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# First order homogenous equations 2 : Another example of using substitution to solve a first order homogeneous differential equations.

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- Let's do one more homogeneous differential equation, or
- first order homogeneous differential equation, to
- differentiate it from the homogeneous linear
- differential equations we'll do later.
- But anyway, the problem we have here.
- It's the derivative of y with respect to x is equal to--
- that x looks like a y-- 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 2xy.
- So in all of these homogeneous equations-- and obviously, we
- don't know it's homogeneous 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-- let me do it in a different
- color-- 1 over x squared, or x to the negative 2,
- over 1 over x squared.
- We're essentially just multiplying by 1.
- And then, that is equal to what?
- 1 plus 3y squared over x squared divided by 2-- if you
- divide x, divide 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 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 can make the substitution that v is equal
- to y over x, or another way of writing that is that y is
- equal to xv.
- And then, of course, the derivative of y with respect
- to x, or if we take the derivative with respect to x
- of 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 can 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 3v squared, we're making the
- substitution v is equal to y over x.
- All of that over 2v.
- Now, let's see what we can do.
- This is where we just get 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 2v.
- So we'll get 2v squared plus 2xv v prime-- 2v times x, yep,
- that's 2xv v prime-- is equal to 1 plus 3v squared.
- Now let's see, let's subtract 2v squared from
- both sides of this.
- And we will be left with 2xv v prime is equal to 1 plus--
- let's see, we're subtracting 2v squared from both sides.
- So we're just left with a 1 plus v squared here, right?
- 3v squared minus 2v squared is just v squared.
- And let's see, we want it to be separable, so let's put all
- the v's on the left hand side.
- So we get 2xv 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 2v-- and I'll now switch back
- to the other notation.
- Instead of v prime, I'll write dv dx.
- 2v times the derivative of v 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 dx,
- 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 to a different color, so you know I'm working
- on a different column now.
- So multiply both sides by dx.
- I get 2v 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 in terms of v and x.
- And what's the integral of this?
- At first, you might think, oh boy, this is complicated.
- This is difficult, maybe some type of trig function.
- But you'll see that it's kind of just the
- reverse chain rule.
- We have a function here, 1 plus v squared,
- an expression here.
- And we have its derivative sitting right there.
- So the antiderivative of this, and you can make a
- substitution if you like.
- You could say u is equal to 1 plus v squared, then du is
- equal to 2v dv.
- And then, well, you would 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 of 1 plus v squared.
- We don't even have to 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 I hope I didn't confuse you.
- That's how I think about it.
- I say, if I have an 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 will make a little bit more sense.
- But anyway, that's the left hand side, and then that
- equals-- Well, this one's easy.
- That's the natural log, the absolute value of x.
- 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 the natural log of the absolute value of some
- constant c.
- I mean, this is still some arbitrary constant c.
- So we can 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 you're
- taking the natural log of-- the natural log of, we could
- say, the absolute value of cx.
- And so the natural log of this is equal to the
- natural log of this.
- So we could say that 1 plus v squared is equal to cx.
- And now we can unsubstitute it.
- So we know v is equal to y over x, so let's do that.
- So we get 1 plus y over x squared is equal to cx.
- Let me scroll this down a little bit.
- Let's multiply both sides of the equation times x squared.
- We could rewrite this as y squared over x squared.
- So we multiply both sides times x squared, you get x
- squared plus y squared is equal to cx to the third.
- And we're essentially done.
- If we want to put all of the variable terms on left hand
- side, we could say that this is equal to x squared plus y
- squared minus cx to the third is equal to 0.
- And this implicitly defined function, or curve, or however
- 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, and 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.

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