Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. A homogeneous first order differential equation is one that can be written in the form DYDX equaling F of y / X. To solve this kind of an equation, we're going to substitute V equaling y / X, or we know that Y would equal VX. So taking the derivative of each side, we'd get DYDX. And on the right hand side we're going to use the product rule. So the first times the derivative of the second plus the second times the derivative of the 1st, and the derivative of X is just in the derivative of X in terms of X is just one. So that's going to simplify to v + X DVDX. If we wanted to solve this now, we could. First of all, we're just going to rewrite it with the XDVDX coming 1st and then the DYDX. Remember DYDX was just F of y / X, but with our substitution that y / X is just V So now we get F of V equaling XDVD X + v and then if we divided everything through by the X we end up with a separable equation. Our F of v / X is equaling DVDX plus 1 / X * v Remember, in a separable equation, we needed the coefficient of our DVDX to be 1, and then there was a function for whatever that was being multiplied by our V. If we look at an equation 2X YY prime equaling X ^2 + 2 Y squared, we'd let Y equal VX. So dy DX is going to equal V plus DV DX times X. Now we're just going to do straight substitutions. SO2X instead of Y, we're going to put in VX. Instead of Y prime, we're going to put in v + X DVDX on the right hand side. That turns into X ^2 + 2, the quantity VX squared. So I'm going to do some distribution and then we're going to realize that the two v ^2, X ^2 on each side is going to cancel. So now I'm just going to take all the VS to one side and all the XS to the other. So we get the integral of two VDV equaling the integral of one over XDX or v ^2 equals the natural log of the absolute value of X + C. Replacing our V with what our V equaled, which was y / X, we can see y / X quantity squared equal natural log of absolute value of X + C. If we multiply everything through by that X ^2, we see that y ^2 equal X ^2 natural log of X plus CX squared. If we look at a different equation X -, y * y prime equaling X + y, we get Y equaling VX. So our DYD X = v + X DVDX. Now we just do a straight substitution again, X minus VX times the quantity v + X DVDX equaling X plus VX, distributing things out, combining like terms, factoring again. So just a bunch of algebra in that. In between step we end up with the X ^2 times the quantity 1 -, v DVDX equaling X times the quantity 1 + v ^2. We're going to take all the VS to one side and all the XS to the other. So we get the integral of 1 -, v / 1 + v ^2 DV. Then we're going to separate that into two separate integrals because we know that the first integral 1 / 1 + v ^2 is really just our tangent inverse of V identity. If we're not familiar with that one, remember we could do a trig substitution here to help us figure it out. Whenever we have something that looks like a Pythagorean theorem, we can think about which side is going to be the variable. In this case would be the opposite, and then we would have our tangent inverse. If we did some straight substitution, we'd realize tangent inverse of V. This next one v / 1 + v ^2. If you let U equal 1 + v ^2, take the derivative of each side, you can then see that this is really just equivalent to 1/2 natural log of U. But EU is 1 + v ^2, so 1/2 natural log of 1 + v ^2. The right hand side 1/2 natural log of 1 / X. No, wait, what did I just do? This one was the 1/2 natural log of 1 + v ^2. This right hand side, the integral one over XDX is just natural log of X. We need a + C in there somewhere. So now all we're going to do is we're going to substitute for the V put in y / X. So our final answer is tangent inverse y / X -, 1/2 natural of the absolute value of 1 + y / X ^2 equaling the natural log of X + C Sometimes the methods we use don't work and we have to look for something different. In this case, we're going to let V equaling X + y so that our DVDX is just going to be 1 plus dy DX. So then the X + y was V Our dy DX we could think of as DV DX -1 equaling 1. So now V DV DX equal 1 + v just doing distributive property and taking the V over. So now we're going to get all the VS to one side 1 + V / 1 + v DV equaling DX if we take the integral of each side. There we could do AU substitution if we wanted. Let U equal 1 + v, so DU is going to equal DV. So we get AU in the denominator Adu. We have AV up in the numerator, so V