Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Non homogeneous equations. Suppose that no term appearing either in F of X or in any of its derivatives satisfies the associated homogeneous equation L * y equaling 0. Then take as a trial solution for Y of PA linear combination of all linearly independent terms and their derivatives. So if we look at an example, Y double prime plus 3, Y prime plus 4Y equal three X + 2, we're going to introduce some new terms. Y sub C is going to be a complementary solution. Y of T is going to be our trial solution. AY of P is going to be the particular solution. Our goal is to find Y of P. So the first thing we're going to do is we're going to solve for what the R would be. So if we have Y double prime plus 3, Y prime plus 4Y, we know that r ^2 + 3 R Plus 4 = 0, or R equals by the quadratic formula -3 plus or minus square root 9 -, 16 all over 2. So R equal -3 halves plus or minus sqrt 7 / 2 I. So the complementary solution here would be Y sub C equaling E to the -3 halves X times the quantity C1 cosine squared of seven over 2X plus C2 sine squared of 7 / 2 X. Now to get our Y trial, we're actually going to look at the portion down here and see it's a polynomial. So our Y trial is going to be AX plus B where we actually want to find A&B. The first important thing that we need to think about is if our Y complementary and our Y trial have anything overlapping or any duplicates. And in this case we don't because in the Y complementary we have E to a power times cosine and times sine, whereas our Y of T is just a polynomial. So YT prime would equal A and YT double prime equals 0. So if we substitute that back into the original, we get 0 + 3 A plus four times the quantity AX plus B equaling the three X + 2. Then we just solve for A&B. So 3A plus 4B equal to our constants on one side has to equal our constant on the other. The coefficient on the X4A has to equal the coefficient on the X3. So we can see a is 3/4 and if a is 3/4, B is negative 116th. So our Y particular is going to be 3/4 X minus 116th. Let's look at another example. Y double prime -4 Y equal 2 E to the 3X. So we start with finding our Y complementary, our Y sub C, so we get r ^2 -, 4 equaling 0R plus two r -, 2, so R equal -2 and 2. So our Y sub C is just C1 E to the negative 2X plus C2 E to the 2X. Now our YT is based off of what it equaled up here. So our YT is going to be AE to the 3X. When we look at our YT and our YC, once again, we can see there's no overlap. This is E to the 3X and this is E to the -2 X&E to the 2X. So we're going to let our YT stand as what it is we're going to do. YT prime equaling 3AE to the three XYT, double prime equal 9 AE to the 3X, sticking it back into the original solution so we can find our A. We get 9AE to the three X -, 4 AE to the 3X, or 5AE to the 3X equaling the two E to the 3X. So we can see A is 2/5. So our Y particular is just 2/5 E to the 3X. Our next example is going to have overlap. So if we have Y double prime -4 Y equaling 2E to the 2X, it's very similar to the last one. We start out the same r ^2 -, 4 = 0, R +2, R -2 R equal -2 and 2. So our Y complement is C1 E to the negative 2X plus C2 E to the 2X. Our YT, once again looking back here, would be a * E to the 2X. Now let's just for a minute solve it like we did the last one. So if we have Y of T equal AE to the two XYT, prime would be two AE to the two XYT. Double prime would equal 4 AE to the 2X. When we stick it back into the original, we'd have 4AE to the two X -, 4 AE to the 2X equaling 2E to the 2X. So we'd get 0 equaling 2 AE to the 2X. That would make a = 0. Now we don't have a solution because zero times anything, so that's not possible. So what we do when we have overlap, if we go back and we look at our YC and our YT, we're needing to make the overlap go away. So what we're going to do is we're going to take this Y of T and we're going to multiply by X to some power until the overlap no longer exists. So if I took this Y of T and I multiplied it by XAE to the two X * X, now this Y of T in relationship to the Y of C no longer has overlap because xe to the 2X is not the same as anything up here. So if we find our YT prime and our YT double prime and then substitute it back into this equation. So we're going to get our YT equaling AE to the two X * X, our YT prime doing the product rule 2 AE to the two ** plus AE to the 2X, our YT double prime doing the product rule again, 4AE to the two X * X + 2 AE to the 2X. And then the derivative of this last term was just two AE to the 2X. Combining our like terms, we can then see that our Y double prime is going to be 4 AE to the two ** plus 4AE to the two X -, 4 YS. So -4 AE to the two X * X is going to still equal that 2 E to the 2X. So our AE2X to the X are going to cancel. There's four of them here and -4 of them here. So we end up with 4A equaling 2 or a equaling 1/2. So our Y particular is 1/2 E to the two X * X that took into account the overlap. Looking at another example, Y to the 4th -2 Y double prime plus Y equal xe to the X, so our Y to our R to the fourth minus two r ^2 + 1 = 0. Factoring that, we can see that R equal 1 and -1 are double roots. So our YC is going to equal E to the X times the quantity C1 plus C2X plus E to the negative X * C three plus C4X. Now our YT based on up here would be E to the X * a plus BX, and we can see that we have overlap. So we need to multiply that Y of T by X to some power so that we don't have overlap. If we did just plain X instead of if we did just plain X, we're going to need X ^2. But if we did X, we could see that we get E to the X times AX plus BX squared and we would still have overlap because this AX would be an overlap with the C sub 2X. So plain old X isn't going to be enough. We're going to multiply by X ^2, and if I multiply by X ^2, I get E to the xax squared plus BX cubed. Now with our YC, we don't have any overlap. This was C1 plus C2X. But now we have AX squared plus BX cubed. So our YT is going to be this equation E to the xax squared plus BX cubed. Now the fact that we need Y to the 4th is going to take some manipulation. So we do our Y of T equaling our E to the xax squared plus BX cubed. And we have to do product rows. So the derivative of E to the X is just E to the X times the second plus the derivative of the second, which would be two AX plus three BX squared times the first. Those both have an E to the X in them. So I'm going to pull it out and then I'm just going to combine any like terms. The second derivative E to the X derivative of E to the X is E to the X. So it's E to the X times that whole parenthesis plus the derivative of the second, IE the inside times the 1st which was E to the X. So we get 2A plus 2AX plus 6 BX plus three BX squared. We can see E to the X is in both of them. So I'm going to factor it out and I'm going to combine like terms. The third derivative E to the X times that whole first parenthesis plus the derivative of the inside times E to the X. So here's the derivative of the second part, combining our like terms again, and Y to the fourth T or YT to the 4th prime E to the X. That entire parenthesis plus E to the X times the derivative of the inside. By the product rule. Combining our like terms, we get this whole last line. So now we're going to have our Y to the 4th -2 Y double prime plus Y equaling xe to the X. So our Y to the fourth was our Y to the 4th derivative was E to the X times this whole quantity. And then we're going to subtract 2 of our Y double primes, and then we're going to add our Y and that's going to all equal xe to the X. So when we combine our like terms, we can see 8A plus 24B equals 0 and 24 B is going to equal 1. So B is 124th A equal -1 eighth. So our Y particular is E to the X * -1 eighth X ^2 plus 124th X ^3. Thank you and have a wonderful day.