Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. General solutions of linear equations. An NTH order linear differential equation of the form P sub zero of XYN plus P sub 1X of Y to the north -1 plus... P to the north -1 XY prime plus P to the NX times Y equals some function in terms of X. If we take that equation and we divide everything through by the function that was multiplied by our Y to the north, we get a new equation and let's have it equaling little F of X. So to make that into a homogeneous linear equation, our little F of X would just equal 0. Now remember, these are all the Y to the north. Y to the north -1 Y prime Y are all linear, IE there's not AY squared in there. Those are all representatives of derivatives. If we let Y one through YN be N solutions of the homogeneous linear equation on the interval I, then if we have all of C1C2 through CN being constants, then the linear combination y = C one Y1 plus C2Y2 through CN YN is also a solution of the equation. The N functions F1 through FN are said to be linearly dependent on the interval I provided that there exists constants C1 through CN, not all zero, such that C1F1 plus C2F2 plus through CNFN equals 0 on I for all XS in the I. The function is said to be linearly independent for the same function in the trivial case where C 1 = C two through CN which equals 0 one way. One way to look at this is to think about that if we have Y1 through YN are solutions of the homogeneous NTH order linear equation on an open interval I where each piece of I is continuous and we let our W equaling WY one through YN. If Y1Y2 through YN are linearly dependent, then the determinant W would equal 0 on the interval I. If they're linearly independent, then that determinant doesn't equal 0 at each point of I. Thus, there's two possibilities. Either the determinant equals 0 everywhere on I or it doesn't equal 0 everywhere on I. Let's look at some examples. We want to prove that something is linearly independent. To prove that it's independent, we're going to look at its determinant, and we're going to have E to the X and then we're going to have X to the -2 and X to the -2 natural log of X. The next line, we're going to actually take their derivatives, so E to the X -2, X to the -3. This last one, remember we have to do the product role. So -2 X to the -3 natural log of X plus X to the -3. Then we're going to do the determinant one more time, E to the X6X to the -4 six X to the -4 natural log of X -, 5 X to the -4. When we figure out what this determinant is, we're going to multiply the diagonals. So we're going to get E to the X times -2 X times that. And then we're going to go to the next diagonal, E to the X is on all of them. So I'm just going to leave it factored out in front. And then finally, the other diagonal. So that's my 3 original diagonals, the Ed, the X is going to be in all the minor diagonals also. So now we're going to subtract and then we're going to combine our like terms. So once I multiply, OK, so diagonal, diagonal, diagonal, subtract. OK, that looks good. Oops, where'd it go? OK, so that's going to equal when we start combining things. This one here is going to combine with this one here. So those are going to actually cancel. Then we're going to have this one here with that one there and those are going to cancel. Then we're going to have this one here with that one there. So we're going to end up with combining any like terms E to the X times 4X to the -7 + 5, X to the -6 + X to the negative 5th. We could pull out an X to the negative 7th in order to get positive coefficients. So if we pull out an X to the negative 7th, we'd end up with X ^2 + 5 X +4, and we can see that that's actually going to factor. And that would be X + 1 * X + 4. So this is linearly independent because when we did the determinant, it didn't equal 0. Given 1/3 order homogeneous linear equation and three linearly independent solutions, we want to find a particular solution satisfying the given initial conditions. So given all of this information, we're going to come up with some equations. We're going to start with Y equaling C1 times our Y1 plus C2 times our Y 2 + C three times our Y-3. So our Y prime would be C 1 -, 2 C 2X to the -3 -, 2 C 3X to the -3 LNX plus C3X to the -3. Remember I have to do the product rule there. And then our Y double prime 6C2X to the -4. We lost the C1 because the derivative of a constant 0 + 6 C 3X to the -4 LNX -2 C 3X to the -4 -, 3 C 3X to the -4. So now if we look at our initial conditions, our Y 1 is oh. Our Y1 should have been oh, Y one right there. Y one is 1, so we're going to have one equaling C1 plus C2. The natural log of one is 0, so that last term's going to go away. Our Y double prime of one equaling a -11. Oh, I left out one, that's why I'm confused. Y prime of 1 is going to equal five. There we go. So now we get 5 equaling C 1 -, 2 C twos. Natural log of one again is going to go to 0 + C three and then for the last condition -11 is going to equal 6C twos. The natural log of one's going to 0 -, 2 C threes -3 C threes, so -5 C threes. If I take these top 2 equations and combine them, let's just subtract to get the C1 to go away. So 5 - 1 is 4. The C 1 - C one cancels -2 C twos minus AC2 would give us -3 C threes C twos plus AC3. And then if we multiplied that through by 5, we get 20 equaling -15 C 2 + 5 C 3. If I combine that directly with the other equation, the C threes would cancel and we would get 9 equaling 9 negative 9C twos. So C2 is going to equal -1. If we find C2, we can come back here and we can find C1 is going to be two. And if we have C1 and C2, we can find C3 which is going to be 1. So the equation, our solution that we were actually looking for is going to be Y equal C-12 XC two negative 1X to the -2 plus 1X to the -2 natural log of X. Obviously, you can leave those ones off if you want. Now make sure you keep the sign there. And that's our solution. If we look at this next one, the directions say a non homogeneous differential equation, a complementary solution Y of C and a particular solution Y of P are given find a solution satisfying the given initial conditions. So here we're given both a complementary solution and a particular. So we're going to have our Y equaling C1 E to the 2X plus C2 E to the -2 X minus the three. Basically, we're just combining those two equations and we're going to solve for our C1 and our C2. So we get Y prime equaling 2C1 E to the two X -, 2 C 2E to the -2 X When we put in our initial conditions Y of 0 equaling 0. So we get 0 equaling C1 plus C 2 - 3. Or we could think of that as three equaling C1 plus C2. Our next original condition's going to be 10 equaling 2C ones -2 C twos. If I divided that all through by two, I'd get 5 equaling C 1 -, C two. If I add those directly, I get 8 equaling 2C ones. Or C1 is 4. And if C1 is 4, we can see that C2 is -1. So our solution's going to be Y equaling 4E to the 2X minus E to the -2 X -3. So that's a solution that satisfies the given initial conditions. Thank you and have a wonderful day.