S23 - 2.3 Some Applications In this section we give some...

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2.3 Some Applications In this section we give some examples of applications of linear and separable differential equations. 1. Orthogonal Trajectories The one-parameter family of curves ( x - 2) 2 +( y - 1) 2 = C ( C 0) (a) is a family of circles with center at the point (2 , 1) and radius C . -1 1 2 3 4 5 -2 -1 1 2 3 4 If we differentiate this equation with respect to x , we get 2( x - 2) + 2( y - 1) y ± =0 and y ± = - x - 2 y - 1 (b) This is the differential equation for the family of circles. Note that if we choose a speciFc point ( x 0 ,y 0 ) 0 ± = 1 on one of the circles, then (b) gives the slope of the tangent line at ( x 0 0 ). Now consider the family of straight lines passing through the point (2 , 1): y - 1= K ( x - 2) . (c) -1 1 2 3 4 5 -4 -2 2 4 6 The differential equation for this family is y ± = y - 1 x - 2 (verify this) (d) Comparing equations (b) and (d) we see that right side of (b) is the negative reciprocal of the right side of (d). Therefore, we can conclude that if P ( x 0 0 ) is a point of intersection of one of 34
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the circles and one of the lines, then the line and the circle are perpendicular (orthogonal) to each other at P . The following Fgure shows the two families drawn in the same coordinate system. -1 1 2 3 4 5 -3 -2 -1 1 2 3 4 5 A curve that intersects each member of a given family of curves at right angles (orthogonally) is called an orthogonal trajectory of the family. Each line in (c) is an orthogonal trajectory of the family of circles (a) [and conversely, each circle in (a) is an orthogonal trajectory of the family of F ( x,y,c ) = 0 and G ( x,y,K )=0 are one-parameter families of curves such that each member of one family is an orthogonal trajectory of the other family, then the two families are said to be orthogonal trajectories. A procedure for Fnding a family of orthogonal trajectories G ( ) = 0 for a given family of curves F ( x,y,C ) = 0 is as follows: Step 1. Determine the differential equation for the given family F ( )=0. Step 2. Replace y ± in that equation by - 1 /y ± ; the resulting equation is the differential equation for the family of orthogonal trajectories. Step 3. ±ind the general solution of the new differential equation. This is the family of orthogonal trajectories. Example ±ind the orthogonal trajectories of the family of parabolas y = Cx 2 . SOLUTION You can verify that the differential equation for the family y = 2 can be written as y ± = 2 y x . Replacing y ± by - 1 /y ± , we get the equation - 1 y ± = 2 y x which simpliFes to y ± = - x 2 y a separable equation. Separating the variables, we get 2 yy ± = - x or 2 ydy = - xdx. integrating with respect to x , we have y 2 = - 1 2 x 2 + C or x 2 2 + y 2 = C. 35
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This is a family of ellipses with center at the origin and major axis on the x -axis. ± -4 -2 2 4 -3 -2 -1 1 2 3 Exercises 2.3.1 Find the orthogonal trajectories for the family of curves. 1. y = Cx 3 . 2. x = Cy 4 .
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S23 - 2.3 Some Applications In this section we give some...

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