(x-siny)dy+tanydx=0求通解,
<p>问题：(x-siny)dy+tanydx=0求通解,<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">乔月圆的回答：<div class="content-b">网友采纳 ∵(x-siny)dy+tanydx=0 ==>xcosydy+sinydx-sinycosydy=0(等式两端同乘cosy) ==>d(xsiny)-d((siny)^2/2)=0 ==>xsiny-(siny)^2/2=C(C是常数) ==>x=siny/2+C/siny ∴原方程的通解是x=siny/2+C/siny.<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">万雅竞的回答：<div class="content-b">网友采纳 不对啊，答案是C(siny)^2－2xsiny＝1，用z＝siny换元做<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">乔月圆的回答：<div class="content-b">网友采纳 我的答案是正确的，验证如下： ∵x=siny/2+C/siny(C是常数) ==>dx=cosydy/2-Ccosydy/(siny)^2，x-siny=C/siny-siny/2 ∴(x-siny)dy+tanydx=(C/siny-siny/2)dy+tany(cosydy/2-Ccosydy/(siny)^2) =(Cdy/siny-sinydy/2)+(sinydy/2-Cdy/siny) =0 故x=siny/2+C/siny是方程(x-siny)dy+tanydx=0的通解。
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