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Can be done and written more concisely by separation of variables. show me each step 

To solve this differential equation using separation of variables, let's go through each step.

Given equation:
\[
\frac{dv_y}{dt} + \frac{b}{m} v_y = g
\]

### Step 1: Rearrange the Equation
Rewrite the equation to separate terms involving \( v_y \) from those involving \( t \):
\[
\frac{dv_y}{dt} = g - \frac{b}{m} v_y
\]

### Step 2: Rewrite in Terms of \( dv_y \) and \( dt \)
Separate the variables \( v_y \) and \( t \):
\[
\frac{1}{g - \frac{b}{m} v_y} \, dv_y = dt
\]

### Step 3: Integrate Both Sides
Integrate both sides with respect to their variables.

#### Left side:
\[
\int \frac{1}{g - \frac{b}{m} v_y} \, dv_y
\]

#### Right side:
\[
\int dt
\]

### Step 4: Solve the Integral on the Left Side
To solve \(\int \frac{1}{g - \frac{b}{m} v_y} \, dv_y\), let’s set up a substitution:
Let \( u = g - \frac{b}{m} v_y \), so \( \frac{du}{dv_y} = -\frac{b}{m} \) or \( dv_y = -\frac{m}{b} \, du \).

Substitute \( u \) and \( dv_y \) in the integral:
\[
\int \frac{1}{u} \cdot -\frac{m}{b} \, du = -\frac{m}{b} \int \frac{1}{u} \, du
\]

This becomes:
\[
-\frac{m}{b} \ln |u| = -\frac{m}{b} \ln \left| g - \frac{b}{m} v_y \right|
\]

#### Right side:
The right side integrates to:
\[
t + C
\]

### Step 5: Combine the Results
Now, we have:
\[
-\frac{m}{b} \ln \left| g - \frac{b}{m} v_y \right| = t + C
\]

### Step 6: Solve for \( v_y \)
Isolate \( v_y \) by exponentiating both sides:
\[
\ln \left| g - \frac{b}{m} v_y \right| = -\frac{b}{m} (t + C)
\]

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

Let \( e^{-\frac{b}{m} C} = C' \), where \( C' \) is a constant, then:
\[
g - \frac{b}{m} v_y = C' e^{-\frac{b}{m} t}
\]

Solving for \( v_y \):
\[
v_y = \frac{mg}{b} - C' e^{-\frac{b}{m} t}
\]

This is the solution for \( v_y \) in terms of \( t \), where \( C' \) can be determined by initial conditions if provided.

Would you like to explore further details or have specific questions on any part?

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**Related Questions:**
1. How can the constant \( C' \) be determined given an initial velocity?
2. What does the term \( \frac{mg}{b} \) represent physically in this equation?
3. How does the solution change if \( g = 0 \)?
4. What if \( b = 0 \), removing the drag component?
5. How does this result compare with the horizontal component?

**Tip:** In separation of variables, isolating \( dv_y \) and \( dt \) early helps simplify the integration process.

Can be done and written more concisely by separation of variables. show me each step

Step-by-step solution to solve a vertical motion differential equation using separation of variables. Learn to integrate and apply substitution to solve for velocity as a function of time in a system affected by gravity and drag. Suitable for undergraduate physics or engineering students.

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