 So let's annihilate some of our functions. So remember, the annihilator corresponds to a linear differential operator that will eliminate a function f. In other words, if we apply the annihilator to f, we're going to get zero. So if we know the type of differential equation f solves, we can recover the annihilator. So if we differentiate some familiar functions, we get the following. If y equals e to power ax, then the derivative will be, which is just ay, and rewriting this in operator form gives us. And so the linear differential operator d minus a will annihilate e to power ax. Or if y equals cosine of ax, then our derivative will be, and since this doesn't recover the function, we'll differentiate again. And here we do have our original function, so rewriting this in operator form gives us. And so the linear differential operator d squared plus a squared will annihilate cosine of ax. And similarly, d squared plus a squared will also annihilate sine of ax. And these suggest the corresponding annihilators. For example, let's try to solve this differential equation. So in operator notation, our differential equation becomes, now if the right hand side were zero, we'd have a nice homogeneous equation, but it's not. However, if you look at this 3e to power 2t, we note that if we differentiate it, we get, and if we subtract twice the function, we get zero, and so an annihilator is going to be d minus 2. So we'll annihilate the 3e to power 2t. And now we have a nice homogeneous differential equation, where our linear differential operator applied to a function gives us zero. So we'll factor our characteristic polynomial. So our characteristic polynomial has roots 2, 2, and 1, giving the general solution. Now note the first part comes from solving the homogeneous equation, and since the first part comes from solving the homogeneous equation, the values of c1 and c2 could be anything. And since they could be anything, we'll make them zero, and try to find the undetermined constant, c3. So suppose y is c3t e to power 2t. We'll substitute that into our original differential equation, which means we need a 2y, a minus 3dy dt, and a second derivative of y. Putting those together. Now this part is supposed to be equal to 3e to power 2t. And on the right hand side, if we gather our coefficients, the coefficient of e to power 2t is going to be c3, and the coefficient of te to power 2t is going to be zero. And so comparing our coefficients, we see that c3 is equal to 3, and so our general solution will be c1 e to power 2t plus c2 e to the t plus 3t e to the t. Or consider this differential equation. In operator notation, the differential equation is, since our non-homogeneous term over on here on the right is cos 3t, we can annihilate it with d squared plus 9, and so our characteristic polynomial has roots plus or minus 2. That's from the original homogeneous part, and plus or minus 3i, which comes from the annihilator. And so the solution to the annihilated equation will be, and again these two roots, plus or minus 2, come from the homogeneous equation, and so this portion of the general solution is going to solve the homogeneous equation, and we can ignore it when trying to find the undetermined coefficients, c3 and c4. Or, which is the same, we could just say they're both equal to 0 and focus on the part that came from the annihilator. So y is c3 cos 3t plus c4 sin 3t. We need 4y, and we need the second derivative of y, and our second derivative minus 4y is, now we want that to be cos 3t. So if we compare our coefficients, we see that minus 13c3, the coefficient of cos 3t, has to be equal to 1, the coefficient of cos 3t. Meanwhile, minus 13c4, the coefficient of sin 3t, must be equal to 0, the coefficient of sin 3t. Solving this system gives us, and so we have our general solution.