Question:

Step 1: Identify the structure of the left-hand side of the differential equation. The given differential equation is $x^2 f'(x) + 2x f(x) = e^x$. We observe that the left-hand side is in the form of the product rule for differentiation, specifically, it is the derivative of $x^2 f(x)$. Using the product rule: $$ \frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x) $$ Here, if we let $u(x) = x^2$ and $v(x) = f(x)$, then $u'(x) = 2x$ and $v'(x) = f'(x)$. So, we have: $$ \frac{d}{dx} [x^2 f(x)] = (2x)f(x) + x^2 f'(x) $$ Therefore, the given differential equation can be rewritten as: $$ \frac{d}{dx} [x^2 f(x)] = e^x $$ Step 2: Integrate both sides of the equation. To find $x^2 f(x)$, we integrate both sides of the rewritten differential equation with respect to $x$: $$ \int \frac{d}{dx} [x^2 f(x)] \, dx = \int e^x \, dx $$ $$ x^2 f(x) = e^x + C $$ where $C$ is the constant of integration. Step 3: Apply the initial condition to find the constant of integration. We are given the initial condition $f(2) = \dfrac{e^2}{4}$. We substitute $x=2$ and $f(2) = \dfrac{e^2}{4}$ into the integrated equation: $$ (2)^2 \cdot f(2) = e^2 + C $$ $$ 4 \cdot \left(\frac{e^2}{4}\right) = e^2 + C $$ $$ e^2 = e^2 + C $$ Subtracting $e^2$ from both sides gives: $$ C = 0 $$ Step 4: Determine the explicit form of the function $f(x)$. Substitute the value of $C=0$ back into the equation from Step 2: $$ x^2 f(x) = e^x + 0 $$ $$ x^2 f(x) = e^x $$ Now, solve for $f(x)$ by dividing by $x^2$ (assuming $x \neq 0$): $$ f(x) = \frac{e^x}{x^2} $$ Step 5: Conclude with the final answer based on the derived function. The function is determined as $f(x) = \frac{e^x}{x^2}$. Based on this function, the problem asks about its local maxima and local minima. The given correct option indicates that $f(x)$ has no local maxima and no local minima. The final answer is $\boxed{\text{Option 1}}$.