Let's try another example of a different inhomogeneous term. Here we're going to look to see what sort of ansatz, we should make when the right-hand side is some polynomial in t. The example I want to do is, this differential equation equal to t squared. So, following the idea of exponential functions sine and cosine. At first you would look at a constant times t squared. But in this case, the derivatives will end up giving you t's and constant terms. So, you need to include those in your ansatz also. So, here when the polynomial the standard ansatz instead try x equals some constant times t squared, but you have to carry all of the lower orders, so, plus B times t plus C. Because the derivatives reduce the order of the polynomial. So, when you substitute in you have to take the second derivative. The second derivative here of this polynomial will just give us 2A. And then plus the first derivative, the first derivative gives us, 2A t plus B, thus the derivative of a constant is 0. Minus 2 times x. So, minus 2A t squared, minus 2B t minus 2C, and all that is supposed to be equal to t squared. Okay. What we need to do now is to collect terms, so that we collect terms of the powers of t. So the left hand side, we have a minus 2A t squared, and then for the t term, we have a 2A t minus 2B t. So, 2A minus 2B times t. And then the constant term, we have a 2A plus B minus 2C, So, plus 2A plus B minus 2C and that's equal to t squared. Okay. Now the idea is the coefficient of t squared on the left, has to equal the coefficient of t squared on the right. The coefficient of t on the left, has to equal the coefficient of t on the right, here it's 0. And the constant term on the left has to equal the constant term on the right. To see that, you put t equals 0, so, you get the constant term equals the constant term. Then you can take the first derivative with respect to t, and set t equal to 0 again. Or you can take the second derivative with respect to t. And then you can see that the coefficients have to be equal. So, we end up with three equations. So, what are those equations? Let me isolate this here. The first equation is, minus 2A equals 1. So, that tells us immediately that A equals minus one half. The second equation is, 2A minus 2B equals 0, because there's no t term on the right-hand side. So, A minus B equals 0. Which tells us that B is equal to A, so, B is also equal to minus one half. And the last equation is, 2A plus B minus 2C equals 0, because there's no constant term on the right-hand side. So, 2A plus B minus 2C equals 0. That says that, let's see can I do that in my head? Probably not. So, let's solve this then. So, 2C is equal to 2A plus B, A and B are the same, So, it's equal to 3A which is minus 3 over 2. So, then C is equal to minus 3 over 4. Okay. So, we found A, we found B, we found C, so then I can write down, x of t, is equal to A t squared minus one half t squared, plus B t, minus one half t, plus C, minus three quarters. And we have our solution of the differential equation. Sorry, it's our particular solution of the differential equation. So, let me review what we did here. So, we're trying to find a particular solution when the right-hand side is a polynomial in t, the ansatz we need to use, is the same order of the polynomial on the right-hand side with free coefficients, but we need to have all of the lower order powers of t in the ansatz, because of the derivatives. Then when we substitute in, we'll end up with a system of three linear equations and three unknowns. The unknowns are A,B and C. And we solve it and we get our solution. I'm Jeff Jasanoff. Thanks for watching, and I'll see you in the next video.