y'=cos(x-y)-cos(x+y)的通解,求详解.
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y'=cos(x-y)-cos(x+y)的通解,求详解.
y'=cos(x-y)-cos(x+y)的通解,求详解.
y'=cos(x-y)-cos(x+y)的通解,求详解.
∵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).
y'=(cosxcosy+sinxsiny)-(cosxcosy-sinxsiny)=2sinxsiny
y=-2sinxcosy
:y' +y= cos x y=
cos(x)看到这个 cos(x)-cos(y)=?
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