Properties of Solution to First Order Linear Differential Equation

To find the solution $y(x)$ of the given differential equation, we write: $(1 + e^x)y' + y e^x = 1$

Recognizing the product rule on the left-hand side: $\frac{d}{dx}\left[y(1 + e^x)\right] = 1$

Integrating both sides with respect to $x$: $y(1 + e^x) = x + C$

Using the initial condition $y(0) = 2$: $2(1 + e^0) = 0 + C \implies 2(1 + 1) = C \implies C = 4$

Thus, the particular solution is: $y(x) = \frac{x + 4}{1 + e^x}$

Now, let us evaluate the given options:

1. Checking $y(-4)$ and $y(-2)$:

• For $x = -4$: $y(-4) = \frac{-4 + 4}{1 + e^{-4}} = 0$ Hence, Option (A) is correct.

• For $x = -2$: $y(-2) = \frac{-2 + 4}{1 + e^{-2}} = \frac{2}{1 + e^{-2}} \neq 0$ Hence, Option (B) is incorrect.

2. Checking for critical points in $(-1, 0)$: A critical point occurs where the first derivative $y'(x) = 0$. Differentiating $y(x)$ with respect to $x$: $y'(x) = \frac{(1 + e^x)\cdot 1 - (x + 4)e^x}{(1 + e^x)^2} = \frac{1 - (x + 3)e^x}{(1 + e^x)^2}$

Let $g(x) = 1 - (x + 3)e^x$. The denominator $(1 + e^x)^2$ is strictly positive for all $x \in \mathbb{R}$, so $y'(x) = 0 \iff g(x) = 0$.

Evaluating $g(x)$ at the endpoints of the interval $[-1, 0]$:

• At $x = -1$: $g(-1) = 1 - (-1 + 3)e^{-1} = 1 - \frac{2}{e}$ Since $e \approx 2.718 > 2$, we have $\frac{2}{e} < 1$, which gives: $g(-1) > 0$

• At $x = 0$: $g(0) = 1 - (0 + 3)e^0 = 1 - 3 = -2 < 0$

Since $g(x)$ is continuous on the closed interval $[-1, 0]$ and changes sign ($g(-1) > 0$ and $g(0) < 0$), by the Intermediate Value Theorem, there exists at least one $c \in (-1, 0)$ such that $g(c) = 0$, which implies $y'(c) = 0$.

Therefore, $y(x)$ has a critical point in the interval $(-1, 0)$.

Hence, Option (C) is correct, and Option (D) is incorrect.

Correct Options: (A) and (C)