If $$x\phi(x) = \int_5^x (3t^2 - 2\phi'(t)) dt$$, $$x > -2$$, $$\phi(0) = 4$$, then $$\phi(2)$$ is _________.

We have the functional equation

$$x\,\phi(x)=\int_{5}^{x}\left(3t^{2}-2\,\phi'(t)\right)\,dt,\qquad x>-2,$$

together with the initial value

$$\phi(0)=4.$$

To remove the integral, we differentiate both sides with respect to the variable $$x$$. The Fundamental Theorem of Calculus tells us that

$$\frac{d}{dx}\int_{5}^{x}f(t)\,dt=f(x).$$

Using this rule and the product rule $$\frac{d}{dx}[x\,\phi(x)]=\phi(x)+x\,\phi'(x)$$, we obtain

$$\phi(x)+x\,\phi'(x)=3x^{2}-2\,\phi'(x).$$

Now we collect all $$\phi'(x)$$ terms on the left side:

$$x\,\phi'(x)+2\,\phi'(x)=3x^{2}-\phi(x).$$

This simplifies to

$$(x+2)\,\phi'(x)=3x^{2}-\phi(x).$$

We rewrite it so that all $$\phi(x)$$ terms are on the left:

$$(x+2)\,\phi'(x)+\phi(x)=3x^{2}.$$

Next we divide by the factor $$(x+2)$$ (which is positive because $$x>-2$$). This gives the first-order linear differential equation

$$\phi'(x)+\frac{1}{x+2}\,\phi(x)=\frac{3x^{2}}{x+2}.$$

The standard linear form is $$y'+P(x)y=Q(x)$$ with

$$P(x)=\frac{1}{x+2},\qquad Q(x)=\frac{3x^{2}}{x+2}.$$

We now compute the integrating factor. By definition,

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int\frac{dx}{x+2}}=e^{\ln|x+2|}=x+2,$$

where the absolute value can be dropped because $$x+2>0$$ in our domain.

Multiplying the entire differential equation by this integrating factor, we have

$$(x+2)\,\phi'(x)+\phi(x)=3x^{2}.$$

Notice that the left side is precisely the derivative of the product $$(x+2)\phi(x)$$, because

$$\frac{d}{dx}\big[(x+2)\phi(x)\big]=(x+2)\,\phi'(x)+\phi(x).$$

Therefore

$$\frac{d}{dx}\big[(x+2)\phi(x)\big]=3x^{2}.$$

We integrate both sides with respect to $$x$$:

$$\int\frac{d}{dx}\big[(x+2)\phi(x)\big]\,dx=\int 3x^{2}\,dx.$$

The left integral simply returns the function inside, while the right integral is elementary:

$$(x+2)\phi(x)=x^{3}+C,$$

where $$C$$ is the constant of integration.

To find $$C$$ we use the given initial condition $$\phi(0)=4$$. Substituting $$x=0$$ gives

$$(0+2)\phi(0)=0^{3}+C\;\Longrightarrow\;2\cdot4=C\;\Longrightarrow\;C=8.$$

Hence the explicit form of $$\phi(x)$$ is

$$\phi(x)=\frac{x^{3}+8}{x+2}.$$

Finally we evaluate at $$x=2$$:

$$\phi(2)=\frac{2^{3}+8}{2+2}=\frac{8+8}{4}=\frac{16}{4}=4.$$

So, the answer is $$4$$. Hence, the correct answer is Option A.