求下列一阶线性微分方程的解(1)y'=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'=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'=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'=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'=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'=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'=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'=1/(x+siny)
(2)(x-siny)dy+tanydx=0,y(1)=π/6
(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=[C‘(y)siny-C(y)cosy]/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=[C-cos(2y)/4]/siny
∴dx/dy+xcosy/siny=cosy的通解是x=[C-cos(2y)/4]/siny(C是积分常数)
∵y(1)=π/6
∴(C-1/8)/(1/2)=1==>C=5/8
∴x=[5-2cos(2y)]/(8siny)
故(x-siny)dy+tanydx=0满足初始条件y(1)=π/6的特解是x=[5-2cos(2y)]/(8siny).