1. Given equation

We have

$$x\phi(x)=\int_5^x \left(3t^2-2\phi'(t)\right)\,dt, \qquad x>-2$$

and $\phi(0)=4.$

We need to find $\phi(2)$.

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2. Differentiate both sides with respect to $x$

Using the product rule on the left:

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

Using the Fundamental Theorem of Calculus on the right:

$$\frac{d}{dx}\left(\int_5^x (3t^2-2\phi'(t))dt\right)=3x^2-2\phi'(x).$$

So,

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

Rearrange:

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

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3. Recognize the left side as a derivative

Notice that

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

Hence the differential equation becomes

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

Integrate:

$$(x+2)\phi(x)=\int 3x^2\,dx = x^3+C.$$

Thus,

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

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4. Use the initial condition $\phi(0)=4$

Substitute $x=0$:

$$\phi(0)=\frac{0^3+C}{0+2}=\frac{C}{2}=4.$$

So,

$$C=8.$$

Therefore,

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

Factor the numerator:

$$x^3+8=(x+2)(x^2-2x+4).$$

Since $x>-2$, we can simplify:

$$\phi(x)=x^2-2x+4.$$

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5. Compute $\phi(2)$

$$\phi(2)=2^2-2(2)+4=4-4+4=4.$$

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6. Compare with stored answer

Derived answer: $4$

Stored correct answer: $4$

They agree.