Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. An integrating factor for a differential equation is a function rho XY such that the multiplication of each side of the differential equation rho in terms of X&Y yields an equation in which each side is recognizable as a derivative linear 1st order equation dy DX plus P of X * y of X. Sometimes we just write that as P of X * y equaling Q of X. So we're going to let rho equal E to the integral of PX DX. That's called the integrating factor. So then we're going to multiply the entire linear 1st order equation through by that row. We get E the integral PXDXDYD X + P of X * e to the integral PXD X * y of X equaling E to the integral P of XD X * Q of X. Now if we look at the left hand side, this is actually just the product rule. And so if we thought about here's my dy DX. So that would be my Y times something. Plus if we look here this PXE, then the integral PX DX is the derivative of E to the integral PX DX. So we have our U * v prime here, plus our U prime times V. So it's just the DDX of E to the integral PXDXY of X equaling that right hand side. So the next thing we're going to do is we're going to use the fundamental theorem of calculus that says an integral and a derivative undo themselves. So we're going to take the integral of the derivative in terms of DX. And if we do it to one side, we're going to do it to the other. So this integral and the derivative canceled, leaving us E to the integral PXDXYX equaling the integral of the other side. I left off a DX there plus some constant. So then we just are going to take, oh, where'd it go? We're just going to take that E to the integral to the other side. So we have Y of X equaling E to the negative integral PX DX times the quantity, the integral E to the integral PX DX times the Q of X DX plus C. So let's look at some examples. If we have Y prime plus Y equal to and an original initial condition of Y0 equaling zero, the first thing we're going to realize is that our function for our P of X is just the coefficient of the Y. So in this case, we have to get the Y prime by itself. That's important. And then our coefficient for the Y is going to be what we're going to make our P of X equation. Now in this case, our Q of X is just two, but our P of X is 1. So our row in terms of XY is just going to be E to that coefficient on the Y term, in this case 1 DX. Well, we know that the integral of 1 DX is just X. So now we're going to take this equation here, the original, and we're going to multiply everything through by E to the X. We're going to recall that this is really just the derivative of exy, derivative of E to the X is E to the X and the derivative of Y is Y prime. So if we thought about the product rule, Y prime the derivative of the Y times the Ed to the X plus the derivative of E to the X, which in this case is E to the X times the Y. That's all going to still equal our 2E to the X. Now we're going to take the integral of the derivative, and if we do it to one side, we're going to do it to the other. The integral and the derivative over here are going to cancel, leaving us just E to the XY. Over here, we can actually evaluate what is the integral to E to the XDX. Well, the integral of 2 E to the X is just two E to the X because the derivative of E to the X is itself plus some constant. So now we're going to go back to the original condition. Actually, let's go ahead and divide everything through by E to the X. So we get Y equaling 2 plus CE to the negative X. Now we're going to go back and we're going to stick in the initial condition. So Y is 0. When our X is 0, we're going to solve for C E to the -0 or E to the zero is 1 so we can see our C is -2. So the equation we're looking for is Y equal to plus actually Y equal to -2 E to the negative X. Let's look at a little more interesting one. For this one, we have XY prime plus five y = 7 X squared. So we're going to start by getting A1 coefficient on the Y prime. So we're going to divide everything through by an X. Now we need it to be in the form DYDX plus some function times our Y equaling whatever else is left. So in this case, we're going to have our P of X equation equaling 5 / X. So when we find our row, our row of the XY is just going to be E to the integral of whatever that P of X equation was, in this case 5 / X DX. To evaluate this, we know that the integral of 5 / X DX is just going to be E5 to the natural log of X. Now technically that's an absolute value, but if we look at our initial condition, we're at 2. So we're going to use the positive for these problems. We're going to always consider this to be a positive value. So now we're going to bring the five up into the exponent. So we're going to get eln of X to the fifth and the E to the Ln are going to cancel, leaving us X to the 5th. So we're going to go back to the original equation and multiply everything through by X to the 5th. Now when we look at this left hand side, that's really the derivative of a product. It's the derivative of X to the fifth Y. If we think about this, the derivative of this Y was Y prime. So we'd have X to the fifth Y prime right here. Plus the derivative of X to the fifth is 5X to the fourth times the other term of Y. So the next step is to take the integral of the derivative, which by the fundamental theorem of calculus is going to cancel. And if we take the integral on one side, we're going to take the integral of the other. So we get X to the fifth, Y equaling X to the 7th plus some C divide everything through by X to the 5th. So we get Y equaling X ^2 plus CX to the -5. We go back and we use our initial