(1+y)−(1+x)dx−dy=−x−yd(x−y)
Also, since d2=0,1+y+1+xdx+dy=x+y+2dx+dy+d2=x+y+2d(x+y+2)
Each of the fraction of the given DE is equal to
−x−yd(x−y) and x+y+2d(x+y+2)
Solving
−x−yd(x−y)=zdz
ln(x−y)−1=lnz+c1c1=z(x−y)1
Also, solving
x+y+2d(x+y+2)=zdz
ln(x+y+2)=lnz+c2c2=zx+y+2
So the solution is F(z(x−y)1,zx+y+2)=0.
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