Question

solve
(x³y³+x²y²+xy+1)ydx+(x³y³-x²y²-xy+1)xdy=0​

Answer 1

Answer:

(1/2)xy + (1/2)lnlxyl + 1 / (2xy) + C

Step-by-step explanation:

xx(x2y2 - xy + 1)dy + y(x2y2 + xy + 1)dx = 0

Let u = xy. Then, y = u/x and dy = (xdu - udx) / x2

Substituting, we get: x(u2 - u + 1)[(xdu - udx) / x2] + (u/x)(u2 + u + 1)dx = 0

[(u/x)(u2 + u + 1) - (u/x)(u2 - u + 1)]dx = (-u2 + u - 1)du

(2u2 / x) dx = (-u2 + u - 1)du

(1/x)dx = [-(1/2) + (1/(2u) - 1/(2u2)]du

Integrate both sides to obtain: lnlxl = -(1/2)u + (1/2)lnlul + 1 / (2u) + C

Since u = xy, we have: lnlxl = -(1/2)xy + (1/2)lnlxyl + 1 / (2xy) + C

Answer 2

Step-by-step explanation:

The equation to solve is:

(x3y3+x2y2+xy+1)ydx+(x3y3−x2y2−xy+1)xdy=0

I tried putting xy=t but that just gave me this:

t3−t2−t+1t3+t2+t+1dt=dxx