A function $y = f(x)$ satisfies $y'(x) = 2f(x) = x^4f^2(x), \forall x > 0$ and $f(1) = -6$. Find the value of $f\left(3^{1/3}\right)$.
Step-by-Step Solution
Key Concept: Bernoulli equation is linearized by substitution $t = y^{1-n}$ to reduce it to a first-order linear ODE solvable by integrating factors.
Given $\frac{xdy}{dx} - 2y = x^4y^3$ or $\frac{dy}{dx} - \frac{2}{x}y = x^3y^2$, divide by $y^2$ to get $\frac{1}{y^2}\frac{dy}{dx} - \frac{2}{xy} = x^3$. This is a Bernoulli equation; substitute $t = -\frac{2}{y}$ to transform it into a linear form $\frac{dt}{dx} - \frac{2t}{x} = 2x^3$. The integrating factor is $e^{\int -\frac{2}{x}dx} = x^{-2}$, giving the general solution $x^2t = \frac{2x^6}{6} + C$. Using initial condition $x = 1, y = -6$ yields $C = 0$, so $t = \frac{x^4}{3}$ and $y = \frac{-6}{x^4}$. Finally, $\frac{dy}{dx} = 24x^{-5} = \frac{24}{x^5}$, and $\frac{dy}{dx}\bigg|_{x=1^{1/3}} = 8$.
Correct Answer: 8