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Find the equation of the orthogonal trajectories to the given family of curves. In each case, sketch some curves from each family.$$y=c e^{x}$$

$2 x+y^{2}=D$

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

Differential Equations

Campbell University

Harvey Mudd College

University of Nottingham

Lectures

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In mathematics, integratio…

06:55

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03:04

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Find the orthogonal trajec…

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01:26

01:45

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08:24

Determine the orthogonal t…

09:36

in this problem. We're looking for the thaw gurnal trajectories to the curve given to be why equals C times the exponential function each of the X. The first step is to differentiate both sides of that equation. So we'll have that D y t x will be equal to the derivative of a constant C times an exponential What that would give us C times either the ex itself. Since our exponential is to the base e To find the full expression for this derivative, we need to substitute out the constant C since see, it depends on both X and Y from the original family of curves. We know that C is equal to why divide by either the ex. So we'll substitute that back into our derivative for our next step. D y d X is now equal to in place of see we have why divide by you to the X times, even the X. That gives us a quick cancellation in this problem, or the base of the X is cancelled from the numerator denominator. And now we have that D y d X is equal to why itself next to find orthogonal trajectories, We'll take that D Y DX must be equal to the negative reciprocal of the expression we have just found. So our differential equation becomes D Y D X equals negative one over why and to solve this differential equation, let's collect all portions that involved why toe left inside and all the portions involved x to the right hand side. Upon doing that, we have that why d y will be equal to negative one times DX. Then we can integrate both sides the left hand side with the respect to why the right hand side with respect to X to obtain that 1/2 a y squared is equal to negative X, plus a constant. Let's call this one upper case C. If we add extra pill sides, the equation is now X plus. 1/2 of y squared is equal to upper case C, and then we can simplify this equation further by multiplying both sides by two. If we do that, the equation becomes two. X plus y squared equals a constant will call this one D where d equals two times Upper Casey, and this is the equation for the orthogonal trajectories off the original family

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