is going to equal U -, 1 equaling the X. So now we can see that this would just really be 1 -, 1 over UDU equaling X. So we'd get U minus the natural log of U plus some C equaling X. But U is 1 + v minus the natural log of 1 + v + C equal X. But V is really X + y. So we get 1 + X + y minus the natural log of 1 + X + y plus some C equaling X. So now if we wanted to solve for Y, we could combine first of all this one, and that's C to get a new C of some sort. So we'd get Y equalling, and we can see that the X is here and the X here are going to cancel. So Y equal the natural log of 1 + X + y plus some new C, Let's call it C1. We can't solve this for Y because Y is within the logarithm and Y is over here. We can't solve it easily, so that's just going to be our final answer. Bernoulli's equation dy DX plus P of XY times Y equal the function Q in terms of XY to the north. If n = 0, N equal 1. This is going to turn into a linear equation. Otherwise we're going to let V equal Y to the 1 -, n So then if we take the derivative of each side, we get DV DX equaling 1 -, n what times Y to the negative NDYDX. So if we solve for dy DX, we could see that YN divided by 1 -, n DVDX equal dy DX. Now we're going to do substitution in the original. So we get Y to the n / 1 -, n DVD X + P of XY equaling Q of XY to the N. If I multiply everything through by 1 -, n / y to the NI get DVDX plus the quantity 1 -, n PXY to the 1 -, n equaling 1 -, n * Q of X. So DVDX plus one minus NP of XV equaling 1 minus NQ of X. So if we look at an example, we're going to have X ^2 y prime +2 XY equal 5 Y to the 4th. So we're going to let Y prime +2 over XY equaling 5 / X ^2 y to the 4th. So we're going to let our V equaling our Y to the 1 -, 4 or our Y equal, or our Y to the -3. So our DV DX is going to equal -3 Y to the -4 dy DX. So we would get -1 third Y to the fourth DV DX equaling our dy DX. So we'd have negative 1/3 Y to the fourth DV DX +2 over XY equaling 5 / X ^2 y to the 4th. We want to get the DVDX by itself, so we're going to have DVDX -6 over XY to the -3 equaling -15 / X ^2. So then our DVDX -6 / X, our Y and the -3 back here was just our V. Now this looks like an equation that we can find our row. If we have row equaling E to the exponent of the integral of -6 over XDX, that's E to the -6 natural log of X, which is X to the -6. So now if we multiply everything through by X to the -6 DVDX -6 over X to the 7th V equaling -15 X to the -8, so that this is going to turn into the derivative using our product rule of X to the -6 V equaling the -15 X to the -8. Now, if we integrate DDXX to the -6 V in terms of DX integrate one side, we're going to integrate the other. So that's going to turn into X to the -6, V equaling 15 sevenths, X to the -7 plus some C. So if we multiply by X to the 6th, we get V equaling 15 sevenths, X to the -1 plus CX to the 6th. Now our V was really our Y to the -3. So now we're going to have Y to the -3 equaling 15 sevenths, X to the -1 plus CX to the 6th. So we could think of this as wide in the -3 equaling 15 over seven, X + 7 CX to the 7th over 7X. Just getting a common denominator there. So y ^3 is going to equal seven X / 15 plus that 7 C could be replaced by some new constant. So y ^3 equal seven X / 15 plus C1X to the 7th. We could take the cube root if we wanted Y equal the cube root of seven X / 15 + C sub 1X to the 7th exact differential equations M the function M in terms of X&YDX plus the function N in terms of XYD y = 0 when DF equal F XDX plus FYDY or the function The derivative of the function in terms of X plus the derivative of the function in terms of Y of F of XY is exactly MDX plus ND Y, where M equal DF DX. The delta of F over the delta of X&N equal the delta of F over the delta of Y. Thus the delta in terms of Y of M is just the delta DY of the delta F delta X, which notation wise is just the delta in terms of Y of F of X or written fxy same concept with the delta of the delta of X of N the function N. So that's just the delta in terms of X of the delta F over delta Y or the delta in terms of Y of F of Y, which is just FYX. If Fxy and FYX are continuous in an open set in the XY plane, then Fxy and FYX so that the delta M over delta Y equal the delta N over delta X is a necessary condition that MDX plus ndy equals 0 be exact. If MY does not equal NX, then the differential equation is not exact. Thus there is no function F of XY where F of X equal M and FY equal N. Criterion for exactness. Suppose that the function the functions