Let’s do one more homogenous differential equation or first order homogenous differential equation to differentiate from the homogenous linear differential equations will do later. But anyway, the problem we have here it’s the derivative of y with respect to x is equal to the x instead of y, as with respect to x is equal to x2 + 3y2. I’m running a little bit small say I’m running of space divided by 2xy.
So, all of these homogenous equations and as if we don’t know homogenous yet so we have to try to write it as a function of y divided by x. It looks like we can do that if divide the top and the bottom by x2 so if we just multiply with the different color.
1/x2 or x-2/1/x2, all right? We’re just multiplying by one and then we get that is equal to 1 + 3y2/x2/2 if you divide x by x2 you just get a 1/x or 2 x y/x or we can just rewrite the whole thing and we get to this is just equal to 1 + 3y/x2/2 x y/x, so yes this is a homogenous 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 interview so we can make the substitution that v is equal to y/x or another way of writing is that y is equal to xv, and then of course the derivative of y with respect to x if we take the derivative sort of x of both of these that’s equal to the derivative of x is 1 x v so it’s 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 we can substitute v for y/x, so let’s do that and so we get v + x sort of dvdvx alright the prime for now just so I don’t take up too much space v prime is equal to 1 + 3 v2 we’re making the substitution v is equal to y/x, 3v2 all of that over 2v.
Now, let’s see what we can do. This is where we get our put our Algebra head on and try to simplify until it’s separable equation in v, so let’s do that so let’s multiply both sides with this equation by 2v, so we’ll get 2v2+ 2 xv, v prime alright 2v times x yeah that’s 2x v prime is equal to 1 + 3 v2.
Now, let’s see let’s subtract 2v2 from both sides of this and we will be left with 2xv, v prime is equal to 1 + see we’re subtracting 2v2 for both sides so we’re just left with the 1 + v2 here right 3v2 - 2v2 just the v2, and let’s see we want it to be separable, so let’s put all the vs on the left hand side, so we get 2xv v prime divided by 1 + v2 is equal to 1, and let’s divide both sides by x so we get x is on the other side, and then we get 2v and now I’ll switch back to the other notation so that v prime alright dv, dx 2v times the derivative of u with respect to x divided by 1 + v2 is equal to I’m adding both sides by x. Notice I wrote the x on the side so that is equal to 1/x, and then if we just multiply both sides of times dx we separated the two variables and we can integrate both side so let’s do that.
Let’s go up here I’ll switch in different colors so that you know I’m dealing with different column now, so multiply both sides by dx I get 2v/1 + v2, dv is equal to 1/x dx and now we can just integrate both sides with this equation, this is inseparable equation in terms of v and x, and what’s the integral of this at first you might think of well, it’s complicated, this is difficult maybe some type of trig function but you’ll see that this is kind of just reverse chain role.
We have a function here 1 + v2 an expression here we have its derivative setting right there, so the anti-derivative of this and you can actually make a substitution if you like because the u is equal to 1 + v2 then the u is equal to 2v dv and then well the anti derivative is just the natural log of u, or in this case the anti-derivative of this is just the natural log of 1 + v2. We don’t have to read or write an absolute value there because that’s always going to be a positive values, so the natural log of 1 + v2 and that’s how I think about it. Let say if I have an expression and I have its derivative multiplied there then I can just take the anti-derivative of the whole expression and I don’t have to worry about what’s inside of it.
So, if this was a 1/x or 1/u it’s just the natural log of it, so that’s how I knew that this was the anti-derivative. If you don’t believe me use the chain rule and take the derivative of this and you’ll get this. I hope it will make a sense, but anyway that’s the left hand side and then that equals–well, this one is easy that’s the natural log, so the absolute value of x plus we could say plus c but just so that we can simplify a little bit. An arbitrary constant c we can really just write that is a natural log of the absolute value of some constant c right? I mean this is still some arbitrary constant c, so we can rewrite this whole equation as the natural log of 1 + v2 is equal to when you add natural log you can in such you just multiply the two numbers that you’re taking the natural log of.
The natural log of 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 can say that 1 + v2 is equal to cx and now we can un substitute so we know v is equal to y/x, so let’s do that, so we get 1 + y/x2 = cx. I’ll scroll this down a little bit. Let’s see let’s multiply both sides of equation times x2. We can rewrite this as y2/x2, so we multiply both sides times x2 you’ll get x2 + y2 = cx3 and we are 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 x2 + y2 - cx3 = 0,
And this implicitly defined function or curve or whatever you want to call it. It’s the solution to our original homogenous first order differential equation, so there you go. I will see you on the next video. An hour I should introduce something we’re going to start and barking on higher order of differential and these are more useful in some ways easier to do than the homogenous and the exact equations that we’ve been doing so far and I’ll see you in the next video.