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Step-by-step Solution
Step 1: Rewrite the given condition in differential equation form
The problem states that for the curve passing through the point $(-2, 2)$, the slope of the tangent satisfies
$f(x) + x\,f'(x) = x^2.$
Let $y = f(x).$ Then the given relation becomes
$y + x \frac{dy}{dx} = x^2.$
Step 2: Express as a standard linear differential equation
Rewrite the equation in terms of $dy/dx$:
\[
x \frac{dy}{dx} + y = x^2 \quad \Longrightarrow \quad \frac{dy}{dx} + \frac{y}{x} = x.
\]
This is a first-order linear differential equation of the form
\[
\frac{dy}{dx} + P(x)\,y = Q(x),
\]
where $P(x) = \frac{1}{x}$ and $Q(x) = x.$
Step 3: Find the integrating factor
The integrating factor (IF) is given by
\[
\text{IF} \;=\; e^{\int P(x)\,dx} \;=\; e^{\int \frac{1}{x}\,dx} \;=\; e^{\ln x} \;=\; x,
\]
assuming $x \neq 0.$
Step 4: Multiply both sides by the integrating factor and integrate
Multiply the original differential equation $\frac{dy}{dx} + \frac{y}{x} = x$ by $x$:
\[
x \frac{dy}{dx} + y = x^2.
\]
Notice this is the same as the original format, but now we recognize that the left side is the derivative of $x\,y$:
\[
\frac{d}{dx}\bigl(x\,y\bigr) = x^2.
\]
Integrate both sides with respect to $x$:
\[
x\,y = \int x^2\,dx = \frac{x^3}{3} + C,
\]
where $C$ is the constant of integration.
Step 5: Apply the initial condition to find the constant
We are given that the curve passes through $(-2, 2)$, so $x = -2$ and $y = 2.$ Substitute these into
$x\,y = \frac{x^3}{3} + C:$
\[
(-2)\cdot 2 = \frac{(-2)^3}{3} + C
\quad\Longrightarrow\quad
-4 = \frac{-8}{3} + C
\quad\Longrightarrow\quad
C = -4 + \frac{8}{3} = \frac{-12 + 8}{3} = -\frac{4}{3}.
\]
Step 6: Write the final form and match with the given options
Substituting $C = -\frac{4}{3}$ back in gives:
\[
x\,y = \frac{x^3}{3} - \frac{4}{3}
\quad\Longrightarrow\quad
3\,x\,y = x^3 - 4.
\]
Since $y = f(x),$ we have
\[
3\,x\,f(x) = x^3 - 4
\quad\Longrightarrow\quad
x^3 - 3\,x\,f(x) - 4 = 0.
\]
This matches the correct answer provided:
\[
x^3 - 3\,x\,f(x) - 4 = 0.
\]
