Question

The differential equation corresponding to the family of curves \[ y=\log_e(ax+3), \] where \(a\) is an arbitrary constant, is

• A. \(x\dfrac{dy}{dx}+3e^{-x}=1\)
• B. \(x\dfrac{dy}{dx}+3e^{y}=1\)
• C. \(x\dfrac{dy}{dx}+3e^{-y}=1\)
• D. \(x\dfrac{dy}{dx}+3e^{x}=1\)

Hint:

To form a differential equation from a family of curves: \[ \boxed{ \text{Differentiate the given equation and eliminate the arbitrary constant.} } \]

Answer 1

The Correct Option is C

Step 1: Differentiate the given family of curves. Given, \[ y=\log_e(ax+3). \] Differentiating with respect to \(x\), \[ \frac{dy}{dx} = \frac{a}{ax+3}. \]

Step 2: Eliminate the arbitrary constant \(a\). From \[ y=\log_e(ax+3), \] we get \[ e^y=ax+3. \] Hence, \[ a=\frac{e^y-3}{x}. \] Substituting into \[ \frac{dy}{dx} = \frac{a}{e^y}, \] gives \[ \frac{dy}{dx} = \frac{e^y-3}{xe^y}. \] Multiplying by \(x\), \[ x\frac{dy}{dx} = 1-3e^{-y}. \] Therefore, \[ \boxed{ x\frac{dy}{dx}+3e^{-y}=1. } \] Hence, the correct option is \(\boxed{(C)}\).