Question

Solve the following

dy/dx=(x-y-2)/(x+y+6)

Answer 1

$(x+y+6)dy=(x-y-2)dx$

$(2+y-x)dx+(x+y+6)dy=0$

Let $M=2+y-x, \ N=x+y+6$

$M_y'=1=N_x'$ , so the given equation is exact.

And there is function $F(x,y):\ F'_x=M,\ F'_y=N$

$F'_x=2+y-x,\\ F=\int (2+y-x)dx=2x+yx-x^2/2+h(y)$

$F'_y=(2x+yx-x^2/2+h(y))'_y=x+h'_y$

$x+h'_y=x+y+6,\ h'_y=y+6,\\ h(y)=\int (y+6)dy=y^2/2+6y$

$F(x,y)=2x+xy-x^2/2+y^2/2+6y=C,\ C\ is\ constant$

$y^2+2y(6+x)-x^2+4x-2C=0$

Using the quadratic formula

$y=-6-x\pm \sqrt{(6+x)^2-(-x^2+4x-2C)}=\\ -6-x\pm\sqrt{36+12x+x^2+x^2-4x+2C}=\\ -6-x\pm\sqrt{2x^2+8x+36+2C}$

Since $x+y+6$ is in the denominator

$x+y+6\not= 0, \ 2x^2+8x+36+2C\not=0.\\ x^2+4x+18+C\not=0\\ (x+2)^2+14+C\not=0\\ x\not=\pm\sqrt{-14-C}-2$

Answer: $y=-6-x\pm\sqrt{2x^2+8x+36+2C}$ ,$x\not=\pm\sqrt{-14-C}-2$