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
