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let's do one more homogeneous differential equation or first-order homogeneous differential equation to differentiate from the homogeneous linear differential equations we'll do later but anyway the problem we have here it's the derivative of y respect to X is equal to that X looks like a Y with respect to X is equal to x squared plus 3y squared I'm writing it a little bit small so that I don't run out of space divided by 2 XY so with all of these homogeneous equations and obviously we don't know if the modulus yet so we have to try to write it as a function of Y divided by X so it looks like we could do that if we divide the top and the bottom by x squared so if we just multiply we do it in a different color 1 over x squared or X to the negative 2 over 1 over x squared right we're essentially just multiplying by 1 and then we get that is equal to what 1 plus 3y squared over x squared divided by what is this - if you divide X divided by x squared you just get a 1 over X so 2 times y over X or we could just rewrite the whole thing and we get this is just equal to 1 plus 3y over x squared divided by 2 times y over X so yes this is a homogeneous equation because we were able to write it as a function of Y divided by X so now we can do the substitution with V and hopefully this is starting to become a little bit of second nature to you so we could make the substitution that V is equal to Y over X or another way of writing that is that Y is equal to X V and then of course the derivative of Y with respect to X or if we take the derivative select X on both sides of this that's equal to the derivative of X is 1 times V this is just the product rule plus x times the derivative of V with respect to X and now we can substitute the derivative of Y with respect to X is just this and then the right hand side of the equation is this but we could substitute V for Y over X so let's do that and so we get V Plus X instead of DV DV X I'll write V Prime for now just so that I don't take up too much space V prime is equal to 1 plus 3 V squared we're making the substitution V is equal to Y over X 3 V squared all of that over 2 V - V now let's see what we can do this is where we just get our put our algebra hat on and try to simplify until it's a separable equation in V so let's do that so let's multiply both sides of this equation by 2 V and let's see we'll get 2 V squared + 2 X V V Prime right 2 V times that yep that's 2 X V V prime is equal to 1 + 3 V squared now let's see let's subtract 2 V squared from both sides of this and we will be left with 2x v V prime is equal to 1 + C we're striking 2 V squared from both sides so we just left with a 1 + v squared here right 3 v squared minus 2 V squared is just V squared and let's see if we want it to be separable so let's put all the V's on the left hand side so we get 2x v V prime divided by 1 plus V squared is equal to 1 and let's divide both sides by X so we get the X's on the other side so then we get 2 V and I'll now I'll switch back to the other notation instead of V Prime all right DV DX - V times the derivative of U with respect to X divided by 1 plus V squared is equal to I'm dividing both sides by X notice I didn't write the X on this side so that is equal to 1 over X and then if we just multiply both sides of this times the X where regular we've separated the two variables and we can integrate both sides so let's do that let's go up here I'll switch it in different color so that you know I'm to be working on a different column now so multiply both sides by DX I get to V over 1 plus V squared DV is equal to 1 over X DX and now we can just integrate both sides of this equation this is a separable equation with in terms of V and X and what's the integral of this at first you might think of why this is complicated this is difficult maybe some type of some type of trig function but you'll see that it's kind of just the reverse chain rule we have a a function here 1 plus V squared and expression here we have its derivative sitting right there so the antiderivative of this and you could make you could actually make a substitution if you like you could say U is equal to 1 plus V squared then D U is equal to 2 V DV and then well do you end up saying that the antiderivative is just the natural log of U or in this case the antiderivative of this is just the natural log the natural log of 1 plus V squared we don't have to even write an absolute value there because that's always going to be a positive value so the natural log of 1 plus V squared and just I just want to I mean I hope I didn't confuse you I kind of that's how I think about it I say if I have a expression and I have its derivative multiplied there then I can just take the antiderivative of the whole expression and I don't have to worry about what's inside of it so if this was a 1 over an X or 1 over U it's just the natural log of it so that's how I knew that this was the antiderivative and if you don't believe me use the chain rule and take the derivative of this and you'll get this and hopefully it'll make a little bit more sense but anyway that's the left hand side and then that equals that well this one's easy that's the natural log say the absolute value of x plus we could say plus C but just so that we can simplify it a little bit an arbitrary constant C we can really just write that as a natural log of the absolute value of some constant C right I mean this is still a some arbitrary constant C so we could rewrite this whole equation as the natural log of 1 plus V squared is equal to when you add natural logs you can essentially just multiply the two numbers that are you're taking the natural log of the natural log if we can say the absolute value of CX and so the natural log of this is equal to the natural log of this right so we could say that 1 plus V squared is equal to CX and now we can unsubstituted ovie is equal to Y over X so let's do that so we get 1 plus y over x squared is equal to C X let me scroll this down a little bit let's see let's multiply both sides equation times x squared this is we could just this we could rewrite this as y squared over x squared so we multiply both x times x squared you get x squared plus y squared is equal to C X to the third and we're essentially done if we want to put all of the variable terms on the left hand side we could say that this is equal to x squared plus y squared minus C X to the third is equal to 0 and this implicitly defined function or curve or however whatever you want to call it is the solution to our original homogeneous first order differential equation so there you go I will see you in the next video and now we're actually going to do something we're going to start embarking on higher order differential equations and actually these are more useful in some ways easier to do than the homogeneous and the exact equations that we've been doing so far see you in the next video