Solving an exact differential equation

[etotheipi]

Homework Statement

Solve $4xy + 1 + (2x^2 + \cos{(y)})y' = 0$

Relevant Equations

N/A

I let $M = 4xy + 1$ and $N = 2x^2 + \cos{(y)}$. Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact and we have $$\frac{\partial f(x,y)}{\partial x} = 4xy + 1$$ From inspection, you can tell this has to lead to $$f(x,y) = 2x^2 y + x + h(y)$$ and we could then derive it partially wrt. $y$ and find out what $h(y)$ is to get the solution. My question is how do we do the integration "formally"; is there such a thing as partial integration? $$\int \partial f(x,y) = \int (4xy + 1) \partial x$$ I say this because if instead we write $$\int d f(x,y) = \int (4xy + 1) dx$$ then we cannot evaluate the RHS since $y = y(x)$. Thank you!

 

[PeroK]

In one sense all integration is by inspection or from first principles. There's no process that can be followed, as there is for differentiation. In general, integration is about looking for functions that you already know are the derivative of something. If you have something like:
$$\frac{\partial f(x,y)}{\partial x} = 4xy + 1$$
Then, you could define $f_0(x) = f(x, y_0) = 4xy_0 + 1$, where $y_0$ is fixed. Then:
$$\int f_0(x) dx = 2x^2y_0 + x + C_0 = F_0(x)$$
You could proceed for any $y$ and the process is the same:
$$F_y(x) = 2x^2y + x + C_y$$
For every fixed $y$ (or $y_0$ - it amounts to the same thing), we have a function $F_y(x)$. But, we normally write $F_y(x) = F(x, y)$. And, instead of $C_y$, we write $C(y)$ or $h(y)$.

Note, therefore, that another way to look at a function of two variables is to identify it as a class of functions of one variable, with one function for each of the second variable:
$$f(x, y) = f_y(x)$$
Note that when one variable is an integer this is what we tend to do. We write $f_n(x)$ for a sequence of functions, instead of $f(x, n)$, which amounts to the same thing. Here $f$ is a function of two variables, one of which is an integer variable.

Note: this perspective will become very important when you get to position and momentum eigenfunctions in QM!

 

[PeroK]

PS we do the same for a sequence. Compare $a_n$ with $a(n)$. Same thing. And $f_x$ with $f(x)$. Also the same thing.

What's the difference between an index and a function variable?

 

[etotheipi]

Thanks again, your insight is really valuable!

Note, therefore, that another way to look at a function of two variables is to identify it as a class of functions of one variable, with one function for each of the second variable:
$$f(x, y) = f_y(x)$$

I'm conscious that the notation $f_y$ etc. is also used to denote partial derivatives, though I think the usages are different enough to not worry about it!

PS we do the same for a sequence. Compare $a_n$ with $a(n)$. Same thing. And $f_x$ with $f(x)$. Also the same thing.

What's the difference between an index and a function variable?

I think what you're getting at is that when we write something like $f_n (x)$ it's usually in the context of considering that function with an arbitrary $n$, doing all of the work, and then at the end seeing what happens if you pick certain values of $n$. I think like you said $f_n (x)$ is better described as a class of functions each with different values of $n$.

For every fixed $y$ (or $y_0$ - it amounts to the same thing), we have a function $F_y(x)$. But, we normally write $F_y(x) = F(x, y)$.

I think that's the key part, though the distinction still seems a little arbitrary. I think I need to do a few more questions to get the knack for it.

Note: this perspective will become very important when you get to position and momentum eigenfunctions in QM!

I didn't know these two things were related, but then again I also don't know what an eigenfunction is yet - so I'll take your word for it!

 

[PeroK]

I'm conscious that the notation $f_y$ etc. is also used to denote partial derivatives, though I think the usages are different enough to not worry about it!

I think that's the key part, though the distinction still seems a little arbitrary. I think I need to do a few more questions to get the knack for it.

Yes, I didn't mean $f_y(x)$ as a partial derivative. But, as an alternative notation for $f(x, y)$.

One way to look at partial differentiation is as follows. Assume a function $f(x, y)$. We can look at $f$ as a class of functions of a real variable ($x$, say) indexed by a second real variable ($y$, say). Then we do ordinary differentiation on each of these functions. I.e. for each function $f_y(x)$, have $f'_y(x)$, the ordinary derivative with respect to $x$.

