推 nini851216 : 感謝！我還以為我寫錯了 08/18 16:58
※ 引述《nini851216 (解壓縮)》之銘言：
: https://i.imgur.com/nB6oRph.jpg
: 字太醜見諒
: 再來就不會了..
: 請求大大幫忙
: https://i.imgur.com/BDv93Cp.jpg
: 答案是這個
: https://i.imgur.com/L3KLflc.jpg
按照你的作法
p = y'
y p dp/dy + p^2 = 2
=> (1/2)y [dp^2 / dy] = 2 - p^2
=> (-1/2)ln|2 - p^2| = lny + c
=> |2 - p^2| = A/y^2, A > 0
設p^2 > 2
p = dy/dx = √[2 + A/y^2]
=> (1/4) d(2y^2 + A) = √[2y^2 + A]dx
=> (1/2)√[2y^2 + A] = x + C
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