Calculus

Start by telling us exactly what you mean by "initial value problem."

If I could tell you that I probably wouldn''t be here.

=)

Perhaps you can start by calling it "The Calculus" and not by confusing us with notation that's not even vaguely explained by the person asking (you). And the right end of your equation can mean two different things. Try using the


tags to distinguish superscripts. Is that a multiply, function name or part of the exponent?

[edited by - 63616c68h on March 17, 2003 9:57:10 PM]

By initial-value problem, I''m assuming you mean some simple differential equation:

y'' = f(y) (or y'''', or y'''''', or whatever)
y(0) = 1 (in this case, or at any other value y(x), it should give you the same result)

So to come up with this kind of formula, differentiate y(x) until you get something that is a function of y(x) and that''s your f(y)

If it''s not supposed to be a differential equation then I don''t know what it could be, and you should ignore this post.

Tom

63616C68h: aren''t you being a little anal? What kind of notation were you taught? To me it looks like cos(x) plus the integral on [0,x] of a Gaussian: e to the negative t squared. I don''t see what else it could mean.

And what the hell is "The Calculus"?

Tom

ParadigmShift

Yes, it is something to do with differential equations. But I''m afraid your explination still leaves me a little baffeled.

And yes, you did read the equation right.



Thanks for trying.

Well, I found this.

dy/dx = -sin x + e^(-x^2), y(0) = 1

don't ask me how i got that, because it's probably wrong.

Anyway, from what I remember in calculus 2, the initial value problem was using something like above where you're given the derivative of some function y and you have to the single function that matches it, given an initial value like above.

So, when you take the integral of -sin x, you get cos x. When you take the integral of e^(-x²), you get the integral of it (awful explanation, i know). the reason y(0) = 1 is because cos(0) is 1, and any integral from a to a is 0.

Of course, none of this may even be right...

[edited by - masonium on March 17, 2003 12:26:35 AM]

[edited by - masonium on March 17, 2003 12:31:30 AM]

HINT: Think Fundamental Theorem of Calculus on the second term of the equation (the integral). Then find the IVP from there.

quote:

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Original post by platypusman
HINT: Think Fundamental Theorem of Calculus on the second term of the equation (the integral). Then find the IVP from there.

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That''s the erf function, if you come up with a primitive for that, you may or may not get rich.