Orthogonal Trajectories/Examples/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:
- $x = y - 2 + C e^{-y}$
Proof
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ with respect to $x$ gives:
- $\dfrac {\d y} {\d x} = 1 - C e^{-x}$
Eliminating $C$:
\(\ds C\) | \(=\) | \(\ds \frac {y - x} {e^{-x} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d y} {\d x}\) | \(=\) | \(\ds 1 - \frac {y - x} {e^{-x} } e^{-x}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d y} {\d x} + y\) | \(=\) | \(\ds x + 1\) |
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
- $-\dfrac {\d y} {\d x} = - x = 1 - y$
The integrating factor is $e^y$, giving:
- $\ds e^y x = \int y e^y - e^y \rd y$
Using Primitive of $x e^{a x}$:
- $\ds \int y e^y \rd y = y e^y - e^y$
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
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.10$: Problem $5$