Step 1: Understanding the Topic
The question asks for the orthogonal trajectories of a given family of curves. An orthogonal trajectory is a curve that intersects every curve in the given family at a right angle. The process involves finding the differential equation that describes the given family and then finding a second differential equation for the orthogonal family, which is then solved.
Step 2: Key Approach
1. Differentiate the given equation with respect to $x$ to find $\frac{dy}{dx}$.
2. Eliminate the parameter $m$ from the original and differentiated equations to get a differential equation of the form $\frac{dy}{dx} = F(x, y)$.
3. The slope of the orthogonal trajectories is the negative reciprocal: $(\frac{dy}{dx})_{OT} = -\frac{1}{F(x, y)}$.
4. Solve this new differential equation to find the equation for the orthogonal trajectories.
Step 3: Detailed Calculation
A. Find the differential equation for the given family:
The given family is $y = -3x - 3 + m e^{2x}$.
Differentiating with respect to $x$:
\[
\frac{dy}{dx} = -3 + 2m e^{2x}
\]
To eliminate $m$, we first solve the original equation for $m e^{2x}$:
\[
m e^{2x} = y + 3x + 3
\]
Substitute this into the derivative equation:
\[
\frac{dy}{dx} = -3 + 2(y + 3x + 3) = -3 + 2y + 6x + 6
\]
\[
\frac{dy}{dx} = 6x + 2y + 3
\]
B. Form the differential equation for the orthogonal trajectories:
Replace $\frac{dy}{dx}$ with $-\frac{1}{dy/dx}$ (or more formally, $\frac{dy}{dx}$ becomes $-\frac{dx}{dy}$):
\[
\left(\frac{dy}{dx}\right)_{OT} = -\frac{1}{6x + 2y + 3}
\]
C. Solve the new differential equation:
\[
\frac{dy}{dx} = -\frac{1}{6x + 2y + 3} \implies (6x + 2y + 3)dy = -dx
\]
\[
(6x + 2y + 3)dy + dx = 0
\]
This is not easily separable. Let's use a substitution. Let $u = 6x + 2y + 3$. Then $\frac{du}{dx} = 6 + 2\frac{dy}{dx}$.
From the OT equation, $\frac{dy}{dx} = -\frac{1}{u}$. Substitute this:
\[
\frac{du}{dx} = 6 + 2\left(-\frac{1}{u}\right) = \frac{6u - 2}{u}
\]
Now we can separate variables:
\[
\frac{u}{6u - 2} du = dx
\]
To integrate the left side, we use partial fractions or algebraic manipulation:
\[
\int \frac{u}{6u - 2} du = \int \frac{1}{6}\frac{6u}{6u - 2} du = \frac{1}{6} \int \frac{6u - 2 + 2}{6u - 2} du = \frac{1}{6} \int \left(1 + \frac{2}{6u - 2}\right) du
\]
\[
= \frac{1}{6} \left( u + \frac{2}{6}\ln|6u-2| \right) = \frac{u}{6} + \frac{1}{18}\ln|6u-2|
\]
The integral of the right side is simply $x$. So, the solution is:
\[
\frac{u}{6} + \frac{1}{18}\ln|6u-2| = x + C
\]
D. Substitute back for u:
Replacing $u = 6x + 2y + 3$:
\[
\frac{6x + 2y + 3}{6} + \frac{1}{18}\ln|6(6x + 2y + 3) - 2| = x + C
\]
\[
x + \frac{y}{3} + \frac{1}{2} + \frac{1}{18}\ln|36x + 12y + 16| = x + C
\]
Simplifying gives the final form of the solution.
Step 4: Final Answer
The equation for the orthogonal trajectories is:
\[
\boxed{
\frac{2y + 6x + 3}{6} + \frac{1}{18}\ln |12y + 36x + 16| = x + C
}
\]