Question

大一数学：求微分方程x*dy/dx+x+sin(x+y)=0的通解

我数学很差，麻烦步骤详细点 谢谢

Answer 1

xdy+xdx+sin(x+y)dx=0令z=x+y，则dy=dz-dxxdz+sinzdx=0即1/x
dx=1/sinz
dz两边同时做
不定积分
∫1/x
dx=Ln(x)+C1
C1为常数令t=tan(z/2)
z=2acrtan(t)
即dz=2/(1+t^2)而sinz=2sin(z/2)cos(z/2)/((cos(z/2))^2+(sin(z/2))^2)=
2t
an(z/2)/(1+(tan(z/2))^2)∫1/sinz
dz=∫(1+t^2)/(2t)*2/(1+t^2)
dt=Ln(t)+C2=Ln(tan(z/2))+C2
C2为常数因此，Ln(x)+C1=Ln(tan(z/2))+C2即Ln(x/tan(z/2))=C2-C1
取指数
x/tan(z/2)=e^(C2-C1)
记C=e^(C2-C1)>0
代入z=x+y可得x=C*tan[(x+y)/2]
则y=2acrtan(x/C)-x
C>0即为通解

Answer 2

解：∵(x+y)dy+(x-y)dx=0
==>(1+y/x)dy+(1-y/x)dx=0
设y=xt,则dy=tdx+xdt
∴(x+y)dy+(x-y)dx=0
==>(1+t)(tdx+xdt)+(1-t)dx=0
==>(t??+1)dx+x(t+1)dt=0
==>dx/x+(t+1)/(t??+1)dt=0
==>ln|x|+∫t/(t??+1)dt+∫1/(t??+1)dt=ln|C|
(C是积分常数)
==>ln|x|+1/2∫d(t??+1)/(t??+1)+arctant=ln|C|
==>ln|x|+1/2ln(t??+1)+arctant=ln|C|
==>x√(t??+1)=Ce^(arctant)
==>√(x??+y??)=Ce^(arctant)
(C是积分常数).