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. Thank you and have a wonderful day.