Answer:
ln(x) - e/(2y^2) = constant; in other words, y = 1 / (√(2ln(kx)) )
Step-by1-step explanation:
divide the expression xy^3dx+ex^2dy=0 by (x^2) x (y^3).
we get, dx/x + edy/y^3 = 0
Integrating , ∫(1/x)dx + ∫e(1/y^3)dy , we get
= > ln(x) + ey^(-2)/(-2) = constant
= > ln(x) - e/(2y^2) = constant
= > e/(2y^2) = ln(x) - d (where d is a constant)
Let d = ln(k) (where k is a constant too)
= > e/(2y^2) = ln(x) - ln(k) = ln(x/k)
2y^2 = 1/ln(x/k)
y = 1 / (√(2ln(cx)) ) (where c = 1/k)