Key Concepts and Formulas

• Linear First-Order Differential Equation: A differential equation of the form $\frac{dy}{dt} + P(t)y = Q(t)$.
• Integrating Factor (I.F.): For a linear first-order differential equation, the integrating factor is given by $I.F. = e^{\int P(t) dt}$.
• General Solution: The general solution is given by $y(t) \cdot (I.F.) = \int Q(t) \cdot (I.F.) dt + C$, where $C$ is the constant of integration.
• L'Hôpital's Rule: If $\lim_{x \to a} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$, provided the limit exists.

Step-by-Step Solution

Step 1: Identify the differential equation and its components. The given differential equation is: $\frac{dy}{dt} + \alpha y = \gamma e^{-\beta t}$ Comparing this with the standard form $\frac{dy}{dt} + P(t)y = Q(t)$, we identify:

• $P(t) = \alpha$
• $Q(t) = \gamma e^{-\beta t}$ This step is essential to recognize the type of differential equation and correctly identify the functions needed for the subsequent steps.

Step 2: Calculate the Integrating Factor (I.F.). The integrating factor is given by: $I.F. = e^{\int P(t) dt} = e^{\int \alpha dt}$ Since $\alpha$ is a constant, the integral is straightforward: $I.F. = e^{\alpha t}$ The integrating factor simplifies the differential equation by allowing us to rewrite the left side as the derivative of a product.

Step 3: Find the general solution $y(t)$. The general solution is given by: $y(t) \cdot (I.F.) = \int Q(t) \cdot (I.F.) dt + C$ Substituting the values of $I.F.$ and $Q(t)$: $y(t) \cdot e^{\alpha t} = \int (\gamma e^{-\beta t}) \cdot (e^{\alpha t}) dt + C$ Simplify the integrand: $y(t) \cdot e^{\alpha t} = \int \gamma e^{(\alpha - \beta)t} dt + C$ Now we integrate. We need to consider two cases: $\alpha \neq \beta$ and $\alpha = \beta$.

Case 1: $\alpha \neq \beta$ In this case, we have: $y(t) e^{\alpha t} = \gamma \int e^{(\alpha - \beta)t} dt + C = \gamma \frac{e^{(\alpha - \beta)t}}{\alpha - \beta} + C$ Dividing by $e^{\alpha t}$: $y(t) = \frac{\gamma}{\alpha - \beta} e^{-\beta t} + C e^{-\alpha t}$

Case 2: $\alpha = \beta$ In this case, we have: $y(t) e^{\alpha t} = \gamma \int e^{(\alpha - \alpha)t} dt + C = \gamma \int 1 dt + C = \gamma t + C$ Dividing by $e^{\alpha t}$: $y(t) = \gamma t e^{-\alpha t} + C e^{-\alpha t}$ Finding the general solution correctly is crucial. Considering different cases avoids errors in integration.

Step 4: Evaluate the limit as $t \to \infty$. We need to find $\lim_{t \to \infty} y(t)$. We are given that $\alpha > 0$ and $\beta > 0$.

Case 1: $\alpha \neq \beta$ $\lim_{t \to \infty} y(t) = \lim_{t \to \infty} \left[ \frac{\gamma}{\alpha - \beta} e^{-\beta t} + C e^{-\alpha t} \right]$ Since $\alpha > 0$ and $\beta > 0$, as $t \to \infty$, $e^{-\beta t} \to 0$ and $e^{-\alpha t} \to 0$. Therefore, $\lim_{t \to \infty} y(t) = \frac{\gamma}{\alpha - \beta} \cdot 0 + C \cdot 0 = 0$

Case 2: $\alpha = \beta$ $\lim_{t \to \infty} y(t) = \lim_{t \to \infty} \left[ \gamma t e^{-\alpha t} + C e^{-\alpha t} \right]$ Since $\alpha > 0$, as $t \to \infty$, $e^{-\alpha t} \to 0$. Therefore, $\lim_{t \to \infty} C e^{-\alpha t} = 0$. Now consider $\lim_{t \to \infty} \gamma t e^{-\alpha t} = \lim_{t \to \infty} \frac{\gamma t}{e^{\alpha t}}$. This is an indeterminate form of type $\frac{\infty}{\infty}$, so we can apply L'Hôpital's Rule: $\lim_{t \to \infty} \frac{\gamma t}{e^{\alpha t}} = \lim_{t \to \infty} \frac{\gamma}{\alpha e^{\alpha t}}$ Since $\alpha > 0$, as $t \to \infty$, $\alpha e^{\alpha t} \to \infty$. Therefore, $\lim_{t \to \infty} \frac{\gamma}{\alpha e^{\alpha t}} = 0$ Thus, $\lim_{t \to \infty} y(t) = 0 + 0 = 0$ In both cases, the limit is 0. The conditions $\alpha>0$ and $\beta>0$ are essential for the exponential terms to decay to zero.

Common Mistakes & Tips:

• Remember to include the constant of integration, $C$, when finding the general solution.
• When integrating, pay attention to the case when $\alpha = \beta$ to avoid errors.
• When evaluating limits, recognize indeterminate forms and use L'Hôpital's Rule if necessary.

Summary We solved the given first-order linear differential equation by finding the integrating factor and then integrating. We considered two cases, $\alpha \neq \beta$ and $\alpha = \beta$, to find the general solution. Finally, we evaluated the limit of the solution as $t$ approaches infinity, using the given conditions $\alpha > 0$ and $\beta > 0$. In both cases, the limit was found to be 0.

Final Answer The final answer is \boxed{0}, which corresponds to option (A).