Discussion
The orthogonal trajectories of the family of curves
\(y = -3x - 3 + me^{x}\) for a real parameter \(m\)
are given by ______.
Answer: C
The orthogonal trajectories are \(x = \frac{-y}{3} + \frac{1}{9} + k\,e^{-3y}\).
First find the differential equation of the family. From \(me^{x} = y + 3x + 3\),
\[ \frac{dy}{dx} = -3 + me^{x} = y + 3x. \]
Orthogonal curves have slope \(-\frac{1}{y + 3x}\), which is linear in \(x\) when \(y\) is the variable:
\[ \begin{aligned} \frac{dx}{dy} &= -(y + 3x) \\ \frac{dx}{dy} + 3x &= -y \end{aligned} \]
Multiply by the integrating factor \(e^{3y}\) and integrate by parts:
\[ \begin{aligned} xe^{3y} &= -\int ye^{3y}\,dy \\ &= -\frac{y}{3}e^{3y} + \frac{1}{9}e^{3y} + k \\ x &= -\frac{y}{3} + \frac{1}{9} + ke^{-3y} \end{aligned} \]
Dropping the minus sign in the orthogonal slope gives \(\frac{y}{3} - \frac{1}{9} + ke^{-3y}\). The options with \(e^{3y}\) multiply the constant by the integrating factor instead of dividing by it.
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