Let $$f$$ be a differentiable function such that $$x^2 f(x) - x = 4\int_0^x tf(t) \, dt$$, $$f(1) = \frac{2}{3}$$. Then $$18f(3)$$ is equal to

Given: $$x^2f(x) - x = 4\int_0^x tf(t)\,dt$$, $$f(1) = 2/3$$.

Differentiating both sides with respect to x:

$$2xf(x) + x^2f'(x) - 1 = 4xf(x)$$

$$x^2f'(x) - 2xf(x) = 1$$

$$f'(x) - \frac{2}{x}f(x) = \frac{1}{x^2}$$

This is a linear ODE. Integrating factor: $$e^{-2\ln x} = x^{-2}$$

$$\frac{d}{dx}\left(\frac{f(x)}{x^2}\right) = \frac{1}{x^4}$$

$$\frac{f(x)}{x^2} = -\frac{1}{3x^3} + C$$

$$f(x) = -\frac{1}{3x} + Cx^2$$

Using $$f(1) = 2/3$$: $$-1/3 + C = 2/3$$, so $$C = 1$$.

$$f(x) = x^2 - \frac{1}{3x}$$

$$f(3) = 9 - \frac{1}{9} = \frac{80}{9}$$

$$18f(3) = 18 \times \frac{80}{9} = 160$$

The correct answer is Option 2: 160.