Explanation Given: The orthogonal trajectories to ellipse is 2 x 2 + y 2 = a 2 . The differential equation is y ′ = y 2 x . Calculation: Consider 2 x 2 + y 2 = a 2 . Differentiate with respect to the x . 2 x + 2 y d y d x = 0 2 x + 2 y y ′ = 0 y ′ = − x y Hence, the implicit differentiation of the family of circles x 2 + y 2 = a 2 is y ′ = − x y . The slope of the orthogonal trajectories at a point is the negative multiplicative inverse of the given curve, the orthogonal trajectories satisfies the differential equation of the first order. That is, y ′ = − 1 f ( x , y ) y ′ = − 1 − x y y ′ = y x So, curves are said to be orthogonal if their slope are negative reciprocal to each other

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ISBN: 9781323178522

Author: Briggs

Publisher: PEARSON

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Chapter D1.3, Problem 43E

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To show: The given differential by applying implicit differential to 2x2+y2=a2 ; explain the family of orthogonal trajectories to 2x2+y2=a2 satisfies the differential equation and to solve the given differential equation which satisfies the orthogonal trajectories and to explain y2=kx for y2=ec|x| .

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Chapter D1.3, Problem 42E

Chapter D1.3, Problem 44E

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Solution Summary: The author explains how the orthogonal trajectories to ellipse are satisfied by applying implicit differential to 2x2+y2. The differential equation is

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