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**Unformatted text preview: **( 2 u + v ) du + ( u + v ) dv = 0 dv du = 2 u v u + v = 2 ( v/u ) 1 + ( v/u ) . With z = v/u , we have v = z + uz , and so z + u dz du = 2 z 1 + z u dz du = 2 z 1 + z z = z 2 2 z + 2 1 + z z + 1 z 2 + 2 z 2 dz = du u Z 1 + z z 2 + 2 z 2 dz = Z du u 1 2 ln z 2 + 2 z 2 = ln | u | + C 1 z 2 + 2 z 2 ) u 2 = C 2 . Back substitution, z = v/u = ( y 3) / ( x 1), yields v 2 + 2 uv 2 u 2 = C 2 ( y 3) 2 + 2( x 1)( y 3) 2( x 1) 2 = C 2 y 2 8 y 2 x 2 2 x + 2 xy = C. 95...

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