$y = f(x)$ is a particular solution of differential equation $\left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2} = e^x$ such that $f(0) = 1 = f'(0)$, then the value of $f^2(\ln 2)$ is ____.
Step-by-Step Solution
Key Concept: Recognize that the given differential equation $\left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2} = e^x$ can be rewritten as $\frac{d}{dx}\left(y\frac{dy}{dx}\right) = e^x$ by treating $y\frac{dy}{dx}$ as a composite function, allowing sequential integration to reduce the order.
Given $\frac{d}{dx}(v\frac{dy}{dx}) = e^x$, integrating gives $v\frac{dy}{dx} = e^x + c$. From $1 \times 1 = e^0 + c$, we get $c = 0$. Therefore $ydy = e^x dx$, and integrating: $\frac{1}{2}y^2 = e^x + c'$. Using $y(0) = 1$, we find $c' = -\frac{1}{2}$, giving $y^2 = 2e^x - 1$ or $f^2(x) = 2e^x - 1$.
Correct Answer: 3