mxy and nxy are continuous and have continuous first order partial derivatives in the open rectangle R where a is less than X less than B&C is less than Y less than D. Then the differential equation MXYD X + n XYDY is equal to 0 is exact in R if and only if delta M over delta Y equal delta N over delta X at each point of the rectangle. That is, there exists a function F in terms of X&Y defined on R with delta F over delta X equaling M and delta F over delta Y equaling N if and only if those two things are equal. Proof. If we have our function in terms of XY equaling the integral of MXYDX plus GY, where delta F over delta X is the M, we're going to next choose AG of Y so that the N is just the delta F over delta Y. Or we could think of that as the delta in terms of Y of the integral MXYDX plus GY. That was what we defined our F of a moment ago. So our N is just the delta over delta Y of the integral MXYDX. Plus if we took the delta in terms of Y of GY, that's just G prime Y, the first derivative G prime Y, then if we solved it would equal N minus the delta in delta Y of the integral MXYDX. So now we're going to show that that delta in terms of delta X of N minus the delta in terms of Y of the integral is just really delta N over delta X minus the delta of delta X of the delta of delta Y of the integral. And so we're going to switch the order and we're going to do the delta of delta X first because we know that the derivative of an integral just cancels out. So this last portion is just going to turn into the the function M. So we're going to get Delta N over Delta X minus Delta M. Over delta Y. Now the hypothesis stated we needed those two to be equal, so if we subtract them, they would have to equal 0. So now we're going to substitute. If we have that integral MXYDX plus GY, we could think of that as the integral of MXYDX plus the integral of the derivative, because an integral of a derivative is going to give us back the original function. But a moment ago we showed that this G prime Y was really just N minus delta over delta Y of the integral MXYDX. So now we have the integral of MXYDX plus the integral of NXYDY minus the integral of delta over DY of the integral MXYDXDY, which is the desired function with FX equaling M and GY equaling N. So if we have four X -, y DX plus six y -, X DY equaling zero, we're going to start by letting our F equal the integral of four X -, y DX. So we're going to get this to be two X squared minus XY. Remember that Y right now is just a constant because we're doing it in terms of DX plus some function in terms of Y. Now we're going to take that F equation and we're actually going to find the derivative in terms of Y. So the derivative of this two X squared is going to be 0, the derivative of negative XY in terms of Y is going to be negative X, and the derivative of G of Y is going to be G prime of Y. So when we look at the N equation, we know that the N is whatever was being multiplied by the DY. So the N was six y -, X. So we need the FY to equal the N So we're going to have the negative X + g prime Y equaling six y -, X. Obviously the negative XS are going to go away, and we can see G prime y = 6 Y So if G prime Y is 6, YG of Y is three y ^2 plus some constant. So when we do our final equation, we're going to have F equaling 2X squared minus XY plus three y ^2 + C Frequently we think about having that set equal to the 0, so we could have some constant equaling the two X squared minus XY plus three y ^2. Looking at another one, if we have X plus tangent inverse YD X + X + y / 1 + y ^2 D y = 0, we'd get F equaling the integral of X plus tangent inverse YDX. So that's going to equal 1 half X ^2 + X tangent inverse Y plus GY, so F of Y. Taking the derivative of that in terms of Y, we'd have zero. For the first part, we'd have plus X / 1 + y ^2 plus G prime of Y. Now our N function is the coefficient with that DY, so it was the X + y / 1 + y ^2. Or we could think of that as X / 1 + y ^2 + y / 1 + y ^2. So the F of Y has got to equal the N So we have X / 1 + y ^2 + g prime Y equaling X / 1 + y ^2 + y / 1 + y ^2. So our G prime of Y is going to equal one y / 1 + y ^2. So our G of Y is just going to be 1/2 the natural log of 1 + y ^2 + C If you don't see that, you could do AU sub where U equal 1 + y ^2 du would equal 2 YDY, 1/2 du equal to YDY, and then integrate that. So we're going to end up with our F being one half X ^2 plus X tangent inverse y + 1/2 natural log 1 + y ^2 plus C and we frequently write that with