The equation for the trajectories <em>orthogonal to</em> the family of functions of the form 5 · x² - 2 · y² = C is equal to (1 / 5) · ㏑ x + (1 / 2) · ㏑ y = C.
<h3>How to find the equation for the orthogonal trajectories of a given equation</h3>
In this problem we have a family of functions in <em>implicit</em> form, that is, a function of the form f(x, y, c) = 0. The equation of a <em>orthogonal</em> trajectory is always perpendicular to a <em>particular</em> form of a equation. First, we determine the <em>first</em> derivative of the given expression:
5 · x² - 2 · y² = C
10 · x - 4 · y · y' = 0
4 · y · y' = 10 · x
y' = (10 · x) / (4 · y)
y' = (5 · x) / (2 · y)
If f(x, y) = (5 · x) / (2 · y), then the differential equation for the <em>orthogonal</em> trajectories related to the family of functions is:
y' = - 1 / f(x, y)
y' = - (2 · y) / (5 · x)
dy / (2 · y) = - dx / (5 · x)
By indefinite integration we get the following solution to the <em>ordinary differential</em> equation:
(1 / 2) · ㏑ y = - (1 / 5) · ㏑ x + C
(1 / 5) · ㏑ x + (1 / 2) · ㏑ y = C
The equation for the trajectories <em>orthogonal to</em> the family of functions of the form 5 · x² - 2 · y² = C is equal to (1 / 5) · ㏑ x + (1 / 2) · ㏑ y = C.
To learn more on orthogonal trajectories: brainly.com/question/20308962
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