independent solution, otherwise it would have become part of the homogenous part. And that's the end of this segment.