The given differential equation is:

$\frac{dy}{dt} = -1 - 3y$

Step-by-Step Solution:

This is a first-order linear differential equation. To solve it, follow these steps:

1. Separate variables: We aim to express all terms involving $y$ on one side, and terms involving $t$ on the other side. $\frac{dy}{dt} = -1 - 3y$ Rearranging terms: $\frac{dy}{-1 - 3y} = dt$

2. Integrate both sides: Now, integrate both sides to find $y(t)$.

   On the left side, the integral is: $\int \frac{1}{-1 - 3y} dy$ On the right side, the integral is: $\int dt$

3. Solve the integral: The left side requires a substitution. Let $u = -1 - 3y$, then $du = -3 dy$. The integral becomes: $\int \frac{1}{u} \frac{du}{-3} = -\frac{1}{3} \int \frac{1}{u} du = -\frac{1}{3} \ln|u| = -\frac{1}{3} \ln|-1 - 3y|$ The right side integrates to $t + C$, where $C$ is the constant of integration.

4. Combine results: We now have: $-\frac{1}{3} \ln|-1 - 3y| = t + C$

5. Solve for $y$: Multiply both sides by $-3$: $\ln|-1 - 3y| = -3t - 3C$ Exponentiate both sides to remove the logarithm: $|-1 - 3y| = e^{-3t - 3C} = e^{-3C}e^{-3t}$ Let $A = e^{-3C}$, so: $|-1 - 3y| = A e^{-3t}$

   Now, solve for $y$: $-1 - 3y = \pm A e^{-3t}$ $3y = -1 \mp A e^{-3t}$ $y = \frac{-1 \mp A e^{-3t}}{3}$

Final Solution:

The general solution to the differential equation is:

$y(t) = \frac{-1}{3} \mp \frac{A}{3} e^{-3t}$

where $A$ is a constant determined by initial conditions.

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Do you want further clarification on this? Here are some related questions:

1. How do you solve differential equations with different types of functions on the right-hand side?
2. What is the significance of the constant of integration, and how is it determined?
3. Can you apply this method to nonlinear differential equations?
4. What are real-life applications of first-order linear differential equations?
5. How does the solution behave as $t$ tends to infinity?

Tip: Always remember that constants like $A$ are determined by initial or boundary conditions in differential equations.