求下列一阶线性微分方程的解（1）y#39;=1/(x+siny)（2）(x-siny)dy+tanydx=0,y(1)=π/6求下列一阶线性微分方程的解（1）y#39;=1/(x+siny)（2）(x-siny)dy+tanydx=0,y(1)=π/6
<p>问题：求下列一阶线性微分方程的解（1）y#39;=1/(x+siny)（2）(x-siny)dy+tanydx=0,y(1)=π/6求下列一阶线性微分方程的解（1）y#39;=1/(x+siny)（2）(x-siny)dy+tanydx=0,y(1)=π/6<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">韩泽生的回答：<div class="content-b">网友采纳 （1）y'=1/(x+siny)==>dx/dy=x+siny 先求dx/dy=x的通解 ∵dx/dy=x==>dx/x=dy ==>ln│x│=y+ln│C│(C是积分常数) ==>x=Ce^y ∴dx/dy=x的通解是x=Ce^y 于是,设dx/dy=x+siny的通解为x=C(y)e^y(C(y)是关于y的函数) ∵dx/dy=C'(y)e^y+C(y)e^y 代入得C'(y)e^y+C(y)e^y=C(y)e^y+siny ==>C'(y)e^y=siny ==>C'(y)=siny*e^(-y) ∴C(y)=∫siny*e^(-y)dy =-e^(-y)(siny+cosy)/2+C(应用分部积分法,C是积分常数) x=[-e^(-y)(siny+cosy)/2+C]e^y =Ce^y-(siny+cosy)/2 故原微分方程的通解是x=Ce^y-(siny+cosy)/2（C是积分常数）. （2）(x-siny)dy+tanydx=0==>(x-siny)dy+sinydx/cosy=0 ==>dx/dy+xcosy/siny=cosy 先求dx/dy+xcosy/siny=0的通解 ∵dx/dy+xcosy/siny=0==>dx/x=-cosydy/siny ==>dx/x=-d(siny)/siny ==>ln│x│=-ln│siny│+ln│C│(C是积分常数) ==>x=C/siny ∴dx/dy+xcosy/siny=0的通解是x=C/siny 于是,设dx/dy+xcosy/siny=cosy的通解为x=C(y)/siny(C(y)是关于y的函数) ∵dx/dy=/sin²y 代入得C‘(y)/siny=cosy ==>C‘(y)=sinycosy=sin(2y)/2 ∴C(y)=∫sin(2y)dy/2 =-cos(2y)/4+C(C是积分常数) x=/siny ∴dx/dy+xcosy/siny=cosy的通解是x=/siny(C是积分常数) ∵y(1)=π/6 ∴(C-1/8)/(1/2)=1==>C=5/8 ∴x=/(8siny) 故(x-siny)dy+tanydx=0满足初始条件y(1)=π/6的特解是x=/(8siny).
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