Solution of Bernoulli differential equation: Divide both sides by $x y^6$: $y^{-6} \frac{dy}{dx} + \frac{1}{x} y^{-5} = x^2$. Let $v = y^{-5} \implies \frac{dv}{dx} = -5 y^{-6} \frac{dy}{dx} \implies y^{-6} \frac{dy}{dx} = -\frac{1}{5} \frac{dv}{dx}$. Substituting into the equation: $-\frac{1}{5} \frac{dv}{dx} + \frac{1}{x} v = x^2 \implies \frac{dv}{dx} - \frac{5}{x} v = -5 x^2$. Integrating Factor (I.F.): $\text{I.F.} = e^{\int -\frac{5}{x} dx} = e^{-5\ln x} = x^{-5}$. Solution for $v$: $v \cdot x^{-5} = \int (-5 x^2)(x^{-5}) dx = -5 \int x^{-3} dx = -5 \left( \frac{x^{-2}}{-2} \right) + c = 2.5 x^{-2} + c$. Multiplying by $x^5$: $v = 2.5 x^3 + c x^5 = x^3 (2.5 + c x^2)$. Since $v = y^{-5} = \frac{1}{y^5}$: $\frac{1}{y^5} = x^3 (2.5 + c x^2) \implies x^3 y^5 (2.5 + c x^2) = 1$. Option breakdown:

• $x^3 y^5 (2.5 + c x^2) = 1$ is the exact solution.
• Swaps exponents of $x$ and $y$.
• Has $y^2$ inside the parentheses instead of $x^2$.
• Swaps exponents and has $y^2$ inside parentheses.