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

Find the integration of the following equation:
Attached Images int.JPG (4.4 KB, 94 views) December 31st, 2013, 01:28 PM #2 Global Moderator   Joined: May 2007 Posts: 6,705 Thanks: 670 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: 20,375 Thanks: 2010 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')� and y' = (2-a)B^(1-a)B' gives a(2-a)(B')� = (B^a)y'' + aB^(a-1)(B')(y') = (B^a)y'' + a(2-a)(B')�, 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')�(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. Tags integration Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post rose Calculus 8 October 25th, 2017 01:08 AM LousInMaths Calculus 10 September 27th, 2012 02:26 AM gelatine1 Calculus 5 September 26th, 2012 06:38 PM fantom2012 Calculus 3 June 11th, 2012 05:49 PM rose Calculus 7 April 11th, 2010 12:51 AM

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