Question:

Step 1: Identify the type of differential equation. The given differential equation is $$ \frac{dy}{dt}+\alpha y=\gamma e^{-\beta t} $$ This is a first-order linear differential equation of the form $\frac{dy}{dt} + P(t)y = Q(t)$, where $P(t) = \alpha$ and $Q(t) = \gamma e^{-\beta t}$. Step 2: Calculate the integrating factor (IF). The integrating factor for a differential equation of the form $\frac{dy}{dt} + P(t)y = Q(t)$ is given by $e^{\int P(t) dt}$. In this case, $P(t) = \alpha$, so the integrating factor is: $$ \text{IF} = e^{\int \alpha dt} = e^{\alpha t} $$ Step 3: Solve the differential equation. Multiply the entire differential equation by the integrating factor $e^{\alpha t}$: $$ e^{\alpha t} \frac{dy}{dt} + \alpha e^{\alpha t} y = \gamma e^{-\beta t} e^{\alpha t} $$ The left side of the equation is the derivative of the product $(y \cdot e^{\alpha t})$ with respect to $t$: $$ \frac{d}{dt}(y e^{\alpha t}) = \gamma e^{(\alpha-\beta)t} $$ Now, integrate both sides with respect to $t$: $$ \int \frac{d}{dt}(y e^{\alpha t}) dt = \int \gamma e^{(\alpha-\beta)t} dt $$ $$ y e^{\alpha t} = \gamma \int e^{(\alpha-\beta)t} dt $$ Step 4: Evaluate the integral to find the general solution for $y(t)$. We need to consider two cases for the integral $\int e^{(\alpha-\beta)t} dt$: Case 1: If $\alpha - \beta \neq 0$. $$ y e^{\alpha t} = \gamma \frac{e^{(\alpha-\beta)t}}{\alpha-\beta} + C $$ Divide by $e^{\alpha t}$ to solve for $y(t)$: $$ y(t) = \frac{\gamma}{\alpha-\beta} e^{(\alpha-\beta)t} e^{-\alpha t} + C e^{-\alpha t} $$ $$ y(t) = \frac{\gamma}{\alpha-\beta} e^{-\beta t} + C e^{-\alpha t} $$ Case 2: If $\alpha - \beta = 0$, which means $\alpha = \beta$. In this case, the integral becomes: $$ y e^{\alpha t} = \gamma \int e^{0 \cdot t} dt = \gamma \int 1 dt $$ $$ y e^{\alpha t} = \gamma t + C $$ Divide by $e^{\alpha t}$ to solve for $y(t)$: $$ y(t) = \gamma t e^{-\alpha t} + C e^{-\alpha t} $$ Step 5: Evaluate the limit $\displaystyle\lim_{t\to\infty}y(t)$. We are given that $\alpha > 0$ and $\beta > 0$. For Case 1: If $\alpha - \beta \neq 0$. $$ \lim_{t\to\infty} y(t) = \lim_{t\to\infty} \left( \frac{\gamma}{\alpha-\beta} e^{-\beta t} + C e^{-\alpha t} \right) $$ Since $\alpha > 0$ and $\beta > 0$, we have: $$ \lim_{t\to\infty} e^{-\beta t} = 0 $$ $$ \lim_{t\to\infty} e^{-\alpha t} = 0 $$ Therefore, $$ \lim_{t\to\infty} y(t) = \frac{\gamma}{\alpha-\beta} (0) + C (0) = 0 $$ For Case 2: If $\alpha = \beta$. $$ \lim_{t\to\infty} y(t) = \lim_{t\to\infty} \left( \gamma t e^{-\alpha t} + C e^{-\alpha t} \right) $$ Since $\alpha > 0$, $\lim_{t\to\infty} C e^{-\alpha t} = 0$. For the term $\gamma t e^{-\alpha t}$, we can rewrite it as $\frac{\gamma t}{e^{\alpha t}}$. This is an indeterminate form of type $\frac{\infty}{\infty}$ as $t \to \infty$. We can apply L'Hôpital's Rule: $$ \lim_{t\to\infty} \frac{\gamma t}{e^{\alpha t}} = \lim_{t\to\infty} \frac{\frac{d}{dt}(\gamma t)}{\frac{d}{dt}(e^{\alpha t})} = \lim_{t\to\infty} \frac{\gamma}{\alpha e^{\alpha t}} $$ Since $\alpha > 0$, as $t \to \infty$, $\alpha e^{\alpha t} \to \infty$. So, $$ \lim_{t\to\infty} \frac{\gamma}{\alpha e^{\alpha t}} = 0 $$ Therefore, for Case 2, $$ \lim_{t\to\infty} y(t) = 0 + 0 = 0 $$ Step 6: State the final answer. In both cases (whether $\alpha = \beta$ or $\alpha \neq \beta$), the limit of $y(t)$ as $t \to \infty$ is 0. The final answer is $\boxed{0}$.