 We can apply a variation of parameters to higher-order equations. Suppose we have a higher-order differential equation. We can try to solve it using a variation of parameters approach. First, we'll solve the homogeneous system. Next, we'll assume the undetermined constants are functions. This allows us to set up a system of equations to find the derivatives. And then we can anti-differentiate to find these undetermined functions. For example, say we want to solve the third derivative equal to secant of t. Now, in general, the anti-differentiation process will require us to find the antiderivatives of some very complicated functions. And it might not always be possible to find those antiderivatives in a closed form. So let's leave our answer in integral form. So we want to solve the homogeneous equation. So we'll make our non-y term zero. And so the homogeneous equation is third derivative equal to zero, which has solution, where c1, c2, and c3 are undetermined constants. So we'll assume a solution of the form where c1, c2, and c3 are undetermined functions of t. Now, we do need to find that third derivative. So let's differentiate. Now, so that we can better see the structure of our system of equations, what we're going to do is the following. The solution to the homogeneous equation is a linear combination of three functions. 1, t, and t squared. So we're assuming that the solution to the non-homogeneous differential equation is going to be c1, t times 1, plus c2, t times t, plus c3, t times t squared. And as we differentiate, we're going to leave them in the form 1, t, and t squared, and just indicate the derivatives that we're going to be taking. And now we'll find the derivative. So this is a product c1, t times 1, so the derivative will be c1, t times the derivative of 1, plus the derivative of c1 times 1. Similarly, when we differentiate this c2, t times t, we're going to get and the derivative of c3, t times t squared will be and so now we'll make our ansatz, we'll make an assumption about what our solution looks like. Now looking ahead, we know we'll want to find the second and third derivative. If this expression involving the first derivatives of c1, c2, and c3 is not equal to zero, then we're going to get second and third derivatives of our functions and we don't want that. We'll make the assumption that this expression involving the derivatives of c1, c2, and c3 is equal to zero. And since it's equal to zero, we can ignore it in the higher order derivatives. Speaking of which, we'll find that second derivative. So the second derivative, that'll be the derivative of the derivative, we're going to differentiate this first term, which again is a product. So that's going to be c1, t times the derivative of the derivative of 1, plus the derivative of c1 times the derivative of 1. And similarly for the other two terms in our sum, which gives us our second derivative, and again we'll assume that these terms involving the first derivative of c1, c2, and c3 are going to be equal to zero. And we do need that third derivative, so we'll differentiate again and get... Now this time things are going to be a little bit different. We want this third derivative to equal c count of t. Well this is our third derivative, but notice the third derivative of 1, the third derivative of t, and the third derivative of t squared are all going to be zero. So these terms vanish and that leaves us just the derivatives of c1, c2, and c3. And so in order to solve our differential equation, we require that this expression in the derivatives of c1, c2, and c3 must be equal to c count t. And so this gives us our system of equations. Now we actually want to solve for c1, c2, and c3, so at this point let's go ahead and find those derivatives. Now the first equation is unchanged. The second equation has the derivative of 1, the derivative of t, and the derivative of t squared. So we can find those derivatives and simplify our second equation. Our third equation requires the second derivative of 1, the second derivative of t, and the second derivative of t squared. So we can find those and simplify our third equation. And that gives us a much tamer system of equations to solve. So let's solve those system of equations. Since the third equation only involves c3 prime, let's go ahead and start with that. So we can solve that for c3 prime. And that tells us that c3 is the antiderivative 1 half secant t. And if you want to show off, you can actually find the antiderivative. Now that we know c3, our second equation only involves c2 prime and c3 prime, so we can use that. So we'll substitute and solve. And so that gives us c2 prime is minus t secant t, and so that c2 itself is going to be the antiderivative minus t secant t. And even if you do want to show off, there's no elementary function that will give us derivative minus t secant t, so we have to leave this in integral form. And now that we know c3 prime and c2 prime, we can solve for c1 prime. So substituting those values into our equation for c1 prime, and so that c1 prime, which means that c1 itself is going to be the antiderivative of 1 half t squared secant t. And again, there's no elementary function with derivative 1 half t squared secant t, so we have to leave this in integral form. Now, you notice that it's a lot of work to find the system of equations, let alone solve the system of equations. So let's see if we can make this a little bit easier on ourselves. So go back to how we generated that system of equations. Now, if you look at how we got these equations, we got them by taking every expression involving the first derivative, except for the last one, and setting it equal to zero. Now, we didn't differentiate the 1 or the t or the t squared until we got to the end, and so this allows us to see the structure of these equations. And so the things to note are the coefficients of the ci's are the derivatives of the functions producing the general solution. So here we have the function 1, the derivative, and the second derivative. Here's our function t, its derivative, and second derivative. And t squared, its derivative, and second derivative. So the first couple of equations are set equal to zero, but the last equation is equal to the non-homogeneous term. And this process generalizes, and this allows us a fast way of setting up our system of equations. We'll take a look at some examples next.