(1-x)dx—(1+y)dy=0
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(1-x)dx—(1+y)dy=0(1-x)dx—(1+y)dy=0(1-x)dx—(1+y)dy=0∵(1-x)dx-(1+y)dy=0==>(1+y)dy=(1-x)dx==>(1+y)²
(1-x)dx—(1+y)dy=0
(1-x)dx—(1+y)dy=0
(1-x)dx—(1+y)dy=0
∵(1-x)dx-(1+y)dy=0
==>(1+y)dy=(1-x)dx
==>(1+y)²=C-(1-x)² (C是任意常数)
∴原方程的通解是(1+y)²=C-(1-x)² .
(1-x)dx—(1+y)dy=0
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