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**Unformatted text preview: **t = u + h and x = v + k with h and k satisfying ( h + k + 2 = 0 3 h k 6 = 0 h = 1 , k = 3 . As dt = du and dx = dv , the substitution yields ( u + v ) dv + (3 u v ) du = 0 du dv = u + v 3 u v = ( u/v ) + 1 3( u/v ) 1 . With z = u/v , we have u = vz , u = z + vz , and the equation becomes z + v dz dv = z + 1 3 z 1 v dz dv = 3 z 2 + 1 3 z 1 3 z 1 3 z 2 + 1 dz = 1 v dv Z 3 z 1 3 z 2 + 1 dz = Z 1 v dv 87...

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