CHAPTER 08.01: PRIMER ON ORDINARY DIFFERENTIAL EQUATIONS: Exact Solution of 2nd Order ODE: Repeated Roots of Characteristic Equation

In this segment, we're going to look at a second-order ODE, we're going to find the exact solution to it, so we’re going to look at an example. So let's start with a differential equation which is given as this, y double-prime, plus 4 y prime, plus 4 is equal to 10 times t, and you're given the initial conditions y of 0 is equal to 6, and y prime of 0 is equal to 2. So those are the initial conditions given to you, that's the ordinary differential equation give to you. So in this case, again, in order to find the homogenous part, we will write down the equation in the operator form. So having written down the equation in the operator form, we'll be able to find out what the homogenous solution should be of what the homogenous part of the solution should be, because we can write down the characteristic equation just like the operator looks like, m squared, plus 4 m, plus 4 equal to 0. Again, this is a quadratic equation, you can use factors also, and you will find out that the roots which you're getting is -2 and -2. Now, keep in mind that now what we are getting is, for this particular example, we are getting two roots which are exactly the same, so these are repeated roots, so you're getting repeated roots. You are getting repeated roots for the quadratic equation. So what that implies is that the homogenous part of the solution of course will be k1 e to the power -2 x, but the second part of the homogenous part of the solution cannot be k2 e to the power -2 x, because that'll form a part of this then. So that's why what you have to do is you have to . . . so you cannot write down plus k2 e to the power -2 x right here, because this part then just looks like just this part here. So what you have to do is, when you have these repeated root, you have to choose the next independent solution for the next part, so the next independent solution will be k2 times x e to the power -2 x. So in this case, since I am using t as my independent variable, so let me just change this to t here, so I'm getting t here, t here, and t here. So these are the things which you have to understand, that you're getting repeated roots here, so your first homogenous part of the solution is k1 e to the power -2 t, the second one will be k2 times t e to the power -2 t, in order to be able to take care of these repeated roots. Now, the particular part of the solution, again, the right-hand side is 10 t, or this is the forcing function for the differential equation which we chose. So in this case, your particular part will be the form of the right-hand side, and all its possible derivatives, so it will be A times t, plus B. So you're going to substitute it back into your . . . your differential equation, in order to find A and B, so it will be the second derivative of A t, plus B, 4 times the first derivative of A t, plus B, plus 4 times A t, plus B equal to 0. You're going to get 0 from here, you're going to get just A from here, that's 4 times A, plus 4 times A times t, plus B equal to 10 t of course, equal to 10 t. So you're going to separate out the terms of t, you get 4 A times t, plus 4 A, plus 4 B. So you're separating out the terms of t, separating out the terms of the constant, you get 10 times t, plus 0, of course, so as to be able to write down the equations properly. So you get 4 A equal to 10, and you get 4 A, plus 4 B equal to 0. So from here, you get A is 2.5, and B is -2.5, so solving these two equations, two unknowns, you will be able to find what A and B are. So that gives you what the particular part of the solution is. So the particular part of the solution is going to look like 2.5 times t, minus 2.5. So having found the particular part of the solution, having found the form of the homogenous part of the solution, we can now write down the full solution, and substitute our initial conditions to be able to find out what the coefficients are. So what I mean by saying that is that you have y is equal to yH plus yP. The homogenous part of the solution is k1 e to the power -2 t, plus k2 times t e to the power -2 t. The particular part is 2.5 times t, minus 2.5. So in this case, what you have to do now is to apply the initial conditions. So y of 0 will be equal to 5. So that gives us k1 e to the power -2 times 0, plus k2 times 0 e to the power -2 times 0, plus 2.5 times 0, minus 2.5 equal to 5, so all you are doing is substituting the value of t equal to 0 into your full expression of the homogenous part plus particular part, and you get k1, you get 0 from here, you get 0 from here, minus 2.5 equal to 5, you get k1 equal to 7.5. Similarly, as part of your homework, what I would like you to do is you're going to apply the second boundary condition, which is y prime of 0 equal to 2, and show that the value of k2 which you're going to get now for this will be 14.5, you're going to get the value of k2 to be 14.5. How are we going to find the value of k2? It is you're going to first take the derivative of y with respect to time, so you will have to take this expression which you have here. You will have to take this particular expression which you have right here, take the derivative with respect to time, and then you're going to put the value of time, t equal to 0 into the expression which you're going to get, put that equal to 2, and since you already know what k1 is, you're going to find out what k2 is, and that's what you're going to get, 14.5. So we have found k1 equal to 7.5, we found k2 equal to 14.5. So that gives us the complete solution to be y is k1 e to the power -2 t, plus k2 t e to the power -2 t, plus 2.5 times t, minus 2.5. So I'm going to substitute the values of k1, which is 7.5, the value of k2, which is 14.5, plus 2.5 t, minus 2.5. So that is the solution to your second-order ordinary differential equation, and the reason why we took this particular differential equation is because we got repeated roots of the homogenous part of the . . . when we looked at the characteristic equation of the homogenous part solution, and that's why we had to choose the next independent solution. And of course, the particular part was just based on taking the form of the right-hand side and all its derivatives. And that is the end of this segment.