Re: [理工] 一階微分方程

看板Grad-ProbAsk作者Honor1984 (希望願望成真)時間10年前 (2015/07/26 18:38)推噓3(3推 0噓 3→)

留言6則, 4人參與討論串2/2 (看更多)

※ 引述《JLintopPG (天下第一控)》之銘言： : 1. Solve dy/dx = [y+√(y^2-x^2)] / x 令y = vx y' = xv' + v xv' + v = v + √[v^2 - 1] => v' ----------- = 1/x √[v^2 - 1] => ln|v + √[v^2 - 1]| = lnx + c => |(y/x) + √[(y/x)^2 - 1]| = Ax : 2. Solve dy/dx = [2√(xy)-y] / x 令y = vx y' = xv' + v xv' + v = 2√v - v => xv' = 2√v [1 - √v] => -ln[1 - √v] = lnx + c => x - √(xy) = A -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 111.249.203.124 ※ 文章網址: https://www.ptt.cc/bbs/Grad-ProbAsk/M.1437907100.A.3E1.html

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rdchen1124

07/27 14:30, , 1^F

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rdchen1124

07/27 14:30, , 2^F

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rdchen1124

07/27 14:41, , 3^F

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