93_pdfsam_math 54 differential equation solutions odd

93_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.6 When C = 0, the above formula gives θ 0or y 0, which were lost in separating variables. Also, we lost another solution, u +1 0or y = θ .Thu s ,th ean sw e ri s θy 2 = C ( y + θ ) 2 and y = θ, where C is an arbitrary constant. 41. The right-hand side of (8) from Example 2 of the text can be written as y x 1+( x y +2) 1 = ( x y +2)+1+( x y +2) 1 = G ( x y +2) with G ( v )= v + v 1 +1. With v = x y +2, we have y 0 =1 v 0 , and the equation becomes 1 dv dx = v + v 1 +1 dv dx = v 2 1 v v v 2 1 dv = dx ln | v 2 1 | =2 x + C 1 v 2 1= Ce 2 x ,C 6 =0 . Dividing by v 2 1, we lost constant solutions v = ± 1, which can be obtained by taking C =0 in the above formula. Therefore, a general solution to the given equation is ( x y +2) 2 =
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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