Question

解这个微分方程 cos(x+y)dy=dx

Answer 1

令x+y=z,则dz/dx=1+dy/dx=1+1/cos(x+y)=1+1/cosz=(cosz+1)/cosz

故

cosz/(1+cosz)*dz=dx

[1-1/(1+cosz)]dz=dx

{1-1/[(1+2cos^2 (z/2)-1}dz=dx

[1-1/2*sec^2 (z/2)]dz=dx

两边分别积分得

z-tan(z/2)=x+C

也即

(x+y)-tan[(x+y)/2]-x=C

或

(x+y)-sin(x+y)/[1+cos(x+y)]-x=C