86_pdfsam_math 54 differential equation solutions odd

# 86_pdfsam_math 54 differential equation solutions odd - 2...

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Chapter 2 15. This equation is homogeneous because dy dx = x 2 y 2 3 xy = 1 ( y/x ) 2 3( y/x ) . Thus, we substitute u = y/x ( y = xu and so y 0 = u + xu 0 )toget u + x du dx = 1 u 2 3 u x du dx = 1 4 u 2 3 u 3 udu 1 4 u 2 = dx x Z 3 udu 1 4 u 2 = Z dx x ⇒− 3 8 ln ± ± 1 4 u 2 ± ± =ln | x | + C 1 ⇒− 3ln ± ± ± ± 1 4 ² y x ³ 2 ± ± ± ± =8ln | x | + C 2 3ln( x 2 ) 3ln ± ± x 2 4 y 2 ± ± =8ln | x | + C 2 , which, after some algebra, gives ( x 2 4 y 2 ) 3 x 2 = C . 17. With the substitutions z = x + y and dz/dx =1+ dy/dx or dy/dx = dz/dx 1th isequat ion becomes the separable equation dz dx 1= z 1 dz dx = z z 1 / 2 dz = dx 2 z 1 / 2 = x + C. Substituting x + y for z in this solution gives the solution of the original equation
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Unformatted text preview: 2 √ x + y = x + C which, on solving for y , yields y = ´ x 2 + C 2 µ 2 − x. Thus, we have y = ( x + C ) 2 4 − x. 19. The right-hand side of this equation has the form G ( x − y ) with G ( t ) = ( t + 5) 2 . Thus we substitute t = x − y ⇒ y = x − t ⇒ y = 1 − t , 82...
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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 Berkeley.

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