求∫dy/(2sin(y/2))=-∫sin(x/2)dxps:原来的问题是求解微分方程dy/dx+cos(x-y)/2=cos(x+y)/2,求解微分方程dy/dx+cos((x-y)/2)=(cos(x+y)/2),原来那个少打了两个括号
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求∫dy/(2sin(y/2))=-∫sin(x/2)dxps:原来的问题是求解微分方程dy/dx+cos(x-y)/2=cos(x+y)/2,求解微分方程dy/dx+cos((x-y)/2)=(cos(x+y)/2),原来那个少打了两个括号
求∫dy/(2sin(y/2))=-∫sin(x/2)dx
ps:原来的问题是求解微分方程dy/dx+cos(x-y)/2=cos(x+y)/2,
求解微分方程dy/dx+cos((x-y)/2)=(cos(x+y)/2),原来那个少打了两个括号
求∫dy/(2sin(y/2))=-∫sin(x/2)dxps:原来的问题是求解微分方程dy/dx+cos(x-y)/2=cos(x+y)/2,求解微分方程dy/dx+cos((x-y)/2)=(cos(x+y)/2),原来那个少打了两个括号
你只要知道arcsinx和sinx的原函数不就能求解了吗
cos(x+y)/2-cos(x-y)/2=0
所以y=c(c为任意常数)
dy/dx+cos(x-y)/2=cos(x+y)/2
dy/dx=cos(x+y)/2-cos(x-y)/2
=cosx\2cosy\2-sinx\2siny\2-cosx\2cosy\2-sinx\2siny\2
=-2sinx\2siny\2
y' =-2sinx\2siny\2
dy\2siny\2=-sinx\2dx
dy\2siny\2=2cosx\2+C
dy\2siny\2 =2cosx\2+C
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求∫dy/(2sin(y/2))=-∫sin(x/2)dxps:原来的问题是求解微分方程dy/dx+cos(x-y)/2=cos(x+y)/2,求解微分方程dy/dx+cos((x-y)/2)=(cos(x+y)/2),原来那个少打了两个括号
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