[理工] [工數] 1階 微分......求解...!!
y^2
xy' = ── + y
x
dy y^2
x ── = ── + y
dx x
x(xdy - ydx) -y^2dx = 0
x
-x d(──) - dx = 0
y
x
── + lnx = c
y
請問這樣算....答案對嗎??
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03/10 21:12, , 1F
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03/10 21:25, 5F
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03/10 22:12, , 6F
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03/10 22:21, , 7F
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03/10 23:07, 8F