 So last time we came to the following conclusion. Suppose we have an nth order non-homogeneous linear differential equation with constant coefficients. If the general solution to the homogeneous equation is going to be given by some linear combination of functions, then a solution to the non-homogeneous equation can be given by assuming each of these constants is actually some function of t, where the derivatives of these parameters satisfy the system of equations, where the coefficients will be the successive derivatives of the functions that solve the homogeneous equation, and all of these equations, except for the last one, are set equal to zero, with the last one equal to the non-homogeneous component. So for example, let's try and solve this differential equation. So first, we'll solve the homogeneous system. So we'll make the terms in our equation that do not contain y or its derivative equal to zero. The characteristic polynomial is d squared minus four with roots plus or minus two. And so the general solution will be, but that's for the homogeneous equation. So we assume that our non-homogeneous equation has a solution of the form. So now we want to find the equations that c1 prime and c2 prime satisfy. So we can get those equations by taking our general solution and differentiating the functions that come from the homogeneous case, e to the two t and e to the minus two t. So our first equation, c1 prime e to the two t plus c2 prime e to the minus two t equal to zero. To get our second equation, we'll differentiate e to the two t and e to the minus two t and keep everything else the same. But since this is our second and last equation, this will be equal to the non-homogeneous portion of our differential equation. And now we can try to solve. If we multiply our first equation by two and add it to the second we get, which tells us that c1 prime is one-quarter e to minus two cosine t. If we multiply the first equation by minus two and add it to the second we get, which tells us that c2 prime is minus a quarter e to the power two t cosine t. Well, those are the derivatives c1 prime and c2 prime. We actually need to find c1 and c2, so we'll find the antiderivatives. And unfortunately, finding this antiderivative is a long and painful process, but we do find them eventually. And don't forget that constant of anti-differentiation. And so our general solution, c1 t, that's this mess, times e to the power two t, plus c2 t, that's this mess, times e to the power minus two t. And we can simplify. Let's set up a more complicated equation. Let's go ahead and leave our solutions in integral form. So first our homogeneous equation is obtained by letting the non-y terms be equal to zero. Our characteristic polynomial, d3 plus 2d squared minus d minus two, and the roots are plus or minus one and minus two, so the general solution to the homogeneous equation will have the form, which is to say a linear combination of the functions e to the t, e to the minus t, and e to the minus two t. So to solve the non-homogeneous equation, we'll assume a solution of the form where we've replaced our constant coefficients with functions of t, and we find c1 prime, c2 prime, and c3 prime will solve a system of equations, which we obtain by finding the derivatives of the functions e to the t, e to the minus t, and e to the minus two t. So our first equation will use those functions equal to zero. Our other equations will be formed by differentiating e to the t, e to the minus t, and e to the minus two t, and keeping everything else the same. So our second equation will look like, still equal to zero, and our third equation will look like, and since this is the third and last equation in this third-order differential equation, this last equation will be equal to the non-homogeneous term tangent t. And now we do some algebra. If we subtract the first equation from the second and the third we get, and this is a system of equations, it's much easier to solve. The third equation only involves c3 prime. So we'll solve it for c3 prime. Let's multiply this second equation by minus one to clear out the negative signs, and we'll solve this for c2 prime using our value for c3 prime that we just found. And now that we know the value of c2 prime and c3 prime, we can use our first equation to solve for c1 prime. And these are the values of c1 prime, c2 prime, and c3 prime, so we can find c1, c2, and c3 by anti-differentiation. And we'll leave our solutions in integral form because we can.