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?

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.