https://file.aiquickdraw.com/mathbot/file/172948687308482cu7us3.jpg

The equation shown in the image is a first-order linear differential equation:

\[
y' + 3y = 1, \quad y(0) = 1
\]

### Steps to solve:

1. **Identify the equation:**
   \[
   y' + 3y = 1
   \]
   This is a linear first-order ordinary differential equation (ODE).

2. **Find the integrating factor:**
   The standard form of a linear ODE is \( y' + P(x)y = Q(x) \). Here, \( P(x) = 3 \) and \( Q(x) = 1 \).
   
   The integrating factor \( \mu(x) \) is given by:
   \[
   \mu(x) = e^{\int P(x) \, dx} = e^{\int 3 \, dx} = e^{3x}
   \]

3. **Multiply through by the integrating factor:**
   Multiply both sides of the equation by \( e^{3x} \):
   \[
   e^{3x}y' + 3e^{3x}y = e^{3x}
   \]
   The left-hand side is the derivative of \( e^{3x}y \):
   \[
   \frac{d}{dx}(e^{3x}y) = e^{3x}
   \]

4. **Integrate both sides:**
   \[
   e^{3x}y = \int e^{3x} \, dx = \frac{1}{3}e^{3x} + C
   \]
   where \( C \) is the constant of integration.

5. **Solve for \( y \):**
   \[
   y = \frac{1}{3} + Ce^{-3x}
   \]

6. **Use the initial condition \( y(0) = 1 \):**
   Substitute \( x = 0 \) and \( y(0) = 1 \) into the equation:
   \[
   1 = \frac{1}{3} + C
   \]
   Solving for \( C \):
   \[
   C = 1 - \frac{1}{3} = \frac{2}{3}
   \]

7. **Final solution:**
   \[
   y(x) = \frac{1}{3} + \frac{2}{3}e^{-3x}
   \]

This is the solution to the differential equation with the given initial condition.

---

Would you like more details on any part of the process? Here are some related questions to expand on:

1. How do we derive the integrating factor for linear differential equations?
2. Can this type of equation be solved using another method?
3. What does the solution \( y(x) = \frac{1}{3} + \frac{2}{3}e^{-3x} \) represent graphically?
4. How does the initial condition affect the general solution?
5. What happens if we change the coefficient \( 3 \) in the equation?

**Tip:** For linear ODEs, finding the integrating factor is a powerful method for solving them efficiently.

Solve the differential equation y' + 3y = 1 with the initial condition y(0) = 1.

Learn how to solve the first-order linear differential equation y' + 3y = 1 with the initial condition y(0) = 1 using the integrating factor method. This problem involves finding the solution for the function y(x) and is suitable for college-level students.

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