→ doom8199:回太晚了 OTZ 10/12 18:49
推 kagato:感激不盡 10/12 18:54
※ 引述《kagato (包)》之銘言：
: [xy(x^2 - y^2)^(1/2) + x]y'=y - x^2 *(x^2 - y^2)^(1/2)
: 拜託版上高手了@@
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[xy√(x^2 - y^2) + x]dy = [y - x^2√(x^2 - y^2)]dx
→ x√(x^2 - y^2) * [ydy + xdx] + [xdy - ydx] = 0
→ (x/2)√(x^2 - y^2) *d(x^2 + y^2) + x^2d(y/x) = 0
→ (1/2)√[1 - (y/x)^2] *d(x^2 + y^2) + d(y/x) = 0
1
→ _______________ d(y/x) = (-1/2) d(x^2 + y^2)
√[1 - (y/x)^2]
-1
→ sin (y/x) = -(x^2 + y^2)/2 + C
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◆ From: 140.113.141.151
※ 編輯: doom8199 來自: 140.113.141.151 (10/12 18:48)