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10、y'=sin(x/2 -y/2) - sin(x/2+y/2)
=sin(x/2)cos(y/2)-cos(x/2)sin(y/2) -sin(x/2)cos(y/2)-cos(x/2)sin(y/2)
=-2cos(x/2)sin(y/2)
dy/sin(y/2) =-2cos(x/2) dx
dy/[2sin(y/4)cos(y/4)] =-2cos(x/2)dx
dy/[2tan(y/4)cos²(y/4)]=-2cos(x/2)dx
d(y/4)/[tan(y/4)cos²(y/4)]=-cos(x/2)dx
d[tan(y/4)]/ tan(y/4)=-2cos(x/2)d(x/2)
ln|tan(y/4)|=-2sin(x/2)+C
tan(y/4)=C e^[-2sin(x/2)]
3、(lny)^3 dy /y =dx/x²
(lny)^3 d(lny)=dx/x²
1/4 (lny)^4 =-1/x +c1
(lny)^4 =-4/x +C
将x=2，y=1代入，得0=-2+C，C=2
故所求特解为(lny)^4 =-4/x +2
=sin(x/2)cos(y/2)-cos(x/2)sin(y/2) -sin(x/2)cos(y/2)-cos(x/2)sin(y/2)
=-2cos(x/2)sin(y/2)
dy/sin(y/2) =-2cos(x/2) dx
dy/[2sin(y/4)cos(y/4)] =-2cos(x/2)dx
dy/[2tan(y/4)cos²(y/4)]=-2cos(x/2)dx
d(y/4)/[tan(y/4)cos²(y/4)]=-cos(x/2)dx
d[tan(y/4)]/ tan(y/4)=-2cos(x/2)d(x/2)
ln|tan(y/4)|=-2sin(x/2)+C
tan(y/4)=C e^[-2sin(x/2)]
3、(lny)^3 dy /y =dx/x²
(lny)^3 d(lny)=dx/x²
1/4 (lny)^4 =-1/x +c1
(lny)^4 =-4/x +C
将x=2，y=1代入，得0=-2+C，C=2
故所求特解为(lny)^4 =-4/x +2
追问
抱歉我没说清楚 第三题是求 满足初值条件的特解
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