求微分方程y✀+sin[(x+y)⼀2]=sin[(x-y)⼀2]通解

2025-02-27 08:14:31
推荐回答(2个)
回答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)

回答2:

y'+sin[(x+y)/2]=sin[(x-y)/2]
y'+sinx/2cosy/2+cosx/2siny/2=sinx/2cosy/2-cosx/2siny/2
y'=-2cosx/2siny/2
dy/siny/2=-2cosx/2dx
左右积分得到
1/2*ln|tgy/4|=-4sinx/2 + C
....