This gives a new class of functions of $x$, indexed by $y$. We call this class of functions the partial derivative of $f$ with respect to $x$. And, we write:
$$\frac{\partial f}{\partial x}(x, y) = f'_y(x)$$
The point of this is not to obscure a simple matter, but to help resolve conceptual difficulties with derivatives and integrals involving more than one variable. In your original problem, you really just wanted to do ordinary integration on a class of single-variable functions. There really was nothing but ordinary integration going on there. Nothing new, really.

And, for me anyway, the analogy with sequences (where the second variable is discrete) is helpful. No one has conceptual difficulties with differentiating a sequence of functions: $f'_n(x)$. Noticing that you can extend this concept to multi-real-variable functions has clarified a number of things for me - like the example from QM.

 

[Physics lover]

There is a much simpler way of solving this problem.

 

[etotheipi]

One way to look at partial differentiation is as follows. Assume a function $f(x, y)$. We can look at $f$ as a class of functions of a real variable ($x$, say) indexed by a second real variable ($y$, say). Then we do ordinary differentiation on each of these functions. I.e. for each function $f_y(x)$, have $f'_y(x)$, the ordinary derivative with respect to $x$.

This gives a new class of functions of $x$, indexed by $y$. We call this class of functions the partial derivative of $f$ with respect to $x$. And, we write:
$$\frac{\partial f}{\partial x}(x, y) = f'_y(x)$$

I think I sort of already do this but hadn't put it together like that. To hold e.g. $y$ constant is to consider a slice running in the $x$ direction, and the derivative wrt $x$ at this $y$ is then the slope of that slice.

The point of this is not to obscure a simple matter, but to help resolve conceptual difficulties with derivatives and integrals involving more than one variable. In your original problem, you really just wanted to do ordinary integration on a class of single-variable functions. There really was nothing but ordinary integration going on there. Nothing new, really.

I think I understand better now, it helps having the new context. The difficulty came in that the result of something like $\int (4xy + 1) dx$ depends on its context, it could be $2x^2 y + x + C(y)$ or undefined depending on how we treat $y$. You would think that any individual unit of maths would have to evaluate to the same thing in any context, but in this case not necessarily.

 

[nrqed]

Homework Statement:: Solve $4xy + 1 + (2x^2 + \cos{(y)})y' = 0$
Relevant Equations:: N/A

I let $M = 4xy + 1$ and $N = 2x^2 + \cos{(y)}$. Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact and we have $$\frac{\partial f(x,y)}{\partial x} = 4xy + 1$$ From inspection, you can tell this has to lead to $$f(x,y) = 2x^2 y + x + h(y)$$ and we could then derive it partially wrt. $y$ and find out what $h(y)$ is to get the solution. My question is how do we do the integration "formally"; is there such a thing as partial integration? $$\int \partial f(x,y) = \int (4xy + 1) \partial x$$

It is simply the usual integral over x with the specification that $y$ is a constant. So if you want, it is
$\int (4xy+1)\bigr|_{y=ct}\, dx =2 x^2 y + x + h(y)$.

 

[etotheipi]

In your original problem, you really just wanted to do ordinary integration on a class of single-variable functions. There really was nothing but ordinary integration going on there. Nothing new, really.

Yep, I stand corrected, you're right - it's just normal integration. $$\frac{\partial f(x,y)}{\partial x} = 4xy + 1$$ $$\int \frac{\partial f(x,y)}{\partial x} dx = \int (4xy + 1) dx$$ With the additional stipulation that $x$ and $y$ are independent here since we're treating a class of functions of fixed arbitrary $y$ (e.g. $4xy + 1 = f_{y}(x)$).

Strangely though, $y$ and $x$ are no longer independent when we finish the question and solve for $y = y(x)$. It's still a bit odd but I don't think it poses a problem. When the partial derivatives are involved its implied that the two are independent, whilst the final solution relies on them being dependent. So the two cases can sort of be treated separately it seems.

Makes my head hurt... but thinking of them as classes of functions like the analogue of sequences makes it easier to understand!

 

[etotheipi]

Just pinning this insights article for future reference: https://www.physicsforums.com/insights/how-to-find-potential-functions-a-10-minute-introduction/

The concept of an anti-partial derivative is mentioned and the notation $\int \dots \partial x$ is used.

 

[MidgetDwarf]

I think reading the relevant definitions and the notation used , the integrand and its "neighbors", would clarify some of the issues you have. I think its more notational on your part then anything. Any modern calculus book has this.