Let $f:[1,\infty)\to[2,\infty)$ be a differentiable function such that $f(1)=2$. If $6\int_{1}^{x}f(t)\,dt=3xf(x)-x^{3}$ for all $x\geq 1$, then the value of $f(2)$ is ________.

Let f:[1,∞)→[2,∞) be a differentiable function such that f(1)=2. If 6∫₁ˣf(t) dt=3xf(x)-x³ for all x≥ 1, then the value of f(2) is ________. $f:[1,\infty) \to [2,\infty),\quad f(1) = 2$ $6\int_{1}^{x} f(t)\,dt = 3x f(x) - x^{3},\qquad x \geq 1$ $\text{To find: } f(2)$ $\text{Plan: }1.\ \text{differentiate both sides}$ $2.\ \text{get a linear ODE } f' - \dfrac{f}{x} = x$ $3.\ \text{use } f(1)=2 \Rightarrow f(2)$ $\dfrac{d}{dx}\!\left[6\int_{1}^{x} f(t)\,dt\right] = 6f(x)$ $\dfrac{d}{dx}\!\left[3x f(x) - x^{3}\right] = 3f(x) + 3x f'(x) - 3x^{2}$ $6f(x) = 3f(x) + 3x f'(x) - 3x^{2}$ $f(x) = x f'(x) - x^{2} \Rightarrow x f'(x) - f(x) = x^{2}$ $f'(x) - \dfrac{1}{x} f(x) = x$ $\text{Linear: } f' + P f = Q,\qquad P = -\dfrac{1}{x}$ $\mu = e^{\int P\,dx} = e^{-\ln x} = \dfrac{1}{x}$ $\dfrac{d}{dx}\!\left(\dfrac{f}{x}\right) = 1$ $\dfrac{f}{x} = x + C$ $f(x) = x^{2} + Cx$ $f(1) = 1 + C = 2 \Rightarrow C = 1$ $f(x) = x^{2} + x$ $f(2) = 2^{2} + 2 = 4 + 2 = 6$ $\boxed{6}$