Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. A second order differential equation in the unknown function Y of X is one of the form G of XYY prime Y double prime equaling 0. This is linear provided that G is linear in the dependent variable Y and its derivatives Y prime and Y double prime. A linear 2nd order derivative can be written in the form AX times Y double prime plus bxy prime plus cxy equaling F of X where ABC and F are continuous on an open interval I. If F of X = 0, then the above equation is a homogeneous equation. Axy double prime plus bxy prime plus CX times Y equal F of X can be rewritten. If we thought about dividing everything through by our a of X and then if we thought about making the equation homogeneous, IE it has to equal 0, we have a new format of Y double prime plus P of XY prime plus Q of XY equaling 0. The sum of any 22 solutions to this equation is also a solution, and any constant multiple of a solution is also a solution. So if Y1 and Y2 are solution of our Y double prime plus P of XY prime plus Q of XY equals 0, then Y equaling C1Y1 plus C2Y2 is also a solution. Let's look at an example. Y double prime -3 Y prime equaling 0Y1 of 1 equal 1 Y 2 equal E to the three XY of 0 equal 4, Y prime of 0 equal -2. So the first thing we're going to do is we're going to have Y equaling C1 times our Y1 plus our C2 times our Y2. Our Y is going to equal C 1 * 1 plus C 2 * e to the three XYC 1 + C two E to the 3X. Now we have initial values for the original and for the first derivative. So let's go ahead and take the first derivative. The first derivative of a constant is 0, but the first derivative of the second part of this equation is going to give us 3C2 E to the 3X. So now when we stick in the initial conditions we get 4 equaling C1 plus C2 E to the 3 * 0. E to the 0 is just one. So that equation's going to turn into 4 equaling C1 plus C2. The second equation -2 = 3 C 2E to the 3 * 0 or -2 equal 3C2. So we can see C2 is going to equal negative 2/3 and thus C1 has got to equal 14 thirds. So a solution to this is going to be y = 14 thirds -2 thirds E to the 3X. Now if we take our first derivative, we can see that the first derivative is -2 E to the 3X. So our second derivative is going to be -6 E to the 3X. If we check this in the original given condition -6 E to the three X -, 3 * -2 E to the 3X, we can see that that does indeed give us 0 Wronskian determinant. If we have W equaling FGF prime G prime which equals the FG prime minus F prime G, If W = 0 then the equation is linear dependent, and if W doesn't equal 0 then it's linear independent. So if we look at something like F of X equaling sine squared X&G of X equaling 1 minus cosine 2X. Let's start with remembering that one minus cosine 2X could really be thought of AS2 sine squared X by one of our Pythagorean identities. This is our angle reduction formula. So instead of 1 minus cosine 2X, let's rewrite G of X AS2 sine squared X. So now we want to know are these two equations dependent or independent? So we're going to use our determinant that says sine squared X and then two sine squared X. The derivative of sine squared X is going to be two sine X cosine X, and the derivative 2 sine squared X is going to be 4 sine X cosine X. So now when we actually find this determinant, we're going to get 4 sine cubed X cosine X -, 4 sine cubed X cosine X. This is going to equal 0. So these two equations are linear, linearly dependent, distinct real roots. If the roots R1 and R2 are real and distinct, then Y of X = C one E to the R1X plus C2 E to the R2X is a general solution of AY double prime plus BY prime plus Cy equals 0. If the roots repeat, then y = C one ER 1X plus C2 xer 1X. So on this equation we want to find the general solution of the differential equations. So if we have Y double prime +2 Y prime minus fifteen y = 0, we're going to say let's re write this with R-squared being Y double prime and R being Y prime and the plain old Y just being a one. So now we're going to factor this r + 5 R -3, so R equal -5 and R equal 3. So Y is going to equal C1 E to the negative 5X plus C2 E to the 3X. Now technically our C1 and C2 can be switched because variables are just variables, but that's a generic equation that would satisfy this original. If we wanted to, we could show it why prime would be -5 C 1E to the -5 X plus 3C2 E to the three XY. Double prime would equal 25 C 1E to the negative 5X plus 9C2 E to the three X SO25C1 E to the negative 5X plus 9C2 E to the 3X +2 of my Y primes -5 C 1E to the negative 5X plus 3C2 E to the three X -. 15 of our C1 E to the negative 5X plus C2 E to the 3X needs to equal 0. So if we look at this, we can see that 25 of these -5 X ones, and here's -10 and here is a -15 of them. So 25 negative 10 negative 15 is indeed 0. If we look at the C twos E to the 3X, there's nine of them, and here is a +6, so 9 + 6 is 15. And then here's a -, 15. So the original equation, the original condition is indeed met by this generic general solution. If we look at this next one, we're going to have nine r ^2 -, 12 R Plus 4 = 0. We get three r -, 2 ^2 equaling 0. So our generic equation ours 2/3 and it's a repeated root. So we're going to have y = C one east to the two thirds X + C two xe to the 2/3 X. If we look at our first derivative, we get 2/3 C 1 E to the 2/3 X plus we have to do the product rule here. So C2 E to the two thirds X + 2/3 C two xe to the 2/3 X and our Y double prime is going to be 4 ninths C1 E to the two thirds X + 2/3 C 2 E to the 2/3 X Plus we have to do the product rule again down here. So 2/3 C 2E to the two thirds X + 4 ninths C2 xe to the 2/3 X. So we can see that these two combine and we get 2/3 + 2/3 or 4 thirds here E to the 2/3 XC 2 E to the 2/3 X. So 4 thirds C2 E to the 2/3 X. So then when we stick this in, we get 9 times our Y double prime -12 YS primes, and then +4 YS, and that needs to equal 0. So when we look here 9 times, this 4 ninth is going to give me 4C1 E to the 2/3 here -12 times the 2/3 three goes into twelve -4, so that's going to give me -8. So a +4 and a -8 and down here is another +4. So that part does indeed go to 0. We can go ahead and check the rest in the same kind of fashion. So 9 and 4 thirds is going to give us 12 of the C2 E to the 2/3 X. And then down here we have -12 times that, and hence that one does go to zero. And finally we get 4 of the C2 xe to the 2/3 X we get -8 down here and +4. So we can see that this y = C one E to the two thirds X + C two xe to the 2/3 X is indeed a solution to that 9Y double prime -12 Y prime plus 4Y equaling 0. Thank you and have a wonderful day.