作业帮求微分方程y'+sin[(x+y)/2]=sin[(x-y)/2]通解答案为当siny/2≠0时，通解为㏑|tany/4|=C-2sinx/2当siny/2=0时，特解y=2kπ
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 20:38:32
![求微分方程y'+sin[(x+y)/2]=sin[(x-y)/2]通解答案为当siny/2≠0时，通解为㏑|tany/4|=C-2sinx/2当siny/2=0时，特解y=2kπ](/uploads/image/z/5183262-54-2.jpg?t=%E6%B1%82%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8By%27%2Bsin%5B%28x%2By%29%2F2%5D%3Dsin%5B%28x-y%29%2F2%5D%E9%80%9A%E8%A7%A3%E7%AD%94%E6%A1%88%E4%B8%BA%E5%BD%93siny%2F2%E2%89%A00%E6%97%B6%EF%BC%8C%E9%80%9A%E8%A7%A3%E4%B8%BA%E3%8F%91%7Ctany%2F4%7C%3DC-2sinx%2F2%E5%BD%93siny%2F2%3D0%E6%97%B6%EF%BC%8C%E7%89%B9%E8%A7%A3y%3D2k%CF%80)
求微分方程y'+sin[(x+y)/2]=sin[(x-y)/2]通解答案为当siny/2≠0时，通解为㏑|tany/4|=C-2sinx/2当siny/2=0时，特解y=2kπ
求微分方程y'+sin[(x+y)/2]=sin[(x-y)/2]通解
答案为当siny/2≠0时，通解为㏑|tany/4|=C-2sinx/2
当siny/2=0时，特解y=2kπ
求微分方程y'+sin[(x+y)/2]=sin[(x-y)/2]通解答案为当siny/2≠0时，通解为㏑|tany/4|=C-2sinx/2当siny/2=0时，特解y=2kπ
(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)
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
....