五 43+2c12 +c dx 2t 3 d y dt 3 +c1)2 dx 2t 故(y")2=t2=x-1, 故所求微分方程为()2=x-1. K心五. dt dx dt dy dx dy = t t c t 2 2 3 4 1 4 + = 1 3 3 2 = t + c dt dx t c dt d dx d y ) 3 2 ( 1 3 2 2 + = t t t = = 2 2 2 2 2 故 ( y) = t = x −1, ( ) 1. 2 故所求微分方程为 y  = x −