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The image contains differential equations that need to be solved for \( y(t) \) with the initial condition \( y(0) = y_0 \). Let's break down and solve each one:

### Part a)
\[
a \cdot \dot{y} + p \cdot y = 0
\]

This is a first-order linear differential equation. Let's solve it step by step:

1. Rewrite the equation:
   \[
   \dot{y} = -\frac{p}{a} \cdot y
   \]

2. This can be solved using the separation of variables technique. Separate \( y \) and \( t \) as follows:
   \[
   \frac{1}{y} \, dy = -\frac{p}{a} \, dt
   \]

3. Integrating both sides:
   \[
   \int \frac{1}{y} \, dy = \int -\frac{p}{a} \, dt
   \]
   \[
   \ln |y| = -\frac{p}{a} t + C
   \]

4. Solve for \( y \):
   \[
   y(t) = C_1 e^{-\frac{p}{a} t}
   \]

5. Apply the initial condition \( y(0) = y_0 \):
   \[
   y_0 = C_1 e^0 = C_1
   \]
   Therefore, \( C_1 = y_0 \), and the final solution is:
   \[
   y(t) = y_0 e^{-\frac{p}{a} t}
   \]

### Part b)
\[
m \cdot \dot{y} + b \cdot y = 0
\]

This is of the same form as part (a). Following the same steps:

1. Rewrite:
   \[
   \dot{y} = -\frac{b}{m} \cdot y
   \]

2. Separate variables and integrate:
   \[
   \frac{1}{y} \, dy = -\frac{b}{m} \, dt
   \]
   \[
   \ln |y| = -\frac{b}{m} t + C
   \]
   \[
   y(t) = C_2 e^{-\frac{b}{m} t}
   \]

3. Apply the initial condition \( y(0) = y_0 \):
   \[
   y(t) = y_0 e^{-\frac{b}{m} t}
   \]

### Part c)
\[
R \cdot C \cdot \dot{y} + y = 0
\]

Again, this is a first-order linear differential equation. Follow the same method as in parts (a) and (b):

1. Rewrite:
   \[
   \dot{y} = -\frac{1}{R \cdot C} \cdot y
   \]

2. Separate variables and integrate:
   \[
   \frac{1}{y} \, dy = -\frac{1}{R \cdot C} \, dt
   \]
   \[
   \ln |y| = -\frac{1}{R \cdot C} t + C
   \]
   \[
   y(t) = C_3 e^{-\frac{1}{R \cdot C} t}
   \]

3. Apply the initial condition \( y(0) = y_0 \):
   \[
   y(t) = y_0 e^{-\frac{1}{R \cdot C} t}
   \]

### Summary of Solutions:
- **(a)** \( y(t) = y_0 e^{-\frac{p}{a} t} \)
- **(b)** \( y(t) = y_0 e^{-\frac{b}{m} t} \)
- **(c)** \( y(t) = y_0 e^{-\frac{1}{R \cdot C} t} \)

These are standard solutions for first-order linear differential equations with constant coefficients.

Would you like further details or explanations on any of these solutions?

Here are 5 follow-up questions to expand your understanding:
1. How do you solve non-homogeneous first-order differential equations?
2. What is the physical meaning of these differential equations in electrical circuits?
3. How would you approach solving a second-order differential equation?
4. What is the significance of the exponential term in the solutions?
5. How does the method of integrating factors apply to solving linear differential equations?

**Tip:** When solving first-order differential equations, always try separation of variables first—it simplifies many common problems!

Solve the differential equation for y(t) with y(0) = y_0. a) a·y' + p·y = 0, b) m·y' + b·y = 0, c) R·C·y' + y = 0

Step-by-step solutions to solve first-order linear differential equations with constant coefficients. Equations include: a·y' + p·y = 0, m·y' + b·y = 0, and R·C·y' + y = 0. Learn the method of separation of variables and apply initial conditions.

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