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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.$$y=c x^{5}$$

$\frac{5 y}{x}$$x^{2}+5 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 thaw Grenell trajectories to the family of curves, described as why equals C Times X to the power five. To find those families. Let's differentiate both sides of that equation. With respect to X, we'll obtain that D Y. D X is equal to five times. See Time's X to the power for the constant C needs to be expressed in terms of X and y, and that could be done from going to the original equation in dividing both sides by excessive power. Five. We obtained that C is equal to white divide by X to the fifth power than are derivative. De Y T X can now be re expressed as it's equal to five times why divide by X to the power five times X to the power for upon simplifying we obtained. Now that D Y D X is equal to five times why divide by X. Now we can take this expression for the derivative and obtained the following differential equation. Orthogonal trajectories will satisfy the differential equation. That D Y D X is the negative reciprocal of the expression just found. So are negative. Our right hand side will be negative. X divide by five y And to solve this differential equation, I'm going to collect all terms involving why the D Y in the five y toe, left hand side and all other terms involving X to the right hand side. So our differential equation now takes the form that five y de y is equal to negative X dx. We can integrate both sides of this equation the left side with respecto Why and the right site with respect to X. To obtain that five halves of Weiss word is equal to 1/2. Excuse me, negative 1/2 of X squared, plus a constant of integration. Let's call that upper case C. We bring ah half of X squared to both sides of this equation. We obtain 1/2 X squared plus five halves of y squared is equal to that constant C, and upon clearing denominators, the family of orthogonal trajectories is X squared. Plus five y squared equals a new constant D where de is equal to two times upper Casey. And that's the expression for the orthogonal trajectories to the original given family

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