# Orthogonal Trajectories/x + C exp -x

## Theorem

Consider the one-parameter family of curves:

$(1): \quad y = x + C e^{-x}$

Its family of orthogonal trajectories is given by the equation:

$x = y - 2 + C e^{-y}$ ## Proof

Differentiating $(1)$ with respect to $x$ gives:

$\dfrac {\mathrm d y} {\mathrm d x} = 1 - C e^{-x}$

Eliminating $C$:

 $\displaystyle C$ $=$ $\displaystyle \frac {y - x} {e^{-x} }$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d y} {\mathrm d x}$ $=$ $\displaystyle 1 - \frac {y - x} {e^{-x} } e^{-x}$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d y} {\mathrm d x} + y$ $=$ $\displaystyle x + 1$

Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:

$-\dfrac {\mathrm d y} {\mathrm d x} = - x = 1 - y$

The integrating factor is $e^y$, giving:

$\displaystyle e^y x = \int y e^y - e^y \, \mathrm d y$

Using Primitive of $x e^{a x}$:

$\displaystyle \int y e^y \, \mathrm d y = y e^y - e^y$

Thus we get:

$e^y x = y e^y - e^y - e^y + C$

which gives us:

$x = y - 2 + C e^{-y}$

$\blacksquare$