How do you find the general solution of differential equations?

[Fritz]

I really don't understand the concept of differential equations! I'm confused about how you would go about solving them.

if (x^2).dy/dx = y^2,

integral(y^2)dy = integral(x^2)

1/3y^3 = 1/3x^3 + A

^ What have I done wrong here?


I don't get the whole thing where you separate the variables etc.

Am I right in saying that integral(dy/dx)dx = y?



Moderators, sorry I posted this twice. It was an accident.
Please can someone help me understand all of this? I understand basic differentiation and integration.

[marlon]

I really don't understand the concept of differential equations! I'm confused about how you would go about solving them.

if (x^2).dy/dx = y^2,

integral(y^2)dy = integral(x^2)

1/3y^3 = 1/3x^3 + A

^ What have I done wrong here?


I don't get the whole thing where you separate the variables etc.

Am I right in saying that integral(dy/dx)dx = y?



Moderators, sorry I posted this twice. It was an accident.
Please can someone help me understand all of this? I understand basic differentiation and integration.
Just put all variables of one sort at one side : x²dy/dx = y². This means that (1/y²)dy = (1/x²)dx and just integrate --> -(1/y) + C = -(1/x) + C'

marlon

[devious_]

x^{2}\frac{dy}{dx} = y^{2}
If you rearrange that you'd see that it equals:
\frac{1}{y^2}dy = \frac{1}{x^2}dx
And not:
y^2 dy = x^2 dx

To solve the equation, simply take the integral of both sides.

[Fritz]

So you substitute y for x and vice versa, take the reciprocal of both variables and then integrate?

I don't understand how you got from the first step to the second step (and then to the third)!

[marlon]

So you substitute y for x and vice versa, take the reciprocal of both variables and then integrate?

I don't understand how you got from the first step to the second step (and then to the third)!

well the integral of (1/x²) is -(1/x), that's all...


use the rule integral of x^n = (x^(n+1)/n+1) and n = -2 here...
marlon

[Fritz]

I understand that, but I don't get why you substituted x^2 for 1/y^2 and y^2 for 1/x^2.

It's not integrating of individual terms I'm confused about...

[devious_]

So you substitute y for x and vice versa, take the reciprocal of both variables and then integrate?

I don't understand how you got from the first step to the second step (and then to the third)!
It's simple manipulation:

a\;\frac{b}{c}=d \Rightarrow \frac{1}{d}\;b=\frac{1}{a}\;c

[Fritz]

I thought that dy/dx is not a fraction (and therefore cannot be rearranged)?

[matt grime]

dy/dx is indeed not a fraction, but there are *some* circumstances where it can be treated as one, and this is one of them. We're just simply taking a practical approach to how to solve the equation, that is omitting some steps that always occur in the same way, to clear up the presentation.

What it is is a short hand wayo of saying that if:

f(y)dy/dx = g(x)

after rearranging, then

int f dy = int g dx

[Fritz]

What do f dy and g dx mean? Don't you mean int f(y)dy = int f(x)dx?

I don't get what f and g on their own mean.

[matt grime]

yes f on its own means f(y), but on the right hand side it is int g(x)dx

normally, where there is no room for confusion, it is acceptable, for example, to drop the letter t from h(t) and refer to the function simply as h.

[JonF]

You can treat the dy/dx like a fraction because it essentially is one. It is the ratio of two differential operators. So when you multiply both sides by the differential operator dx, you get cancellation on one side and a dx on the other.

[Fritz]

^ Thank you. That's what I thought!

[matt grime]

dx isn't a differential operator (it doesn't take a function and yield another function, but you don't want to get too technical about that) they are referred to as infinitesimals occasionally.

[JonF]

It’s not? What is it classified as then? It’s some kind of operator right?

[gazzo]

just a differential arn't they ?

depending on whether the indeterminant is a dependent or independent variable

dx = \Delta x
dy = f'(x) dx

:bugeye:

[matt grime]

It’s not? What is it classified as then? It’s some kind of operator right?

