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Unformatted text preview: Chapter 2
To ﬁnd the orthogonal trajectories, we must solve this diﬀerential equation. To this end, note that this equation is separable and thus 1 1 dx = dy ⇒ x y ⇒ eln x−C = eln y ln x = ln y  + C ⇒ y = kx, where k = ±e−C . Therefore, the orthogonal trajectories are lines through the origin. (c) Let F (x, y ) = xy . Then we have Fx (x, y ) = y and Fy (x, y ) = x. Plugging these expressions into the ﬁnal result of part (a) gives x dx − y dy = 0. To ﬁnd the orthogonal trajectories, we must solve this diﬀerential equation. To this end, note that this equation is separable and thus x dx = y dy ⇒ x2 y2 = +C 2 2 ⇒ x2 − y 2 = k , where k := 2C . Therefore, the orthogonal trajectories are hyperbolas. 33. We use notations and results of Problem 32, that is, for a family of curves given by F (x, y ) = k , the orthogonal trajectories satisfy the diﬀerential equation ∂F (x, y ) ∂F (x, y ) dx − dy = 0. ∂y ∂x (a) In this problem, F (x, y ) = 2x2 + y 2 and the equation (2.16) becomes ∂ (2x2 + y 2) ∂ (2x2 + y 2 ) dx − dy = 0 ∂y ∂x Separating variables and integrating yield 2y dx = 4x dy ⇒ ⇒ 70 2dy dx = x y ⇒ ln x = 2 ln y  + c1 ⇒ x = ec1 y 2 = c2 y 2 ⇒ ⇒ dx = x 2dy y ⇒ 2y dx − 4x dy = 0. (2.17) (2.16) eln x = e2 ln y+c1 x = ±c2 y 2 = cy 2 , ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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