Key Concepts and Formulas

• Linear First-Order Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$. The integrating factor is given by $IF = e^{\int P(x) dx}$, and the solution is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$.
• Tangent Line Equation: The equation of the tangent line to a curve $y = f(x)$ at a point $(x_0, y_0)$ is given by $y - y_0 = f'(x_0)(x - x_0)$.
• Intercepts: The $x$-intercept is the point where $y=0$, and the $y$-intercept is the point where $x=0$.

Step-by-Step Solution

Step 1: Identify and Solve the Differential Equation

We are given the differential equation $\frac{dy}{dx} - y = 2 - e^{-x}$. This is a linear first-order differential equation. We need to find the integrating factor and solve for $y(x)$.

Comparing the given equation with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we have $P(x) = -1$ and $Q(x) = 2 - e^{-x}$.

The integrating factor (IF) is given by: $IF = e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x}$

Multiplying the differential equation by the integrating factor: $e^{-x}\frac{dy}{dx} - e^{-x}y = e^{-x}(2 - e^{-x})$ $\frac{d}{dx}(ye^{-x}) = 2e^{-x} - e^{-2x}$

Integrating both sides with respect to $x$: $\int \frac{d}{dx}(ye^{-x}) dx = \int (2e^{-x} - e^{-2x}) dx$ $ye^{-x} = -2e^{-x} + \frac{1}{2}e^{-2x} + C$

Multiplying by $e^x$ to isolate $y$: $y(x) = -2 + \frac{1}{2}e^{-x} + Ce^x$

Step 2: Apply the Limit Condition to Find C

We are given that $\lim_{x \to \infty} y(x)$ is finite. $\lim_{x \to \infty} y(x) = \lim_{x \to \infty} \left(-2 + \frac{1}{2}e^{-x} + Ce^x\right)$

As $x \to \infty$, $e^{-x} \to 0$ and $e^x \to \infty$. For the limit to be finite, the term $Ce^x$ must vanish, which implies $C = 0$.

Therefore, the particular solution is: $y(x) = -2 + \frac{1}{2}e^{-x}$

Step 3: Find the Equation of the Tangent at x = 0

First, find the y-coordinate at $x=0$: $y(0) = -2 + \frac{1}{2}e^{-0} = -2 + \frac{1}{2} = -\frac{3}{2}$ So the point of tangency is $(0, -\frac{3}{2})$.

Next, find the derivative of $y(x)$: $\frac{dy}{dx} = \frac{d}{dx}\left(-2 + \frac{1}{2}e^{-x}\right) = -\frac{1}{2}e^{-x}$

Evaluate the derivative at $x=0$ to find the slope of the tangent: $m = \frac{dy}{dx}\Big|_{x=0} = -\frac{1}{2}e^{-0} = -\frac{1}{2}$

The equation of the tangent line is: $y - \left(-\frac{3}{2}\right) = -\frac{1}{2}(x - 0)$ $y + \frac{3}{2} = -\frac{1}{2}x$ $2y + 3 = -x$ $x + 2y + 3 = 0$

Step 4: Find the x and y Intercepts

To find the $x$-intercept ($a$), set $y=0$: $x + 2(0) + 3 = 0$ $x = -3$ So, $a = -3$.

To find the $y$-intercept ($b$), set $x=0$: $0 + 2y + 3 = 0$ $2y = -3$ $y = -\frac{3}{2}$ So, $b = -\frac{3}{2}$.

Step 5: Calculate a - 4b

$a - 4b = -3 - 4\left(-\frac{3}{2}\right) = -3 + 6 = 3$

Common Mistakes & Tips

• Sign Error in IF: Be careful with the sign of $P(x)$ when calculating the integrating factor.
• Limit Evaluation: Remember that $\lim_{x \to \infty} e^{-x} = 0$ and $\lim_{x \to \infty} e^x = \infty$. This is crucial for determining the constant of integration.
• Algebraic Errors: Double-check all algebraic manipulations, especially with fractions and negative signs.

Summary

We solved the given linear first-order differential equation, applied the limit condition to find the particular solution, determined the equation of the tangent line at $x=0$, found the $x$ and $y$ intercepts of the tangent line, and finally calculated $a - 4b$.

The final answer is \boxed{3}.