Question 1025715
<pre>
dy/dx =(x+y+4)/(x-y-6)

Cross-multiply

(x+y+4)dx = (x-y-6)dy 

Since all terms are linear,
assume a general quadratic = 0
as a solution:

Ax²+Bxy+Cy²+Dx+Ey+F = 0

Take differentials of both sides

2Axdx+Bxdy+Bydx+2Cydy+Ddx+Edy = 0

(2Ax+By+D)dx+(Bx+2Cy+E)dy = 0  

Equate coefficients of like terms:

(x+y+4)dx = (x-y-6)dy

2A=1,   B=1, D=4, B=1, 2C=-1,  E=-6
A=1/2,                 C=-1/2

So the assumed solution: 

Ax²+Bxy+Cy²+Dx+Ey+F = 0

becomes

(1/2)x²+1xy-(1/2)y²+4x-6y+F = 0

F is the arbitrary constant.  If you like
you can multiply through by 2 and replace
2F by arbitrary constant little c:

x²+2xy-y²+8x-12y+c = 0

Edwin</pre>