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Differential equations
Course: Differential equations > Unit 1
Lesson 8: Homogeneous equationsFirst order homogeneous equations 2
Another example of using substitution to solve a first order homogeneous differential equations. Created by Sal Khan.
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At7:07how come we can just drop the absolute value function from the cx term?(25 votes)
I'm sure Sal just forgot to put an absolute value here. We can not drop it. The reason is that C is an ARBITRARY constant. Arbitrary means that it can be ANYTHING and it should NOT depend on whether x is positive or negative. Let me explain. If we drop the absolute value then the choice of C, positive or negative, puts restrictions on a domain of the function. If C>0 then x must be >0 too, if C<0 then x must be <0. It's a huge problem. Doing so we voluntary deprive yourself from a half of solutions! But if we keep the absolute value then sign of x does not depend on choice of C. Here is an example: suppose a pair (a,b) is a solution to our diff. equation. If we save the absolute value, then pair (-a,b) is also solution, but if we drop the absolute value it's not.(8 votes)
I have been struggling with understanding exactly what we are doing in these videos on diff equations. It si like I see how to do it, but I'm bot sure what I am doing. But is it correct to say that we are presented with some expression containing the derivative of a function, and the solution is an expression with no derivative, only the functions themselves? So once I have reached an expression without any dx/dy, y' etc, I am done?(10 votes)
Hey Tommy, you have it right.
Heres an analogy that may be helpful. In algebra we have algebraic equations, e.g. 2x+5=-1 , the solution of which is a number. In this case x=-3.
In differential calculus we have differential equations, e.g., y'(x)=y(x), the solution of which is a function.
The ultimate goal is to write that function in terms of simple operations (+,-,*,/,sin,cos,log,...).
Sometimes there is no solution. Sometimes there is a solution but it can't be represented in terms of simple operations. In an introductory class you will only get a glimpse of the universe of differential equations.(11 votes)
Why don't we solve for Y to complete the problem? I understand we can easily do that and I also understand the final result of this video, but why do we leave it in the form someFunction(x,y) = 0 rather than solving Y explicitly?(4 votes)
It's not as easy to "solve for y" as you may think. The final answer in this video (x^2 + y^2 − cx^3 = 0) is expressed as an implicit function of y. Clearly, we have y^2 = cx^3 − x^2, which is still valid, but not quite "solved for y." You might be tempted to say that y = sqrt(cx^3 − x^2), but that's only the positive square root, so you're losing half the values!
To demonstrate, I plotted both the explicit form and the not-quite-correct explicit form (https://i.imgur.com/y50vYNL.png). You can see that the curves are the same, but that the explicit form only has the positive values.
Now, in this example, you could say (hand-wavily) that y = ± sqrt(cx^3 − x^2). This isn't really a function, because there are two y-values for each x-value in the domain. On the other hand, you could say that y_1 = sqrt(cx^3 − x^2) and y_2 = −sqrt(cx^3 − x^2); together, these two functions represent the entire solution set.
In general, the implicit function theorem (https://en.wikipedia.org/wiki/Implicit_function_theorem) states (loosely speaking) that any implicit function (i.e., f(x,y) = 0) can be expressed as the "combination" of some set of functions {y_1(x), y_2(x), …}. In this case, we only needed two. In other cases, we might need more. For example, the implicit function sin(y) = 0 is true for y = 0, 2π, −2π, 4π, −4π, …, so we would need infinitely many explicit functions: {y_1 = 0, y_2 = 2π, y_3 = −2π, y_4 = 4π, y_5 = −4π, …}.
Finally, for some applications, you don't need a (set of) explicit function(s). For example, you may just want to test whether a given value is on the curve. In this case, you could easily just evaluate f(x, y) for a given point and test whether it's zero.
(Bonus: imagine trying to come up with the explicit functions for this graph! https://i.imgur.com/G20kEbR.png)(8 votes)
At6:56couldn't he have taken everything as a power of e. to cancel the natural log. Or would it have become e^(ln(x) + C) which is the same as e^(ln(x))*e^c which is cx.
So it would be the same solution, so, nvm. Just figured that in the fly. Was my logic correct?(5 votes)- Your logic is correct, I regualary use the approach you suggested in my Differential Equations class and the teachers does it too. I think Sal just took a different approach(5 votes)
- Do we always substitute v = y/x and write the function in terms of y/x?(3 votes)
If we have a differential equation of this form, then this is the first thing you should try (if no other obvious and simpler method is apparent).(4 votes)
where is the homogeneous equations practice ??(4 votes)
i'm having trouble defining the proper method to solve this eq: y'-1=e^(x+2y)
any help would be appreciated(3 votes)
Mostafa, check out example #4 of this page: http://tutorial.math.lamar.edu/Classes/DE/Substitutions.aspx. It's very similar to the problem you are working on.(3 votes)
at3:50when putting the xs and vs together why is the equation =1 [2xv v'] / [1+v^2]=1 <<<< where does this 1 come frome if you are just dividing the right side?(3 votes)- define an integrating factor(1 vote)
- An integrating factor is a function you multiply by to make the integral obvious. It accomplishes this by making one side of the equation look like the result of using the product rule.(3 votes)
at7:45, why solve it like that? Shouldn't the solution to this d.e. be y=x* sqrt(cx-1) ?(2 votes)
7:07
Video transcript
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.