Step 1: Rearranging.
We have:
\[
\frac{dy}{dx} = \frac{x + e^x}{y}
\]
Multiply both sides by $y$:
\[
y \frac{dy}{dx} = x + e^x
\]
Step 2: Recognizing derivative form.
This suggests:
\[
\frac{d}{dx}\left( \frac{y^2}{2} \right) = x + e^x
\]
Step 3: Integration.
Integrating both sides w.r.t. $x$:
\[
\frac{y^2}{2} = \int (x + e^x) \, dx
\]
\[
\frac{y^2}{2} = \frac{x^2}{2} + e^x + C
\]
Step 4: Simplify.
\[
y^2 = x^2 + 2e^x + C'
\]
Final Answer: \[ \boxed{y^2 = x^2 + 2e^x + C} \]
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f(x + y) = f(x) f(y) \) for all \( x, y \in \mathbb{R} \). If \( f'(0) = 4a \) and \( f \) satisfies \( f''(x) - 3a f'(x) - f(x) = 0 \), where \( a > 0 \), then the area of the region R = {(x, y) | 0 \(\leq\) y \(\leq\) f(ax), 0 \(\leq\) x \(\leq\) 2 is :