$$ \text{Let } f \text{ be a differentiable function such that } 2(x+2)^2f(x)-3(x+2)^2 = 10\int_0^x (t+2)f(t)\,dt,\quad x\ge0. \text{ Then } f(2) \text{ is equal to:}\underline{\hspace{1cm}} $$

Given $$2(x+2)^2f(x) - 3(x+2)^2 = 10\int_0^x(t+2)f(t)dt$$, find f(2).

Differentiate both sides with respect to x

$$4(x+2)f(x) + 2(x+2)^2f'(x) - 6(x+2) = 10(x+2)f(x)$$

$$2(x+2)^2f'(x) = 6(x+2)f(x) + 6(x+2)$$

$$2(x+2)f'(x) = 6f(x) + 6$$

$$(x+2)f'(x) - 3f(x) = 3$$

Find f(0) from original equation

At x=0: $$2(4)f(0) - 3(4) = 0$$

$$8f(0) = 12 \Rightarrow f(0) = 3/2$$

Solve the ODE

$$f'(x) - \frac{3}{x+2}f(x) = \frac{3}{x+2}$$

IF = $$e^{-3\ln(x+2)} = \frac{1}{(x+2)^3}$$

$$\frac{d}{dx}\left[\frac{f(x)}{(x+2)^3}\right] = \frac{3}{(x+2)^4}$$

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

$$f(x) = -1 + C(x+2)^3$$

Apply f(0) = 3/2

$$3/2 = -1 + C(8) \Rightarrow C = \frac{5/2}{8} = \frac{5}{16}$$

$$f(x) = -1 + \frac{5}{16}(x+2)^3$$

Find f(2)

$$f(2) = -1 + \frac{5}{16}(4)^3 = -1 + \frac{5 \times 64}{16} = -1 + 20 = 19$$

The answer is 19.