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(C) The given differential equation is $f(x)+x f'(x)=x^2$,which can be written as $y+x \frac{dy}{dx}=x^2$.Dividing by $x$ (assuming $x 
eq 0$),we get

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$x^3-3x f(x)-4=0$

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(C) The given differential equation is $f(x)+x f'(x)=x^2$,which can be written as $y+x \frac{dy}{dx}=x^2$.Dividing by $x$ (assuming $x 
eq 0$),we get $\frac{dy}{dx}+\frac{y}{x}=x$.This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=\frac{1}{x}$ and $Q=x$.The integrating factor $IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$.The solution is given by $y \cdot IF = \int Q \cdot IF dx + C$.$y \cdot x = \int x \cdot x dx + C = \int x^2 dx + C = \frac{x^3}{3} + C$.Since the curve passes through $(-2, 2)$,we substitute $x=-2$ and $y=2$:$2(-2) = \frac{(-2)^3}{3} + C \Rightarrow -4 = -\frac{8}{3} + C$.$C = -4 + \frac{8}{3} = -\frac{4}{3}$.Substituting $C$ back into the equation: $xy = \frac{x^3}{3} - \frac{4}{3}$.Multiplying by $3$,we get $3xy = x^3 - 4$,or $x^3 - 3x f(x) - 4 = 0$.

Let us consider a curve $y=f(x)$ passing through the point $(-2, 2)$ and the slope of the tangent to the curve at any point $(x, f(x))$ is given by $f(x)+x f'(x)=x^2$. Then:

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