Question

Solve the following differential equation: \[ (\tan^{-1}y - x) \, dy = (1 + y^2) \, dx. \]

Hint:

To solve separable differential equations, isolate variables, integrate each side, and simplify the result accordingly.

Answer 1

Rearrange the given equation to separate the variables \( x \) and \( y \): \[ \frac{\tan^{-1}y - x}{1 + y^2} \, dy = dx. \] Rewrite as: \[ \frac{\tan^{-1}y}{1 + y^2} \, dy - x \, dx = 0. \] Step 1: Split the equation. The first term involves \( y \), and the second term involves \( x \). Separate these: \[ \frac{\tan^{-1}y}{1 + y^2} \, dy = x \, dx. \] Step 2: Integrate both sides. For the left-hand side: \[ \int \frac{\tan^{-1}y}{1 + y^2} \, dy. \] Let \( u = \tan^{-1}y \), so \( du = \frac{1}{1 + y^2} \, dy \): \[ \int \frac{\tan^{-1}y}{1 + y^2} \, dy = \int u \, du = \frac{u^2}{2} + C_1 = \frac{(\tan^{-1}y)^2}{2} + C_1. \] For the right-hand side: \[ \int x \, dx = \frac{x^2}{2} + C_2. \] Step 3: Combine the results. \[ \frac{(\tan^{-1}y)^2}{2} = \frac{x^2}{2} + C, \] where \( C = C_2 - C_1 \). Multiply through by 2 to simplify: \[ (\tan^{-1}y)^2 = x^2 + 2C. \] Let \( 2C = C_1 \), so: \[ (\tan^{-1}y)^2 - x^2 = C_1. \] Final Answer: \[ \boxed{(\tan^{-1}y)^2 - x^2 = C_1} \]