December 31st, 2013, 02:02 AM  #1 
Newbie Joined: Dec 2013 Posts: 8 Thanks: 0  Integration
Find the integration of the following equation:

December 31st, 2013, 01:28 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,705 Thanks: 670  Re: Integration
BB'' + (B')^2(1a)=0, a=9k/2 (BB')' = a(B')^2 I'll let you try from here. 
December 31st, 2013, 01:31 PM  #3 
Newbie Joined: Dec 2013 Posts: 3 Thanks: 0  Re: Integration
Are we to assume k is an unknown constant?

December 31st, 2013, 02:36 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,375 Thanks: 2010 
If k is constant and a is not 2, let y = B^(2a), then y' = (2a)B^(1a)B', and so (2a)BB' = (B^a)y'. Differentiating again and using (BB')' = a(B')² and y' = (2a)B^(1a)B' gives a(2a)(B')² = (B^a)y'' + aB^(a1)(B')(y') = (B^a)y'' + a(2a)(B')², so (B^a)y'' = 0. The original equation is undefined for B = 0, so y'' = 0, which can be solved easily. Strictly speaking, the domain of B should be chosen to prevent B = 0. If a = 2, BB'' + (B')²(1a) = 0 implies B''/B' = B'/B, so lnB' = lncB, where c is a nonzero constant. Hence B' = cB, and that's easily solved. The resulting exponential function is conveniently never zero. Choose the (second) constant of integration to be nonzero, so that B' is nonzero. 

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