Question

Given \[ y = -3x - 3 + m e^{2x} \] Find the Orthogonal Trajectories (O.T.).

Hint:

To find orthogonal trajectories: first obtain the differential equation of the given family, then replace \( \frac{dy}{dx} \) by its negative reciprocal and solve.

Answer 1

Step 1: Differentiate the given family of curves.
\[ y = -3x - 3 + m e^{2x} \] Differentiate with respect to \( x \):
\[ \frac{dy}{dx} = -3 + 2m e^{2x} \] Step 2: Eliminate the parameter \( m \).
From the original equation, \[ m e^{2x} = y + 3x + 3 \] Substitute into derivative: \[ \frac{dy}{dx} = -3 + 2(y + 3x + 3) \] \[ = -3 + 2y + 6x + 6 \] \[ = 2y + 6x + 3 \] Thus, the differential equation of the given family is: \[ \frac{dy}{dx} = 2y + 6x + 3 \] Step 3: Form differential equation of orthogonal trajectories.
For orthogonal trajectories: \[ \left( \frac{dy}{dx} \right)_{O.T.} = - \frac{1}{2y + 6x + 3} \] Step 4: Solve the equation.
\[ \frac{dy}{dx} = -\frac{1}{2y + 6x + 3} \] Rearrange: \[ (2y + 6x + 3) dy = - dx \] Let \[ u = 2y + 6x + 3 \] Then \[ \frac{du}{dx} = 2\frac{dy}{dx} + 6 \] Substitute \( \frac{dy}{dx} = -\frac{1}{u} \): \[ \frac{du}{dx} = 2\left(-\frac{1}{u}\right) + 6 \] \[ = 6 - \frac{2}{u} \] This gives: \[ \frac{du}{dx} = \frac{6u - 2}{u} \] Separate variables: \[ \frac{u}{6u - 2} du = dx \] Integrate: \[ \int \frac{u}{6u - 2} du = \int dx \] Simplify: \[ \frac{1}{6} \int \left( 1 + \frac{2}{6u - 2} \right) du = x + C \] After integration: \[ \frac{u}{6} + \frac{1}{18} \ln |6u - 2| = x + C \] Substitute back \( u = 2y + 6x + 3 \): \[ \frac{2y + 6x + 3}{6} + \frac{1}{18} \ln |6(2y + 6x + 3) - 2| = x + C \] Final Answer: \[ \boxed{ \frac{2y + 6x + 3}{6} + \frac{1}{18} \ln |12y + 36x + 16| = x + C } \]