Question

求微分方程y'+sin[(x+y)/2]=sin[(x-y)/2]通解

Answer 1

(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)