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 November 25th, 2010, 03:58 AM #1 Newbie   Joined: Nov 2010 Posts: 3 Thanks: 0 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
 November 26th, 2010, 01:08 AM #2 Member   Joined: Nov 2009 Posts: 72 Thanks: 0 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).
 November 26th, 2010, 02:34 AM #3 Newbie   Joined: Nov 2010 Posts: 3 Thanks: 0 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
 November 26th, 2010, 05:47 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,941 Thanks: 2210 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 d²y/du² = (dy/du)'/u' = -A(dy/du)' = A²y'' + AA'y', and so you need to solve d²y/du² + y = 0, which is easy to do.
November 26th, 2010, 12:01 PM   #5
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Re: 2nd differential equation

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 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).

November 26th, 2010, 12:05 PM   #6
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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 d²y/du² = (dy/du)'/u' = -A(dy/du)' = A²y'' + AA'y', and so you need to solve d²y/du² + 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?

 November 26th, 2010, 03:56 PM #7 Newbie   Joined: Nov 2010 Posts: 3 Thanks: 0 Re: 2nd differential equation thanks skipjack for your help thanks martexel
November 27th, 2010, 01:36 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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