The given problem is an initial value problem, which can be solved by solving the first-order linear differential equation:

$y' + 6y = 1$ with the initial condition: $y(0) = 1.$

Step-by-step solution:

1. Rewrite the equation: $y' + 6y = 1.$ This is a linear differential equation in standard form $y' + p(t)y = g(t)$, where $p(t) = 6$ and $g(t) = 1$.

2. Find the integrating factor: The integrating factor $\mu(t)$ is given by: $\mu(t) = e^{\int p(t) \, dt} = e^{\int 6 \, dt} = e^{6t}.$

3. Multiply the equation by the integrating factor: $e^{6t}y' + 6e^{6t}y = e^{6t}.$ The left-hand side is the derivative of $e^{6t}y$: $\frac{d}{dt}(e^{6t}y) = e^{6t}.$

4. Integrate both sides: Integrating both sides with respect to $t$: $e^{6t}y = \int e^{6t} \, dt = \frac{1}{6}e^{6t} + C,$ where $C$ is the constant of integration.

5. Solve for $y(t)$: $y(t) = \frac{1}{6} + Ce^{-6t}.$

6. Apply the initial condition $y(0) = 1$: $1 = \frac{1}{6} + C e^{0} = \frac{1}{6} + C.$ Solving for $C$: $C = 1 - \frac{1}{6} = \frac{5}{6}.$

7. Final solution: $y(t) = \frac{1}{6} + \frac{5}{6}e^{-6t}.$

Thus, the solution to the initial value problem is:

$y(t) = \frac{1}{6} + \frac{5}{6}e^{-6t}.$

Would you like further clarification or details? Here are five related questions to expand your understanding:

1. How would you solve a similar initial value problem with a different initial condition?
2. Can you explain how the integrating factor helps in solving linear differential equations?
3. How would the solution change if the equation were non-homogeneous, i.e., $y' + 6y = f(t)$ for some function $f(t)$?
4. What are the properties of the solution $y(t)$ as $t \to \infty$?
5. How can you verify the solution by substituting it back into the original differential equation?

Tip: When solving differential equations, always check your solution by plugging it back into the original equation to ensure it satisfies both the equation and initial condition.