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December 31st, 2013, 03:02 AM   #1
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Integration

Find the integration of the following equation:
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December 31st, 2013, 02:28 PM   #2
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Re: Integration

BB'' + (B')^2(1-a)=0, a=9k/2

(BB')' = a(B')^2

I'll let you try from here.
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December 31st, 2013, 02:31 PM   #3
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Re: Integration

Are we to assume k is an unknown constant?
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December 31st, 2013, 03:36 PM   #4
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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.
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