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Differential equations
Course: Differential equations > Unit 1
Lesson 8: Homogeneous equationsFirst order homogenous equations
Introduction to first order homogenous equations. Created by Sal Khan.
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Y=(x)Ln|x| +Cx
if C is an arbitrary constant, wouldn't Cx=C?(11 votes)
x is not a constant, so cx cannot equal a constant.(48 votes)
- Does the domain of the answer have to be restricted so that x=/=0 because v=y/x?(14 votes)
Exactly. That's why it's so important to be careful if you change the original terms. Always check against the original.(7 votes)
- Suppose a term in the differential equation is 'xy' , then its degree is 2. Now what is the degree of the term tan(xy) or log(xy) ?(6 votes)
The term "degree" can only be used to qualify polynomials. Expressions that include trigonometric functions are no longer polynomials, and therefore don't have a "degree".
I think you are confusing the term "degree" of a polynomial with a differential equation "linearity". A "linear" differential equation (that has no relation to a "linear" polynomial) is an equation that can be written as:dⁿ dⁿ⁻¹ dⁿ⁻² dy
――y + A₁(x)――――y + A₂(x)――――y + ⋯ + A[n-1](x)―― + A[n](x)y
dx dx dx dx
In your example, sincedy/dx = tan(xy)cannot be rewritten in that form, then it would be a non-linear differential equation (and thus also non-homogenous, as only linear differential equation can be homogenous).(8 votes)
- dy/dx=y^2 this equetion is linear or not(4 votes)
Variables Separable as it just becomes dy/y^2 = dx(3 votes)
Sal is preventing me from hating maths. It's all clear now, but video quality is pretty low, r u guys thinking of redoing it?(6 votes)- I was not clear of what exactly is a homogenous equation or homogenous function. What is it in simple terms ?(3 votes)
- I wouldn't get caught up in the terminology cause it's a bit weird. Really there are 2 types of homogenous functions or 2 definitions. One, that is mostly used, is when the equation is in the form:
ay" + by' + cy = 0
(where a b c and d are functions of some variable, usually t, or constants)
the fact that it equals 0 makes it homogenous. If the equation was
ay" + by' + cy = d
then you'd end up with a result that was the same as the homogenous result PLUS a particular solution.
Now those are mostly what we talk about when we call something homogenous. They are nice because you can usually solve them with a characteristic equation. There's a COMPLETELY DIFFERENT meaning of homogenous also used in DE (just to confuse us I guess), where y' = F(y/x). One example of this type of homogenous function is y' = y/x + 1, which doesn't really equal 0, but this form is called homogenous and you use a change of variables (let v = y/x) to solve it. Again, I wouldn't get caught up in the terminology and what it means to be homogenous, since there are really 2 different definitions. But mostly if something is homogenous, it's a good thing because we have methods to solve it. And if it's not, then we solve it as if it were homogenous ( set it equal to zero instead of whatever it's equal to) and then do some additional calculations for a particular solution based on what it equals.(6 votes)
- At5:00Sal says that the derivative of v with respect to x is equivalent to 1/x. But isn't dv/dx = d(y/x)/dx which is (-y/x^2), not 1/x? I think Sal was thinking of dv/dy.
dv/dy would be d(y/x)/dy equaling 1/x as the y would be derived to 1.(4 votes)
He was not thinking of dv/dy. He differentiated both sides of y=vx with respect to x and solve for dv/dx.
You can also compute dv/dx directly, however dv/dx = d(y/x)dx is not (-y/x^2) like you suggested since y is a function of x. So dv/dx = d(y/x)/dx = [(x(dy/dx)-y]/x^2, by substituting dy/dx with the expression from the original problem, the right hand side simplifies to 1/x.(4 votes)
So how does this relate to the method of exact equations? Is it that when rearranged where you have Mdx+Ndy=0 and then when you check for exactness; (partial of M)/(dy) = (partial of N)/(dx), this chosen function of (x+y)/(x) fails that part so you need to do substitution? I learned the exact method so far and don't really recognize this method yet.(2 votes)- The method that Sal used to solve this particular homogenous differential equation is "separation of variables". But the main focus of the video was to define what a "Homogenous Differential Equation" is, not a particular method to solve them.(4 votes)
Are these the same as first degree equations?(2 votes)- Why does this section not have practice?
