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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.

Solve the initial value problem: y' + 6y = 1, y(0) = 1.

Learn how to solve the initial value problem y' + 6y = 1 with the condition y(0) = 1 using the integrating factor method. This step-by-step solution covers linear first-order differential equations and integrating factors.

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