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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.$$y=c / x$$

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

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 orthogonal trajectories to the family of curves described ABS y equals a constant C divide by X. To find those trajectories will differentiate the left side of the equation, obtaining D Y DX and the right side of the equation. D by D X of C over X. So the left hand side is just the derivative of why. So just call that the white the excess before the right hand side will turn into negative see divide by X squared. So we have an expression for the derivative. But I expression involves a constant C well, so we'll have some work to do to determine what the drifter will look like after we saw for C, we solve for C by going to the original equation. Multiply pull sides by X and we get C is equal to X times y so we'll obtain negative X times. Why divide by X squared, which simplifies into the expression de y DX is equal to negative. Why divide by X So the family of curves like we'll see Divide by X has derivative. Wyche will excuse me. D y d X equals negative white divide by X to now find the orthogonal trajectories will take the negative reciprocal zone. The right hand side network with an expression de y DX equals X divide by Why to solve this differential equation, I'm going to involve bring all parts involving Why, to the left hand side, that would be the y and why and all parts involving X to the right hand side. Our differential equation will then be why times D y is equal to X times DX. Let's next integrate both sides and find the anti dread of each side. The left hand side will be, ah, half of y squared, and likewise, the right hand side is also half of X squared, plus a constant of integration. Let's call that upper Casey. The next step we can take is to subtract both sides by well half X squared to obtain negative 1/2 of X squared, plus 1/2 of Weiss. Word equals the constant C. If we multiply both sides of the equation by negative too, the equation now becomes X squared minus twice. Word equals a new constant D, where de is equal to negative two times see, which is still in arbitrary constant. So this equation represents the family of curves orthogonal to the original family

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