I need to practice. Do you have a suggested resource?(2 votes)
5:00Video transcript
I will now introduce you to
the idea of a homogeneous differential equation. Homogeneous is the same word
that we use for milk, when we say that the milk has been--
that all the fat clumps have been spread out. But the application here,
at least I don't see the connection. Homogeneous differential
equation. And even within differential
equations, we'll learn later there's a different type
of homogeneous differential equation. Those are called homogeneous
linear differential equations, but they mean something actually
quite different. But anyway, for this purpose,
I'm going to show you homogeneous differential
equations. And what we're dealing
with are going to be first order equations. What does a homogeneous
differential equation mean? Well, say I had just a regular
first order differential equation that could be
written like this. So dy dx is equal to some
function of x and y. And let's say we try to do this,
and it's not separable, and it's not exact. What we learn is that if it can
be homogeneous, if this is a homogeneous differential
equation, that we can make a variable substitution. And that variable substitution
allows this equation to turn into a separable one. But before I need to show you
that, I need to tell you, what does it mean to be
homogeneous? Well, if I can algebraically
manipulate this right side of this equation, so that I can
actually rewrite it. Instead of a function x and
y, if I could rewrite this differential equation so that
dy dx is equal to some function, let's call that G,
or we'll call it capital F. If I can rewrite it
algebraically, so it's a function of y divided by x. Then I can make a variable
substitution that makes it separable. So right now, it seems
all confusing. Let me show you an example. And I'll just show you the
examples, show you some items, and then we'll just do
the substitutions. So let's say that my
differential equation is the derivative of y with respect
to x is equal to x plus y over x. And you can, if you'd like,
you can try to make this a separable, but it's not
that trivial to solve. Or at least, I'm looking at an
inspection, and it doesn't seem that trivial to solve. And as we see right here,
we have the derivative. It's equal to some function
of x and y. And my question is to you, can
I just algebraically rewrite this so it becomes a function
of y over x? Well, sure, if we just divide
both of these top terms by x. This is the same thing as
x over x plus y over x. This equation is the same
thing as dy over dx is equal to this. Which is same thing as
rewriting this whole equation-- I'm going to switch
colors arbitrarily-- as this, dy over dx is equal to x divided
by x is equal to 1, if we assume x doesn't equal 0. Plus y over x. So you're probably wondering
what did I mean by a function of y over x? Well, you can see it here. When I just algebraically
manipulated this equation, I got 1 plus y over x. So if I said that y over x is
equal to some third variable, this is just a function of
that third variable. And actually, I'm going
to do that right now. So let's make a substitution
for y over x. Let's say that v-- and I'll do v
in a different color-- let's say that v is equal
to y over x. Or another way, if you just
multiply both sides by x, you could write that y
is equal to xv. And we're going to substitute v
for y over x, but we're also going to have to substitute
dy over dx. So let's figure out what
that is in terms of the derivatives of v. So the derivative of y with
respect to x is equal to-- what's the derivative of
this with respect to x? Well, if we assume that v is
also a function of x, then we're just going to use
the product rule. So the derivative of x is 1
times v plus x times the derivative of v with
respect to x. And now, we can substitute this
and this back into this equation, and we get--
so dy over dx, that is equal to this. So we get v plus x dv dx, the
derivative of v with respect to x, is equal to-- that's just
the left hand side-- it's equal to 1 plus y over x. But we're making this
substitution that v is equal to y over x. So we'll do 1 plus v. And now, this should be pretty
straightforward. So let's see, we can subtract
v from both sides of this equation. And then what do we have left? We have x dv dx is equal to 1. Let's divide both sides by x. And we get the derivative of
v with respect to x is equal to 1 over x. It should maybe start becoming
a little bit clearer what the solution here is, but let's
just keep going forward. So if we multiply both sides by
dx, we get dv is equal to 1 over x times dx. Now, we can take the
antiderivative of both sides, integrate both sides. And we're left with v is equal
to the natural log of the absolute value of x plus c. And we are kind of done, but it
would be nice to get this solution in terms of just y and
x, and not have this third variable v here. Because our original problem was
just in terms of y and x. So let's do that. What was v? We made the substitution that
v is equal to y over x. So let's reverse substitute it
now, or unsubstitute it. So we get y over x is equal to
the natural log of x plus c, some constant. Multiply both sides times x. And you get y is equal to
x times the natural log of x plus c. And we're done. We solved that seemingly
inseparable differential equation by recognizing that it
was homogeneous, and making that variable substitution
v is equal to y over x. That turned it into a separable equation in terms of v. And then we solved it. And then we unsubstituted
it back. And we got the solution to the
differential equation. You can verify it for yourself,
that y is equal to the x natural log of the
absolute value of x plus c. Oh, actually, I made
a mistake. y over x is equal to the natural
log of x plus c. If I multiply both sides of this
equation times x, what's the solution? It's not just x natural
log of x. I have to multiply this
times x, too, right? Distributive property-- that
was an amateur mistake. So the correct solution is y is
equal to x natural log of the absolute value of
x plus x times c. And if you want to figure out
c, I would have to give you some initial conditions. And then you could
solve for c. And that would be the particular
solution, then, for this differential equation. In the next video, I'll just
do a couple more of these problems. I'll see you then.