Exact equations with integrating factor?

(3xy) dx + (y² + 9x²)dy = 0

Assume that an integrating factor depends on y. Then if P = 3xy and Q = y² + 9x², then for the integrating factor u(y), we must have:

[ln(u)] ' = (1 / P)[∂Q / ∂x - ∂P / ∂y]

[ln(u)] ' = (1 / (3xy))[∂{y² + 9x²} / ∂x - ∂{3xy} / ∂y]

[ln(u)] ' = (1 / (3xy))[18x - 3x]

[ln(u)] ' = (1 / (3xy))[15x]

[ln(u)] ' = 5 / y

Integrate both sides:

∫ [ln(u)] ' dy = ∫ 5 / y dy

ln(u) = 5 ln(y)

ln(u) = ln(y⁵)

u = y⁵

Multiply through:

(3x y⁶) dx + (y⁷ + 9x² y⁵)dy = 0

Lo and behold, the equation is exact. Assume that there exists a function f(x, y) such that ∂f/∂y = y⁷ + 9x² y⁵ and ∂f/∂x = 3x y⁶. Integrate both sides of the x equation:

∫ ∂f/∂x = ∫ 3x y⁶ dx

f(x, y) = (3/2)x²y⁶ + h(y)

∂ f(x, y) /∂y = 9x²y⁵ + h' (y)

y⁷ + 9x² y⁵ = 9x²y⁵ + h' (y)

h ' (y) = y⁷

∫ h ' (y) dy = ∫ y⁷ dy

h(y) = (1/8)y⁸

Thus f(x, y) = (3/2)x²y⁶ + (1/8)y⁸. The final solution is then:

(3/2)x²y⁶ + (1/8)y⁸ = C

Done!