If $\phi(x)$ is a differential real-valued function satisfying $\phi'(x) + 2\phi(x) \leq 1$, then the maximum value of $2\phi(x)$, equal to ____.
Step-by-Step Solution
Key Concept: Separate the functional equation into two cases using $\lambda = \pm 1$ and apply boundary conditions to determine the constant of integration.
Given $f'(y) = \frac{1+f(y)}{1+y}$ and $f'(x) = \frac{1+f(x)}{x(1+x)}$, we find $\lambda = \frac{1}{x}$ so $\lambda = \pm 1$. Integrating $f'(x) = \frac{1}{1+f(x)}$ gives $f(x) = C(1+x)^{\pm 1} - 1$. Using $f(-1) = C - 1$ and $f(0) = -1$, we get $C = 0.1$. Therefore $f(x) = (1+x)^{-1} - 1 = \frac{-x}{1+x}$, and $1 + f(2018) = \frac{1}{2019}$, so $(2019)(1 + f(2018)) = 1$.
Correct Answer: 1