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Find the orthogonal trajectories of the family of curves. Use a graphing device to draw several members of each family on a common screen.$ y^2 = kx^3 $

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$2 x^{2}+3 y^{2}=c$

Calculus 2 / BC

Chapter 9

Differential Equations

Section 3

Separable Equations

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Idaho State University

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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for this problem we are asked to find the orthogonal trajectories of the family of curves, Y squared equals K X cubed. And then we're asked you are asked to use graphing device draws several members of each family on a common screen. I will be skipping that step as me flooding that. I won't do much for you. So we have two white or starting out differentiating implicitly will get to Y Y prime equals K X cubed or a three times K X squared, I should say three K X squared on the right hand side. That means then we have that Y prime equals 3/2 times X times K and mr K there 3/2 times K X squared over. Why? Which now means that we have the equation for the slope or the slopes of that family of curves. We can reduced down the way that we're writing this a little bit. So it's three K X squared over two Y. So if we want the orthogonal trajectories. All right. Why with a little perpendicular sign, then the derivative there is going to be negative two, Y over three K X squared. So now we have a differential equation. We want to solve. We'll have that Y prime over why is going to equal negative to over three K X squared? Or we can write that as negative 2/3 K tend to X power of negative two. Now integrating both sides. We'll have that the right hand side. Second here, we still have that negative to over 3K. Then for X power of negative two we'll go to explore of negative two plus one. So that's exposure of negative one over negative two plus one. So over negative one that just can't swaps the sign and needs us with an X. In the denominator. And we want to include our constant integration. Then on the left hand side uh integral of one over Why is going to be one of why? So we have then that's why is going to be equal to some C. Times E. To the power of to over three K. X. To over three K. X. Right there.

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