求微分方程y'+sin[(x+y)/2]=sin[(x-y)/2]通解

(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,...