Simple DE question

[Frodo Baggins]

A slight generalization of the equation dy/dx = F(x), or dy = F(x)dx is an equation which can be brought into the form:

f(x)dx = g(y)dy


How did they get f(x)dx = g(y)dy??

[Frodo Baggins]

comon fellas, I need this to actually get past this chapter!

Where are the math ninjas?

[Gandalf]

I don't know where that 'g' came from.

[Cjwinnit]

Edit: Ignore me I didn't read it properly. I think i got it tho.....

Um, just for fun, work backwards and see what you get..

g(y)dy = f(x)dx

Integrate both sides:

G(y) + A = F(x) <-- A is const. of integration......

so as dy = F(x)dx then:

dy = (G(y) + A)dx

But in this I don't think F(x) is the indefinite integral of f(x) or the first equation would be

dy/dx F(x) = f(x)

Which in this case isn't necessarily true.

???


Edit: Someone give Esau GD access.....

[SinsFeelNatural]

Where's the math smack master when we need him. :D

[Cjwinnit]

Originally posted by SinsFeelNatural
Where's the math smack master when we need him. :D

He is Doc Logic now :-/

[Frodo Baggins]

This is totally confusing, didn't esau give me a proof? God, I can't rmemember what it was, I think it went something like this

Starting with:

dy = f'(x) dx

Integrating both sides:

g(y) + A = f(x)


Er...where did he go from there?

Oh jeez, it looks like I'm gonna have to bring this up with one of my teachers

[Frodo Baggins]

Hmm..I asked my Data Management math teacher at school, she has no clue....

I think it has to do with the fact that the book was originally printed in 1961?

Anybody familiar with the notation?

[Shadow ÒÓ]

Originally posted by Cjwinnit
Edit: Ignore me I didn't read it properly. I think i got it tho.....

Um, just for fun, work backwards and see what you get..

g(y)dy = f(x)dx

Integrate both sides:

G(y) + A = F(x) <-- A is const. of integration......

so as dy = F(x)dx then:

dy = (G(y) + A)dx

But in this I don't think F(x) is the indefinite integral of f(x) or the first equation would be

dy/dx F(x) = f(x)

Which in this case isn't necessarily true.

???


Edit: Someone give Esau GD access.....

I'd like to remind the members, you are NOT allowed to post in any language but English. Since the mods/admins use English as the native language, there's no way to know if are posting flames/warez/whatever.

Now someone please tell me wtf he said? :rolleyes:

[SinsFeelNatural]

It's an inside joke with us non senior, non mods. :D

[skou]

Shadow, that's Calculus. I know, because I failed Calculus 3 times.:D

steve

[Frodo Baggins]

cjw and others,

I got it. It's actually pretty ackward.

Starting with:

(1) dy/dx = F(x)

which implies
(2) dy = F(x) dx

Now, the book coompletely screwed us up cuz they never created an equation using the two givens 1 and 2. They generalized it.

Create a function H of x AND y as independent variables

dy/dx = H(x, y)
dy = H(x, y) dx

Now suppose H(x,y) = f(x) (1/g(y))
In other words, suppose H(x,y) is factorable into two functions f(x) and 1/g(y) (Just follow me on the 1/g(y), you'll see)

In that case:

dy = f(x) (1/ g(y) ) dx

Which leads to:
f(x) dx = g(y) dy

And that is basically it

Anyways, what is the entire point of this?

Solve:
dy/dx = 3x^2 y

Again, let H(x,y) = 3x^2 y

1/y dy = 3x^2 dx

In this case, g(y) = 1/y, and f(x) = 3x^2

Integ. ( 1/y dy ) = Integ. (3x^2 dx)
ln y = x^3 + C
y = e^(x^3 + C)
y = e^(x^3) e^C
y = Ce^(x^3)

What a stupid book, they could have phrased it better...sheeesh

[Cjwinnit]

Not exactly a great one. All that's done is get the line:

g(y)dy = f(x)dx

Divide both sides by g(y):

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

and replace the phrase f(x)/g(y) with one function H(x,y).

And for implicit differentiation of most functions it still doesn't help, for example:

dy/dx H(x,y) where H(x,y) = x^2 + y^2

Comparitively simple but can't be solved using this equation....

Might be helpful where H(x,y) is seperable but there are nicer ways of doing it but slightly more advanced theories....