Question

求微分方程y'+sin[(x+y)/2]=sin[(x-y)/2]通解

Answer 1

解:(1)当y=C时，sin[(x+C)/2]=sin[(x-C)/2]
移项，和差化积有2cos{[(x+C)/2+(x-C)/2]/2}sin{[(x+C)/2-(x-C)/2]/2}=0，即cos(x/2)sin(C/2)=0
要恒成立，只有sin(C/2)=0，即C=2kπ
(k∈Z)
所以此时，y=2kπ
(k∈Z)
(2)当y≠C时，有y'+sin(x/2)cos(y/2)+cos(x/2)sin(y/2)=sin(x/2)cos(y/2)-cos(x/2)sin(y/2)
化简有y'=-2cos(x/2)sin(y/2)
分离变量有dy/sin(y/2)=-2cos(x/2)dx
同步对各自变量积分有(1/2)ln|tan(y/4)|=C-4sin(x/2)，即ln|tan(y/4)|=C-2sin(x/2)
所以此时，tan(y/4)=Ce^[-2sin(x/2)]
(C为任意常数)
综合上述:y=2kπ
(k∈Z)
或
tan(y/4)=Ce^[-2sin(x/2)]
(C为任意常数)