Let $f : (0, \infty) \to \mathbf{R}$ be a function which is differentiable at all points of its domain and satisfies the condition $x^2 f'(x) = 2x f(x) + 3$, with $f(1) = 4$. Then $2 f(2)$ is equal to:

Step 1: Rearrange the differential equation

The given differential equation is $x^2 f'(x) = 2x f(x) + 3$. To solve this, we first rearrange it into a standard form. We move the term $2x f(x)$ to the left side of the equation.

Step 2: Divide by $x^4$ to form a derivative of a quotient

The left side of the equation now resembles the numerator of the quotient rule for differentiation, specifically $\frac{d}{dx}\left(\frac{f(x)}{x^2}\right) = \frac{x^2 f'(x) - f(x)(2x)}{x^4}$. By dividing both sides of our rearranged equation by $x^4$, we transform the left side into the derivative of $\frac{f(x)}{x^2}$.

Step 3: Integrate both sides

Now that the left side is the derivative of a function, we can integrate both sides with respect to $x$. The integral of a derivative simply gives back the original function, plus a constant of integration.

Step 4: Solve the integrals

We perform the integration on both sides. The integral of $\frac{d}{dx}\left(\frac{f(x)}{x^2}\right)$ is $\frac{f(x)}{x^2}$. The integral of $3x^{-4}$ is $3 \frac{x^{-3}}{-3}$, which simplifies to $-\frac{1}{x^3}$. We add the constant of integration $C$.

Step 5: Use the initial condition to find C

We are given the initial condition $f(1) = 4$. We substitute $x=1$ and $f(x)=4$ into our integrated equation to solve for the constant $C$. This gives us $4 = -1 + C$, so $C=5$.

Step 6: Find $f(x)$ and then $2f(2)$

Now that we have the value of $C$, we can write the explicit form of $f(x)$. We then substitute $x=2$ into this function to find $f(2)$, and finally multiply by 2 to get the required value.

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