y'=cos(x-y)-cos(x+y)的通解,求详解.

Question

y'=cos(x-y)-cos(x+y)的通解,求详解.

Answer 1

∵y'=cos(x-y)-cos(x+y)
==>y'=(cosxcosy+sinxsiny)-(cosxcosy-sinxsiny) (应用余弦和差角公式)
==>y'=2sinxsiny
==>dy/siny=sinxdx
==>∫[1/(cosy-1)-1/(cosy+1)]d(cosy)=2∫sinxdx
==>ln│(cosy-1)/(cosy+1)│=-2cosx+ln│C│ (C是积分常数)
==>(cosy-1)/(cosy+1)=Ce^(-2cosx)
==>cosy=[1+Ce^(-2cosx)]/[1-Ce^(-2cosx)]
∴原方程的通解是cosy=[1+Ce^(-2cosx)]/[1-Ce^(-2cosx)] (C是积分常数).