conditions. So 5 equal 2 ^2 + C two to the negative 5th. So 5 equal 4 + 130 second C so we get one equal 132nd C or C is 32. So our final answer is going to be Y equal X ^2 + 32 X to the negative 5th. Looking at another example, we have Y prime equaling 1 -, y cosine X. So we're going to distribute here. First of all, because we need to get it into the right format, we're going to take the Y primes and the YS all to one side. So we get Y prime with a coefficient of one plus a function in terms of X, IE cosine X times our Y value equaling whatever's left. So our P of X equation is always the coefficient on the Y. In this case, it's cosine X, so our row is going to equal E to the integral of the cosine X DX. Well, that's just E to the sine X. When we have this, we're going to now multiply everything through by E to the sine XY prime plus E to the sine X cosine XY equaling E to the sine X times cosine X. So the left hand side is going to really be the derivative of E to the sine XY. Remember, product rules. So the derivative of one term, the derivative of Y, is that Y prime E times the first E to the sine X plus the derivative of the first E to the sine X. The derivative is cosine X * e to the sine X times the second, IE the Y, and that just all equals E sine X times the cosine X. Fundamental theorem of calculus. We're going to take the integral of the derivative and if we do it to one side, we're going to do it to the other. The integral and the derivative are going to cancel. So we get E sine XY equaling over on the right hand side. When we evaluate this, we're just going to get PSI E to the sine X plus some C putting in that original condition. Actually, let's divide everything through by E to the sine X. So we get E equaling one plus CE to the negative sine X. Now putting in the original condition that Y of π equaling 2, so Y is 2, when X is π sine of π is 0, E to the zero is 1. So solving for C, we can see C = 1. So our final equation is going to be Y equal 1 plus the C is 1, so E to the negative sign X. When we look at this next one, we're going to have X ^2 + y prime plus three XY equaling XY of 0 equaling 1. We need to have A1 coefficient on our Y prime. So the first thing we're going to do is we're going to divide everything through by X ^2 + 4. So we're going to have Y prime plus three X / X ^2 + 4 Y equaling X / X ^2 + 4. So my P of X equation is whatever is being multiplied by the Y term, in this case 3X over X ^2 + 4. So my rho is going to equal the integral of E to the. Nope, I wrote that wrong, sorry. E to the integral three X / X ^2 + 4 DX. When we evaluate this, if we look at the denominator of the X ^2 + 4. If we thought about doing AU sub, perhaps if we let U equal X ^2 + 4, we can see that DU is 2 XDX or 1/2 DU equaling XDX. You may or may not need this U sub. So here we're going to get E equaling or E to the 3 * 1/2 for that XDX all over U times the DU. So that's really E to the integral of 3 / 2 U DU, and that's going to equal E to the three halves natural log of U. Now by the power rule, we're going to take that three halves up on top. So we're going to get E to the natural log of U to the three halves. The E and the lane are going to cancel and resubstituting back EU, we're going to get X ^2 + 4 to the three halves so that X ^2 + 4 to the three halves is going to get multiplied all the way through this equation. So we're going to have X ^2 + 4 to the three halves times our Y prime. Plus, if I multiply X ^2 + 4 to the three halves times the three X / X ^2 + 4, we end up with three X * X ^2 + 4 to the 1/2 times the Y equaling X * X ^2 + 4 to the one half. This left hand side is going to be equivalent to the derivative in terms of X of X ^2 + 4 to the three halves times Y prime or times Y. And then we're going to have that equaling X * X ^2 + 4 to the one half. If we look back here, the derivative of the Y is Y prime. So the derivative of the second times the 1st X ^2 + 4 to the three halves times Y prime. Plus the derivative hit here is going to be bring down the three halves to the one less power. So the half derivative of the inside 2X. So 3 halves times 2X would give us the 3X all of that times the X ^2 + 4 to the one less power to the half, and that was all times our Y. So now we're going to take the integral of a derivative fundamental theorem of calculus. Those are going to cancel, and if we do it to one side, we're going to do it to the other. So the left side just turns into X ^2 + 4 to the three halves power times Y equaling the right hand side. If we wanted to do AU sub again, we could have U equaling X ^2 + 4, so our DU would equal 2X DX. So 1/2 DU would equal the XDX. So when we look at this we get the integral here's the X and the DX which is going to get replaced by 1/2 DU and that left us U to the 1/2. So add 1 to the exponent, divide by the new exponent. 1/2 * 2/3 is going to give us 1/3 plus some constant C Substituting our U back, we'd end up with one third X ^2 + 4 to the three halves plus our C So back here 1/3. X ^2 + 4 to the three halves plus C Now we're going to go back and we're going to divide through by the X ^2 + 4 to the three halves. So that's going to give us Y equaling 1/3 + C * X ^2 + 4 to the -3 halves. Putting in our initial condition of Y of 0 equaling one, we get one equaling 1/3 plus C 0 ^2 + 4 to the -3 halves. 1 - 1/3 would give us 2/3 sqrt 4 is 2/2 to the -3 is 1/8 so we get 2/3 equaling 1/8 C Multiply each side by 8 and we get C equaling 16 thirds. So our final answer is going to be Y equal 1/3 + 16 thirds times the quantity X ^2 + 4 to the -3 halves. Thank you and have a wonderful day.