y(1+xy)dx + x(1-xy)dy=0

1.	y(1+xy)dx + x(1-xy)dy=0
Answer» The given differential equation is (1 + xy)ydx + x(1 – xy)dy = 0 ydx + xdy + xy(ydx – xdy) = 0 d(xy) + xy(ydx – xdy) = 0 (d(xy)/(xy)) + (ydx – xdy) = 0 d(xy)/(xy) = (xdy – ydx) d(xy)/(xy)² = (xdy – ydx)/xy = (dy/y) – (dx/x) Integrating, we get – (1/xy) = log \|(y/x)\| + c

Answer»

The given differential equation is

(1 + xy)ydx + x(1 – xy)dy = 0

ydx + xdy + xy(ydx – xdy) = 0

d(xy) + xy(ydx – xdy) = 0

(d(xy)/(xy)) + (ydx – xdy) = 0

d(xy)/(xy) = (xdy – ydx)

d(xy)/(xy)² = (xdy – ydx)/xy = (dy/y) – (dx/x)

Integrating, we get

– (1/xy) = log |(y/x)| + c

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y(1+xy)dx + x(1-xy)dy=0

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