The equation provided in the image is:

$\text{(tan}^{-1} y - x) dy = (1 + y^2) dx$

Let's solve this differential equation.

Step-by-step solution:

1. Rearrange the equation:

$(\tan^{-1} y - x) dy = (1 + y^2) dx$

This is a first-order differential equation. Let's attempt to separate the variables $x$ and $y$ for easier integration.

2. Rewrite the equation in a separable form:

$\frac{\tan^{-1} y - x}{1 + y^2} dy = dx$

Now, integrate both sides.

3. Integration of the left-hand side:

The integral of $\frac{\tan^{-1} y - x}{1 + y^2} dy$ can be split:

$\int \frac{\tan^{-1} y}{1 + y^2} dy - \int x \, dy$

• The first integral $\int \frac{\tan^{-1} y}{1 + y^2} dy$ looks complicated, but upon simplification, it can be solved using advanced integration techniques.
• The second integral $- \int x \, dy = -x \cdot y$, assuming $x$ is independent of $y$.

4. Integration of the right-hand side:

The integral of $dx$ is simply:

$\int dx = x$

5. Combine the results:

We now combine both sides and introduce the constant of integration, $C$, to arrive at the general solution:

$\text{Solution: (Left side expression)} = x + C$

Where the left side involves solving the integration of $\frac{\tan^{-1} y}{1 + y^2}$, and then we can equate it to $x + C$.

Would you like me to go deeper into the integration process for the left-hand side, or explain any other step in detail?

Further Questions:

1. What are first-order differential equations?
2. How do you approach solving separable differential equations?
3. What is the significance of the inverse trigonometric functions like $\tan^{-1} y$ in calculus?
4. Can you give examples of applications where such differential equations appear?
5. How does the method of integration by parts help in more complex integrals?

Tip: When solving integrals involving inverse trigonometric functions, use substitutions to simplify the expression.