the F just being a function. So we're going to have some C equaling one half X ^2 + X tangent inverse y + 1/2 natural log 1 + y ^2 reducible second order equations. The general form F of XYY prime Y double prime equaling 0. If either the dependent variable Y or the independent variable X is missing, then the equation may be reduced to a first order equation through substitution. If Y is missing, then the function in terms of XY prime and Y double prime equals zero. We're going to let P equal Y prime, which is our dy DX. So our Y double prime would equal DP DX. Thus the new function would be in terms of XP and P prime equaling 0. If our X is missing, then our function is in terms of YY prime and Y double prime equaling 0. So we're going to let P equal Y prime, which is our DYDX. So our Y double prime is DPDX, but we didn't have a DX, so we're going to think of that as DPD y * d YDX or that DYDX remember was P so that would turn into PDPDY. So the new function would be in terms of YPPDPDY all equaling 0. So if we have Y double prime plus four y = 0, we can see there is no X. So we're going to let Y prime equal P which is our dy DX. So our Y double prime is going to be DP DX, which is really DP dy times dy DX. But remember that dy DX was P. So we get PDPDY. So when we do our substitution, we'd have PDPDY equaling -4 Y, just taking that to the other side. So now we want to get all the PS on one side, all the YS on the other, and we're going to integrate each side. So now we get one half P ^2 equaling negative, two y ^2 + C. We want to solve for P So we're going to get P ^2 equaling -4 Y squared plus some new C. And then P is going to equal sqrt -4 Y squared plus some C. Now we could factor out a negative. We could factor out of four actually. So let's think about pulling out of four. And we're going to call this C1 into something new. Let's call it AC2. Remember, variables are just variables for the constant portion. So I could make C1 equaling 4C twos. So now we're going to get row equaling or P sorry P equaling 2 square roots of some new C -, y ^2. So now that P was really our dy DX. So I get dy DX equaling 2 square roots C ^2 minus or C sub 2 -, y ^2. So now we're going to take all the dy's to one side and all the DX's to the other. So we're going to get dy divided by two square roots of C sub 2 -, y ^2 equaling DX. If I integrate both sides, I'm going to get X equaling 1/2 sine inverse of Y over some new constant because let's think about C2 being equal to a ^2 and then we'd have a ^2 -, y ^2. So we could think of that as over a plus our C So let's C2C sub 2 equals some new variable a ^2. So then if we wanted to solve for Y, we would get X -, C all times 2 equaling sine inverse y / a. So we could have that be sine two X -, 2 C times a equaling Y. Now this 2C is really really just a constant again. So if we thought about this, we could think of this as a sine 2X cosine two C -, a cosine 2X sine 2C. So this A cosine 2C could be a new constant, let's call it capital A. And this a actually negative A. And sine 2C could be another constant, let's call it capital B. So we would get an equation for Y equaling a sine 2X plus B cosine 2X. If we look at another equation XY double prime plus Y prime equaling 4X, this one doesn't have AY in it. So we're going to let R Y prime be our dy DX which is just our P, and our Y double prime is now going to be P prime because there was no Y in here. So then this is going to turn into XP prime plus Y prime, which was P equaling 4X. If we divide everything through by X, we get P prime plus one over XP equaling 4. So if we let our row equaling E to the integral of one over XDX, this would be E to the natural log of X, or just plain X. So we'd get XP prime plus P equaling 4X. That left side's going to be equivalent to the derivative of XP. So then we're going to take the integral of a derivative. Fundamental theorem of calculus says those are going to cancel. If I take the integral one side, we need to take the integral of the other. So we're going to get XP equaling two X ^2 + C. Now we're going to solve for P. So we're going to get P equaling two X + C / X. Now our P really was our dy DX, so that's going to equal the two X + C / X. We're going to take all of the X's to one side, all of the Y's to the other, and then we're going to integrate each side. So we get Y equaling X ^2 + C natural log of X plus another constant, we'll call it a. So we get Y equaling X ^2 + C Ln X + a as our final for that one. Thank you and.