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December 31st, 2013, 02:02 AM   #1
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Integration

Find the integration of the following equation:
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 December 31st, 2013, 01:28 PM #2 Global Moderator   Joined: May 2007 Posts: 6,581 Thanks: 610 Re: Integration BB'' + (B')^2(1-a)=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: 19,505 Thanks: 1740 If k is constant and a is not 2, let y = B^(2-a), then y' = (2-a)B^(1-a)B', and so (2-a)BB' = (B^a)y'. Differentiating again and using (BB')' = a(B')² and y' = (2-a)B^(1-a)B' gives a(2-a)(B')² = (B^a)y'' + aB^(a-1)(B')(y') = (B^a)y'' + a(2-a)(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(1-a) = 0 implies B''/B' = B'/B, so ln|B'| = ln|cB|, where c is a non-zero 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 non-zero, so that B' is non-zero.

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