https://file.aiquickdraw.com/mathbot/file/1730320783370y7wh6l1w.jpg

The expression in the image is:

\[
\frac{v - 1}{v} \, dv = \frac{du}{u}
\]

This looks like a separable differential equation, where we can integrate both sides with respect to their variables \( v \) and \( u \).

To solve this:

1. Rewrite it as:
   \[
   \frac{v - 1}{v} \, dv = \frac{du}{u}
   \]

2. Simplify \(\frac{v - 1}{v}\) as:
   \[
   \frac{v - 1}{v} = 1 - \frac{1}{v}
   \]
   So the equation becomes:
   \[
   \left(1 - \frac{1}{v}\right) dv = \frac{du}{u}
   \]

3. Now, distribute \( dv \):
   \[
   dv - \frac{1}{v} \, dv = \frac{du}{u}
   \]

4. Integrate both sides:
   \[
   \int \left(dv - \frac{1}{v} \, dv\right) = \int \frac{du}{u}
   \]

   This separates to:
   \[
   \int dv - \int \frac{1}{v} \, dv = \int \frac{du}{u}
   \]

5. Solve each integral:
   \[
   v - \ln |v| = \ln |u| + C
   \]

where \( C \) is the constant of integration.

Would you like further details, or do you have any questions about this solution?

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Here are some related questions to deepen understanding:

1. How can we determine the constant of integration in a practical problem?
2. What are the properties of logarithmic functions useful in simplifying integration results?
3. Can you use a similar approach to solve equations with different fractions on each side?
4. How do separable differential equations differ from exact differential equations?
5. What applications in physics use differential equations similar to this form?

**Tip:** When dealing with fractions in differential equations, try simplifying terms to make integration straightforward.

Rewrite the equation \( \frac{v - 1}{v} \, dv = \frac{du}{u} \) and solve by separation of variables.

Step-by-step solution for solving the differential equation \( \frac{v - 1}{v} dv = \frac{du}{u} \) using separation of variables. Covers core concepts in calculus including integration and separation techniques.

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