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November 25th, 2010, 04:58 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, 02:08 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, 03:34 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, 06:47 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 21,105 Thanks: 2324 
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, 01:01 PM  #5  
Member Joined: Nov 2009 Posts: 72 Thanks: 0  Re: 2nd differential equation Quote:
 
November 26th, 2010, 01:05 PM  #6  
Member Joined: Nov 2009 Posts: 72 Thanks: 0  Re: Quote:
 
November 26th, 2010, 04:56 PM  #7 
Newbie Joined: Nov 2010 Posts: 3 Thanks: 0  Re: 2nd differential equation
thanks skipjack for your help thanks martexel 
November 27th, 2010, 02:36 AM  #8  
Global Moderator Joined: Dec 2006 Posts: 21,105 Thanks: 2324  Quote:
 

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