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1. { e^(x+y)-y }dx + {xe^(x+y)+1}dy = 0 2. 2δu/δx-3δu/δy+2u=2x initial condition is u(x,y)=x^2 for the line 2y+x=0 δ是偏微符號 感激不盡了 >"< -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 61.227.182.90