How can orthogonal trajectories be found for a specific family of curves?

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SUMMARY

The discussion focuses on finding orthogonal trajectories for the family of curves defined by the equation (x-c)² + y² = c². The initial approach involves deriving the first derivative, resulting in y' = -(x-c)/y, and subsequently substituting c with (x² + y²) / 2x. The transformation leads to y' = 2xy / (x² - y²). However, it is established that this is not a separable differential equation but a homogeneous equation, allowing the substitution of v = y/x to convert it into a separable form.

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  • Understanding of differential equations, specifically homogeneous equations.
  • Familiarity with the concept of orthogonal trajectories in calculus.
  • Proficiency in substitution methods for solving differential equations.
  • Knowledge of derivatives and their applications in curve analysis.
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  • Study the method of finding orthogonal trajectories in more detail.
  • Learn about homogeneous differential equations and their properties.
  • Explore substitution techniques in solving differential equations, particularly the v = y/x substitution.
  • Practice solving separable differential equations to enhance problem-solving skills.
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intenzxboi
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The set of orthogonal trajectories for the family indicated by
( x-c)^2 + y^2 = c^2

My work:

y' = -(x-c)/y Since c= ( x^2 + y^2 ) / 2x

plugging back in and doing -1/y' i got

y' = 2xy / ( x^2 - y^2)

Then I am supposed to move the x and y to a side and integrate but i don't see how it is possible.
 
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intenzxboi said:
The set of orthogonal trajectories for the family indicated by
( x-c)^2 + y^2 = c^2

My work:

y' = -(x-c)/y Since c= ( x^2 + y^2 ) / 2x

plugging back in and doing -1/y' i got

y' = 2xy / ( x^2 - y^2)

Then I am supposed to move the x and y to a side and integrate but i don't see how it is possible.
You can't. This is NOT a "separable" differential equation. It is, however, a "homogeneous" equation because replacing x by a x and y by ay on the right gives 2(ax)(ay)((ax)^2- (ay)^2)= 2a^2xy/(a^2(x^- y^2)= 2xy/(x^2- y^2) again. That essentially means that the right hand side can be written as a function of y/x. Let v= y/x so that y= xv, y'= xv'+ v and replace y in the right side by xv to get a separable equation for v as a function of x.
 

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