Question

If \(a\) and \(b\) are arbitrary constants, then the differential equation corresponding to the family of curves \(y = \tan (ax + b)\) is?

• A. \((1 + y^2) y'' - 2y' y = 0\)
• B. \((1 + y^2) y'' - 2y^2 y' = 0\)
• C. \((1 + y^2) y'' + 2y^2 y' = 0\)
• D. \((1 + y^2) y'' - 2y' y = 0\)

Hint:

Differentiate implicitly to eliminate constants in the given family of curves and derive the corresponding differential equation.

Answer 1

The Correct Option is A

Start by differentiating the given curve \(y = \tan (ax + b)\). 1. First derivative: \(y' = a \sec^2 (ax + b)\). 2. Second derivative: \(y'' = 2a^2 \sec^2 (ax + b) \tan (ax + b)\). Now, use these to eliminate \(a\) and \(b\) and express the equation in terms of \(y\) and its derivatives. This leads to the differential equation: \[ (1 + y^2) y'' - 2y' y = 0. \]