Question

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

Hint:

Use substitution for complex differential equations involving inverse trigonometric functions.

Answer 1

Step 1: Rearrange terms. \[ \frac{dy}{dx} = \frac{1 + y^2}{\tan^{-1} y - x} \] Using substitution \( v = \tan^{-1} y \Rightarrow dv = \frac{dy}{1 + y^2} \), \[ \frac{dv}{dx} = \frac{1}{v - x} \] Step 2: Solve using integration. \[ \int \frac{dv}{v - x} = \int dx \] \[ \ln |v - x| = x + C \] Replacing \( v = \tan^{-1} y \), \[ \ln |\tan^{-1} y - x| = x + C \]