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November 25th, 2010, 04:58 AM   #1
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2nd differential equation

Find the general solution of
A(x)y''(x) + A'(x)y'(x) + y(x)/A(x) = 0
where A(x) is a known function and y(x) is unknown.

If any one can help
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November 26th, 2010, 02:08 AM   #2
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Re: 2nd differential equation

hard to say since some of your symbols have been blotted out, but whatever it is, it might be promising to multiply through by A(x) and try to recognize some part of the equation as a derivative of something else: A(x)^2y''(x)+A(x)A'(x)y(x) is almost the derivative of A(x)^2y'(x).
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November 26th, 2010, 03:34 AM   #3
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Re: 2nd differential equation

thanks
but
A(x)^2y''(x)+A(x)A'(x)y(x) is not the derivative of A(x)^2y'(x)
A(x)^2y''(x)+[color=#FF0000]2[/color]A(x)A'(x)y(x) is the derivative of A(x)^2y'(x)
also A(x)y''(x)+A'(x)y(x) is the derivative of A(x)y'(x)
thus (A(x)y'(x))' + y(x)/A(x) = 0 where prime ' denote the differential respect to x
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November 26th, 2010, 06:47 AM   #4
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I'll also use prime to denote differentiation with respect to x.
Let u' = -1/A, so 1/u' = -A and (dy/du)' = (y'/u')' = (-Ay')' = -Ay'' - A'y'.
Hence dy/du = (dy/du)'/u' = -A(dy/du)' = Ay'' + AA'y',
and so you need to solve dy/du + y = 0, which is easy to do.
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November 26th, 2010, 01:01 PM   #5
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Re: 2nd differential equation

Quote:
Originally Posted by rony
thanks
but
[size=150]A(x)^2y''(x)+A(x)A'(x)y(x) is not the derivative of A(x)^2y'(x)
A(x)^2y''(x)+[color=#FF0000]2[/color]A(x)A'(x)y(x) is the derivative of A(x)^2y'(x)
Yes, I'm well aware of that. I thought that perhaps something blotted out in your original equation would make it work. (apparently nothing was blotted out, I guess).
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November 26th, 2010, 01:05 PM   #6
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Re:

Quote:
Originally Posted by skipjack
I'll also use prime to denote differentiation with respect to x.
Let u' = -1/A, so 1/u' = -A and (dy/du)' = (y'/u')' = (-Ay')' = -Ay'' - A'y'.
Hence dy/du = (dy/du)'/u' = -A(dy/du)' = Ay'' + AA'y',
and so you need to solve dy/du + y = 0, which is easy to do.
This might be a petty complaint, but isn't that tacitly making the assumption that A is injective, so that y can be reparametrized in terms of u without losing any information?
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November 26th, 2010, 04:56 PM   #7
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Re: 2nd differential equation

thanks skipjack for your help

thanks martexel
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November 27th, 2010, 02:36 AM   #8
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Quote:
Originally Posted by martexel
. . . isn't that tacitly making the assumption that A is injective?
No. Since y/A and A' appear in the original equation, A is non-zero and differentiable, and 1/A is an injective function of A, which suffice for what is done. At no stage is it assumed that x is a function of u or of A.
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