Hope this doesn't confuse anyone unnecessarily, but dx is the dual vector to the vector \partial_x, and is called a 1-form. It is an element of the cotangent bundle, but it is not a differential operator. (Most things can be thought of as "operators" I suppose, it's one of those frequently used ambiguous labels). A differential operator looks something like:

\mathcal{L}= \partial_{xy} - \partial_{x} +x^2\partial{_y}

[HallsofIvy]

No, its a "differential", not a differential "operator".

I would be very reluctant to tell anyone "You can treat the dy/dx like a fraction because it essentially is one."

In standard calculus, dy/dx is defined as the limit of the fraction Δy/Δx, NOTas the fraction "dy" over "dx". One can then show that, generally, one can treat dy/dx as if it were a fraction by going back before the limit, using the fact that Δy/Δx is a fraction and then going forward again. Typically, one then defines the "differentials" dy and dx separately in a purely symbolic manner.

Yes, it is possible to define the derivative in terms of "infinitesmals", dx and dy, from the start ("non-standard calculus") but to define "infinitesmals" rigorously requires some very deep logical gymnastics!

[matt grime]

Another caveat, and a demonstrable one, is that although dy/dx behaves as a fraction, second and higher derivatives do not (nor do partial derivatives)

everyone knows that (dy/dx)^{-1}= dx/dy but it is certainly not true that

(d^2y/dx^2)^{-1} = d^2x/dy^2

and it is quite instructive to actually try and work out the proper relationship.

[HallsofIvy]

Another caveat, and a demonstrable one, is that although dy/dx behaves as a fraction, second and higher derivatives do not (nor do partial derivatives)

everyone knows that (dy/dx)^{-1}= dx/dy but it is certainly not true that

(d^2y/dx^2)^{-1} = d^2x/dy^2

and it is quite instructive to actually try and work out the proper relationship.

In fact, that's one of the reasons the superscript numbers are place in different positions in the "numerator and denominator": in order that
1/(d2y/dx2)= dx2/d2y makes no sense!

[matt grime]

You'd be surprised at the number of students I have taught who didn't believe me when I told them this fact, though. Then there's the always amusing:

suppose f(x,y,z)=0 is used to define each variable implicitly as a function of the other two, show that:

\frac{\partial y}{\partial x}\frac{\partial x}{\partial z}\frac{\partial z}{\partial y} = -1

[fourier jr]

geez i think you guys are making this way too complicated. the symbol "dx" is simply another way of saying "a teensie bit of x", and of course can be treated as a fraction. (as far as a novice calculus student is concerned) That guy Silvanus Thompson did it well in his "Calculus Made Easy" book.

[matt grime]

So, what's 1/dx then?

[JonF]

So, what's 1/dx then?
I’ll go with undefined…


And thanks guys. Just getting into linear algebra, thought I knew what I was talking about…

[HallsofIvy]

geez i think you guys are making this way too complicated. the symbol "dx" is simply another way of saying "a teensie bit of x", and of course can be treated as a fraction. (as far as a novice calculus student is concerned) That guy Silvanus Thompson did it well in his "Calculus Made Easy" book.

So I guess, being easy, it doesn't matter if it is wrong?

[arildno]

So I guess, being easy, it doesn't matter if it is wrong?
I think great Isaac (N. that is) made a brilliant analogy when he emphasized that "the fluxion" couldn't be regarded as a fraction between two numbers:

He simply noted that if we have a fractional expression where we let both the numerator and denominator go to infinity, but where that expression still converged to some number, no one would say that the infinity in the numerator was a NUMBER, just as little as someone would say the infinity in the denominator was a NUMBER.

[matt grime]

I’ll go with undefined…


And thanks guys. Just getting into linear algebra, thought I knew what I was talking about…


What's this got to do with linear algebra? you can safely learn linear algebra without any reference to what's going on here. Indeed you shouldn't use any of this in *learning* linear algebra. The stuff I talked about, dual vectors of the tangent bundle, and so on, is safely a graduate level subject in the US, and in most of the UK too.