CHAPTER 08.01: PRIMER ON ORDINARY DIFFERENTIAL EQUATIONS: Exact Solution of 2nd Order ODE: Distinct Roots of Characteristic Equation
In this segment, what we're going to do is we're going to take a second-order ODE, and we're going to take an example to find the exact expression. So let's take this example here, that we have second derivative of y, plus 5 dy by dx, plus 6 times y is equal to 12 x. How do we go about solving this particular problem here? So again, what we're going to do is we're going to write down in the operator form. And the reason why we're writing in the operator form, because this allows us to develop the characteristic equation for the homogenous part. So the characteristic equation will look like just like this, it will be m squared, plus 5 m, plus 6 equal to 0. I can use my knowledge of quadratic equations, or factoring, and I will get the values of m to be -3 and -2, those are the two roots of that particular equation, -3 and -2. What that implies is that the homogenous part of the solution is now k1 e to the power -2 x, plus k2 e to the power -3 x. That's what I am going to get as the homogenous part of the solution, k1 e to the power -2 x, plus k2 e to the power -3 x. It doesn't matter to which one it's associated with, k1 and k2, I just took it like that, so it's k1 e to the power -2 x, plus k2 e to the power -3 x, that's the homogenous part of the solutions, and later on, we'll find out what k1 and k2 are, based on the initial conditions, which I unfortunately did not write what the initial conditions are. We have y of 0 equal to 5, and y prime of 0 equal to 2, and that completes the . . . the example itself. So we're given the differential equation, we're also given the initial condition at 0 is 5, and the initial condition of the derivative at 0 is 2. So later on, we'll find k1 and k2 form these two initial conditions which are given to us, because we have to first find the particular part of the solution. Now, we know that the right-hand side, or the forcing function, is 12 x. So what we have to do is, in order to be able to find out what the particular solution will be, that we have to choose the particular part of the solution to be the form of the function which is given to us, and all its possible derivatives. So I know that the form of the right-hand side is A times x, 12 times x is A times x form, and the derivative of 12 times x is just a constant, which will be denoted by B, and the derivative of a constant is 0, so I don't need to really show that up, so that's what the particular part of the solution is, A times x, plus B Now what I have to do is I have to substitute it back into my differential equation. That's what I'll have to do. So I substitute the particular part of the solution back into the differential equation, because I need to find out what A and B is. Now, the second derivative of this whole quantity is going to be 0, so I don't get anything from there, 5, the first derivative is just going to be A, and I'll get 6 A x, plus B equal to 12 x. So what I'm going to do is I'm going to collect the terms which are corresponding to x, which is 5 A plus . . . 6 A, which is how much? So I get 6 A x, so I have 6 A x here, so that is the term corresponding to x, and the term corresponding to the constants is 5 A, plus B, and that's 12 times x. So since A and B can be any numbers to begin this, that means that the coefficients of the x and the constant have to be matching the right-hand side, which in this case will be 0, so that means 6 A has to be 12, and 5 A, plus B has to be 0. So I get two equations, two unknowns, so in this case I get A is equal to 2, and B is equal to -5 by 3. This should be 6 B, so this should be 6 B there, 6 B there, 6 B there, and that's how I get B is equal to -5 by 3. So I missed a 6 here, and then 6 there, 6 there, so B is equal to -5 by 3. So the particular part of the solution will be A times x, which is 2 times x, minus 5 by 3. So what I have to do now is to add the homogenous part and the particular part to be able to come up with the complete solution. So y is equal to yH plus yP. The homogenous part is k1 e to the power -2 x, plus k2 e to the power -3 x, plus the particular part is 2 x, minus 5/3. So in order to be able to find out the values of k1 and k2, what I have to do is I have to apply the initial condition y of 0 is equal to 5, and y prime of 0 is equal to 2. Now I'll apply y of 0 equal to 5, I get 5 is equal to k1, plus k2, minus 5 by 3, and I've got to apply y prime of 0 equal to 2, so first I, in order to do that, I need to find what y prime of x is. y prime of x will be -2 k1, e to the power -2 x, minus 3 k2 e to the power -3 x, plus 2, and y prime of 0 is now 2, it will be -2 k1, minus 3 k2, plus 2. So that's what I'm going to get, this is what I get from the first initial condition of y of 0 equal to 5, and this is what I get from the second initial condition, that y prime of 0 is equal to 2. Once I solve these two equations, two unknowns, I'll be able to find out what the values of k1 and k2 are. k1 turns out to be 20, and k2 turns out to be -40 by 3. So that's the value of k1 and k2 which I get by solving those two equations, two unknowns. So which tells me that the total solution, which is y is equal to k1 e to the power -2 x, plus k2 e to the power -3 x, plus 2 x, minus 5 by 3, will be obtained by the values of k1 and k2 being substituted back in here, because those are the ones which we got by using the initial condition. So I get 20 e to the power -2 x, minus 40 by 3 e to the power -3 x, plus 2 x, minus 5 by 3. So that's how we obtain the solution for the second-order ordinary differential equation in that example. And that's the end of this segment.