"(x+y+6)dy=(x-y-2)dx"
"(2+y-x)dx+(x+y+6)dy=0"
Let "M=2+y-x, \\ N=x+y+6"
"M_y'=1=N_x'" , so the given equation is exact.
And there is function "F(x,y):\\ F'_x=M,\\ F'_y=N"
"F'_x=2+y-x,\\\\\n F=\\int (2+y-x)dx=2x+yx-x^2\/2+h(y)"
"F'_y=(2x+yx-x^2\/2+h(y))'_y=x+h'_y"
"x+h'_y=x+y+6,\\ h'_y=y+6,\\\\\n h(y)=\\int (y+6)dy=y^2\/2+6y"
"F(x,y)=2x+xy-x^2\/2+y^2\/2+6y=C,\\ C\\ is\\ constant"
"y^2+2y(6+x)-x^2+4x-2C=0"
Using the quadratic formula
"y=-6-x\\pm \\sqrt{(6+x)^2-(-x^2+4x-2C)}=\\\\\n-6-x\\pm\\sqrt{36+12x+x^2+x^2-4x+2C}=\\\\\n-6-x\\pm\\sqrt{2x^2+8x+36+2C}"
Since "x+y+6" is in the denominator
"x+y+6\\not= 0, \\ 2x^2+8x+36+2C\\not=0.\\\\\nx^2+4x+18+C\\not=0\\\\\n(x+2)^2+14+C\\not=0\\\\\nx\\not=\\pm\\sqrt{-14-C}-2"
Answer: "y=-6-x\\pm\\sqrt{2x^2+8x+36+2C}" ,"x\\not=\\pm\\sqrt{-14-C}-2"
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