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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.$$y^{2}=2 x+c$$

$y=D e^{-x}$

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

Differential Equations

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in this problem. We're looking for the family of orthogonal trajectories to the curve given to B y squared equals two x plus e. Due to the power of to in this equation, we must differentiate both sides implicitly. If we differentiate the left hand side with respect to X, we'll obtain to why times de y dx equals on the right hand side. We obtain simply a two since the constant C will go to zero in differentiation. We can solve this equation now for D Y dx by divine pull sides by two y so that D Y t X will be to over two. Why or de y? The X is one over Why now? Defying the family of our doggerel trajectories, we formed the differential Equation D y DX, which is equal to the negative reciprocal of the expression just found. So now D Y d X is equal to negative y. Let's divide both sides of this differential equation by why to obtain that one over. Why Times D y. The X is equal to negative one. Then we can use the formula that the derivative with respect to X of the natural logger them of why is equal to one over Why Times de y dx. In other words, what we have over here in our differential equation can be substituted in for D by D. X of the natural log rhythm of why, although that doesn't look like much, progress, will be able to do some cancellations on the step just after Sultry writes that D by D. X of the natural log of why is now equal to negative one. The next step is we can integrate both sides of this equation with respect to X. Let's do that immediately at this step, and we noticed the cancellation immediately. The left hand side is looking for an anti derivative to the derivative, so all that's left is natural log y on the left hand side, giving us a large simplification. So now natural log of why is equal to Negative X plus a constant Let's call it upper case C. We have a log rhythmic equation. If we convert this to a base e exponential equation, we'll obtain that Y is equal to each of the power of negative X plus C, then using properties of exponents, why is now equal to U to the power of C times each of the power of negative X Another, since substitution will allow us to write our answer in the form y equals D times each of the power of negative X, where d equals e to the C. And that's the format for this trajectory of orthogonal curves.

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