Hello wonderful mathematics people, This is Anna Cox from Kellogg Community College. Homogeneous equations with constant coefficients. If the roots are one R2 through R sub N of the characteristic equation, a sub NR to the north plus a sub n -, 1 R to the north -1 plus through a sub 2R to the second plus a sub 1R plus a sub 0 = 0 are real and distinct, then Y of X = C one E to the R1X plus C2, E to the R2X plus through C sub NE to the RNX. Now it can be shown that E to the R1, xe to the R2X, etcetera is a solution to the original characteristic equation. And if each of them is a solution to the characteristic equation, then the sum of any constant times those distinct roots has to also be a solution. So that's a general solution of a sub NY to the n + a sub n -, 1 Y to the n - 1, remembering those are derivatives plus through a sub 2Y double prime plus a sub one Y prime plus a sub not equaling 0. Thus, the N linearly independent functions E to the R1 XE to the R2X through E to the RNX constitutes a basis for the N dimensional solution space or E sub R1 X, E sub R2X through ERNX are the solutions for this repeated roots. If the characteristic equation a sub NR to the north plus a sub n -, 1 R to the north -1 etcetera, has a repeated root of multiplicity K, then the part of the general solution of the differential equation and the A sub NY to the n + a sub n -, 1 to the Y to the n - 1 through a sub 2Y double prime plus a sub one Y prime plus a sub not equaling 0 corresponds to R is of the form C1 plus C2X plus C3 X squared plus... C sub KX to the K -, 1 * e to the RX. That's for the repeated root portion if we if we have complex roots. If the characteristic equation has a unrepeated pair of complex conjugate roots A+ or minus BI, where B is not equal to 0, then the corresponding part of the general solution has the form E to the AX times the quantity C1 cosine BX plus C2 sine BX. The linearly independent solutions E to the AX cosine BX and E to the AX sine BX generate A2 dimensional subspace of the solution space. Finally, if we have repeated complex roots, so if we have the conjugate pair A+ or minus BI, once again with the B not equaling 0 has multiplicity K, then the corresponding part of the general solution has the form A sub 1 + a sub 2X plus all the way out through a sub KX to the K -, 1 E to the quantity A+ bix plus the quantity B1 plus B2X, etcetera. And that's really just X sub P. The summation P goes to zero P = 0 up through K - 1 of X to the PE to the AX quantity C to the P cosine BX plus D sub P sine BX. So looking at some examples, if we start with 2Y double prime -7 Y prime plus three y = 0, the first thing we're going to do is we're going to replace the Y double prime with R-squared, the Y prime with R, and we're going to solve for what R is. So we get two r ^2 - 7, R Plus 3 = 0. We can see that that factors. If we don't see that it factors, we can always do quadratic formula and we get R equaling 1/2 and R equaling 3. So Y is going to equal C1 E to the half, X + C two E to the 3X. If we find our Y prime and our Y double prime, we can then substitute in to the original condition to show that that is truly a valid statement. If we look at another example, Y double prime -6 Y prime plus 13 Y equaling 0, we start with replacing our primes. So we get r ^2 - 6 R Plus 13 equaling 0. So using quadratic formula, we can show that R is 3 ± 2 I if we get y = C one E to the 3X cosine 2X plus C2 E to the 3X sine 2 XY prime equal 3C1 E to the 3X cosine 2 X -2 C 1E to the 3X sine 2X plus 3C2 E to the 3X sine 2X plus 2C2 E to the 3X cosine 2X. If you go ahead and find the Y double prime and simplify, you can see that that's 5C1 E to the 3X cosine two X -, 12 C 1E to the 3X sine 2X plus 5C2 E to the 3X sine 2X plus 12C2 E to the 3X cosine 2X. Substituting that into the original equation, we can see that that gives us a valid initial condition if we look at one of these with initial conditions given. So we're going to start by replacing our primes with Rs. So we get r ^3 + 10. R-squared plus 25R equals 0. Factoring that, we get R times the quantity r + 5 ^2 equaling 0. So we can see our R is zero -5 and -5, IE we have a double root. So Y is going to equal C1 E to the 0X and E to the 0X is really just one plus C2 E to the negative 5X plus C3 xe to the -5 X. When we find our Y prime and our Y double prime, then substituting in the original given conditions, we can now solve for C1C2 and C3. By solving, we can see that Y equal 24 fifths -9 fifths E to the -5 X -5 xe to the -5 X. If we find our Y prime, our Y double prime, and our Y triple prime, we can show that the initial condition is true. I'll leave that for you to do. Finally, if we have Y to the fourth minus Y to the third prime plus Y double prime -3 Y prime -6 Y equaling 0, first thing we replace those primes with Rs, so we get R to the fourth minus r ^3 + r ^2 -, 3 R -6 equaling 0. We don't have an easy way to find R to the fourth other than through some synthetic division and trial and error. Remember to look for roots that are the constant divided by the leading coefficient, the factors of those. So the only numbers I would want to try would be positive -1, positive -2, positive -3, and positive -6 in this particular equation. So I used 2 and I realized that 2 is a factor by doing the synthetic division. Once I pull out the two, I can see that I get r ^3 + r ^2 + 3 R Plus 3 left. Factoring what was left, I can then see we get r - 2 * r + 1 * r - sqrt 3 I times the quantity r + sqrt 3 I. So we get 4 solutions, two -1 root 3 I and negative root 3 I. So our Y is going to equal C1 E to the 2X plus C2 E to the negative X + C three E to the 0X because there was no real portion with the complex portion, and E to the 0X is really just one cosine of root 3 * X + C four E to the 0X sine squared of 3 * X. We can then find our Y prime, Y double prime, Y triple prime, and Y to the 4th prime to show that the initial condition is true. Once again, I will leave that for you to do.