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$m$ denotes a fixed nonzero constant, and $c$ is the constant distinguishing the different curves in the given family. In each case, find the equation of the orthogonal trajectories.$$y^{2}=m x+c$$

$y=D e^{-\frac{2}{m} x}$

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

Differential Equations

Harvey Mudd College

University of Nottingham

Boston College

Lectures

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for this problem were given a non zero constant M. And our goal is to find the orthogonal trajectories to this family of curves y squared equals M X plus e due to the power to on the dependent variable. Why we must differentiate implicitly with both sides on both sides with respect to X. So starting on the left hand side, the derivative of y squared turns into two y times de y dx. The driven of the right hand side will become just m itself, since C is constant and X is raised to the power one. So now if we saw for D y t X, we have that D y. T X is equal to m divide by two y. The next step in this process is now to solve the differential equation that D Y DX is equal to the negative reciprocal of that derivative we just found. So that would be equal to negative two y divide by M. So our goal now is to solve this differential equation. Let's take this next step of dividing both sides just by why Now we obtain that one over. Why Times d y the X equals negative to over em. In our next step, we can write the following expression that the left hand side is equivalent to the derivative with respect to X of the natural log of why itself and they'll be equal to negative two over em. The reason we can make this substitution is that the drift of the natural log brings white to the denominator. And since we're differentiating with X respecto X implicitly, we still multiply by the D Y DX are now. Our next step is to find the anti derivative of both sides of that equation with respect to X. That gives us an immediate cancellation because the left hand side is looking for the anti derivative of a derivative. But that leaves just natural log y. So now the equation is the natural log of y is equal to negative two over em Times X plus a cost of integration. Let's call that one upper case C. Now this log rhythmic equation can be turned into an Exponential Basie equation by writing Why equals he to the power of naked of two over m times X plus C using properties of exponents. This can also be expressed as why equals each of the power of C times each of the power of negative two over em. Times X. Let's make one final substitution to express her answer in the form. Why equals de times each of the power of negative two over M times X, where d equals e to the power? See, then this expression will be the family of orthogonal trajectories that we were looking for.

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