 Another method of solving differential equations is known as the variation of parameters. And this emerges as follows. While we can use the method of undetermined coefficients to solve kth-order linear differential equations, we have to find the appropriate annihilator or ansatz. But sometimes we can't. What can we do then? And the thing to remember is that the problem exists regardless of your ability to solve it. We have to be able to do something. So let's consider a simple case where we have a homogeneous second-order linear differential equation. Now we'll assume we can solve this and assume that this has solutions y equals c1u1 plus c2u2, where c1 and c2 are constants. Now we'll assume the inhomogeneous equation has solutions y equals c1u1 plus c2u2, where c1 and c2 are functions. And remember one of the main rules in solving differential equations is that it is easier to obtain forgiveness than permission. If we have to wonder about whether we're allowed to do this, we might not get anywhere. We can just try it out and see if we do actually get a solution. So let's see how that might work. Suppose u1 and u2 solve the homogeneous equation. And assume the non-homogeneous equation has solutions y equals c1u1 plus c2u2, where c1 and c2 are undetermined functions of t. So we'll differentiate. And again it's easier to obtain forgiveness than permission. We'll assume c1primeu1 plus c2primeu2 equals 0. Why not? If it works, we'll have a solution and all is forgiven. Of course, if it's not written down, it didn't happen, so we'll record that assumption over here. That simplifies this derivative. And because this is a second-order differential equation, we do need to find that second derivative. Of course, we can only get the forgiveness if we actually solve the equation. So let's see if this solves the equation. We'll substitute our values into our expressions. g of ty, f of ty prime, and y double prime. Adding up the left-hand side should give us h of t. Adding up the right-hand side will give us... And the important thing to remember is that since u1 and u2 solve the original homogeneous differential equation, this coefficient of c1 and this coefficient of c2 are both going to be 0. And so our equation simplifies to... And now I have two equations that c1 prime and c2 prime satisfy, and so we can solve for c1 prime and c2 prime. So let's try to solve this differential equation, which has no obvious annihilator. So the solution to the homogeneous equation is going to be c1 sin t plus c2 cosine t. We'll treat c1 and c2 as functions of t, so y prime is going to be... But we'll assume c1 prime sin plus c2 prime cosine is equal to 0. And that simplifies our equation, but we still need to find our second derivative, so we'll find that. So now we want y double prime plus y to equal tangent of t. So take y and add y double prime, and that gives us... tangent equals c1 prime cosine t minus c2 prime sin t. And there's a second equation that involves c1 prime and c2 prime. So let's solve this system for c1 prime and c2 prime. And one way we might do that is we multiply the first equation by sin of t and the second equation by cosine of t we get. And if we add our two equations, and we can simplify this, so c1 prime is sin of t. Well that worked out so well. Let's do it again. This time we'll multiply the first equation by cosine t and the second by minus sin of t, which we'll give us. And if we add them we get, and then simplify, which gives us c2 prime. Of course we need to know what c1 and c2 are, but we can anti-differentiate. So the anti-derivative of c1 prime will give us... c1 is minus cosine, the anti-derivative of c2 prime will give us a rather horrific mess, but we can find the anti-derivative eventually. And so we get y equals cosine t sin t plus this horrible mess, plus the general solution of the homogeneous equation, where we return c1 and c2 to being actual constants.