Orthogonal Trajectories/x + C exp -x

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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:

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


XplusCExpMinusXOrthogonalTrajectories.png


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:

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

$\blacksquare$


Sources