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show mw the next steps with detailed explanations

To proceed with further steps, let’s analyze the given equation and understand the context. The equation provided is:

\[
v_y(t) = \frac{mg}{b} + C e^{-\frac{mg}{b}t}
\]

where:
- \( C \) is a constant of integration.
- \( m \) is likely the mass of an object.
- \( g \) represents gravitational acceleration.
- \( b \) is likely a damping or resistance coefficient.
- \( t \) is time, and \( v_y(t) \) is the velocity of the object in the \( y \)-direction as a function of time.

This equation is typical in differential equations where the forces acting on an object are modeled with gravity and air resistance (or damping) that is proportional to the velocity.

### Steps for the Next Part of the Solution:

1. **Interpretation of the Solution Form**:
   This solution form indicates that the velocity \( v_y(t) \) approaches a terminal velocity over time. The term \( \frac{mg}{b} \) represents the steady-state or terminal velocity, which is the velocity the object approaches as \( t \to \infty \), because \( e^{-\frac{mg}{b}t} \to 0 \) as \( t \) increases.

2. **Determine the Constant \( C \)**:
   To fully specify the solution, you would need to determine the value of \( C \). This is typically done by applying an initial condition. For instance, if the initial velocity at \( t = 0 \) is \( v_y(0) = v_0 \), then we substitute \( t = 0 \) and \( v_y(0) = v_0 \) into the equation:

   \[
   v_y(0) = \frac{mg}{b} + C e^{-\frac{mg}{b} \cdot 0}
   \]
   which simplifies to
   \[
   v_0 = \frac{mg}{b} + C
   \]
   Solving for \( C \):
   \[
   C = v_0 - \frac{mg}{b}
   \]

   Now, you can rewrite the solution as:
   \[
   v_y(t) = \frac{mg}{b} + \left(v_0 - \frac{mg}{b}\right)e^{-\frac{mg}{b}t}
   \]

3. **Interpret the Final Form**:
   This equation shows that \( v_y(t) \) consists of two parts:
   - The term \( \frac{mg}{b} \) represents the terminal velocity (the velocity the object approaches over time).
   - The exponential term \( \left(v_0 - \frac{mg}{b}\right)e^{-\frac{mg}{b}t} \) represents the transient behavior of the velocity, which decays over time due to the damping effect.

4. **Analyze the Behavior Over Time**:
   As \( t \) increases, \( e^{-\frac{mg}{b}t} \) approaches zero, meaning the transient term vanishes, and the velocity \( v_y(t) \) approaches the terminal velocity \( \frac{mg}{b} \).

5. **Conclusion and Interpretation in Physical Context**:
   The equation describes how an object subjected to gravitational force and air resistance (or damping) will eventually reach a constant velocity, known as the terminal velocity, due to the balance of forces. The initial velocity affects only the transient phase.

Would you like more details on each step or have any questions?

Here are some additional questions for exploration:
1. How would you find the position function \( y(t) \) if given this velocity \( v_y(t) \)?
2. What happens if the damping coefficient \( b \) is very small or very large?
3. How does this solution compare to a scenario without damping?
4. Can you find the time \( t \) at which \( v_y(t) \) reaches a certain percentage of the terminal velocity?
5. How would this equation change if there were additional forces acting on the object?

**Tip:** When dealing with differential equations involving exponential terms, identifying initial conditions can simplify and finalize the solution significantly.

Show me the next steps with detailed explanations for solving the differential equation \( v_y(t) = \frac{mg}{b} + C e^{-\frac{mg}{b}t} \).

Learn the detailed steps to solve a differential equation with exponential decay. The problem involves finding the velocity \( v_y(t) \) of an object influenced by gravity and damping, with a solution involving constants of integration and terminal velocity concepts. Suitable for undergraduate students.

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