90_pdfsam_math 54 differential equation solutions odd

# 90_pdfsam_math 54 differential equation solutions odd -...

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Chapter 2 Substituting x +1for u and y +2for v gives the solution to the original equation ( y +2) 2 +2( x +1)( y +2) 3( x +1) 2 = C. 31. In this equation with linear coeﬃcients, we make a substitution x = u + h , y = v + k ,where h and k satisfy ( 2 h k =0 4 h + k =3 ( k =2 h 4 h +2 h =3 k =1 , h =1 / 2 . Thus x = u +1 / 2, y = v +1. As dx = du and dy = dv , substitution yields (2 u v ) du +(4 u + v ) dv =0 du dv = 4 u + v 2 u v = 4( u/v )+1 2( u/v ) 1 z = u v u = vz du dv = z + v dz dv z + v dz dv = 4 z +1 2 z 1 v dz dv = 4 z +1 2 z 1 z = (2 z +1)( z +1) 2 z 1 2 z 1 (2 z +1)( z +1) dz = 1 v dv Z 2 z 1 (2 z +1)( z +1) dz = Z 1 v dv . To Fnd the integral in the left-hand side of the above equation, we use the partial fraction decomposition 2 z 1 (2 z +1)( z +1) = 4 2 z +1 + 3 z +1 . Therefore, the integration yields 2ln | 2 z +1 |
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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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