527solns6

527solns6 - 642:527 SOLUTIONS: ASSIGNMENT 6 FALL 2007 NOTE:...

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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 6 FALL 2007 NOTE: A. There is a Maple worksheet on the Assignments and solutions web page which shows the phase plane portraits for some of these problems. I do not know how to have Maple put arrows on curves, so for many of these problems I have included the direction field in the plots; you can read off the direction of flow from this field.. B. Vectors are denoted either by boldface or, for Greek letters, underlining: . Section 7.2: 1. Let us write (5) as my 2 / 2 + mx 2 / 2 = E , where E is the energy. Then solving for y gives y = dx/dt = radicalbig (2 E kx 2 ) /m , and separating variables we have integraldisplay dt = t + t = radicalbigg m 2 E integraldisplay dx radicalbig (1 ( k/ 2 E ) x 2 ) = radicalbigg m k integraldisplay du radicalbig (1 u 2 ) = 1 sin 1 u, where we have made the change of variables u = radicalbig k/ 2 E x and written = radicalbig k/m . This gives x = E sin( t + ) with A = radicalbig 2 E/k and = t . 4. (a) See solution in text. (b) Here dy/dx = x/y so that y dy = x dx , and integrating we find that the trajectories lie on the curves x 2 + y 2 = c 2 , that is, on circles. From dy/dt = x 2 we see that the flow is always downward along the cirlces, so that the trajectories are in fact semicircles, traced from the positive y axis (at t = ) to the negative y axis ( t = ). All points on the y axis are singular points, since x = y = 0 if x = 0. 5. (a) See solution in text. (b) Here dy/dx = 1 so the solution curves are y = x + C . From dy/dt = y we see that the flow is upward in the upper half plane and downward in the lower; all points on the x axis are singular points. I leave the sketch to you. Note that the solution satisfying x (0) = x , y (0) = y is y = y e t , x = y e t + C = y ( e t 1) + x . (c) Think of x as position so tht y is velocity; then this is a particle of mass 1 moving under a repulsive spring-like force. We have dy/dx = x/y and separation of variables gives x 2 y 2 = C as the trajectories; these are hyperbolas with asymptotes y = x . The flow is always to the right in the upper half plane, to the left in the lower half plane. See the Maple worksheet on the web page (f) This is similar to 4 (b); dy/dx = 4 so the tracjectories are y = 4 x + C , i.e., straight lines of slope 4. From dx/dt = x we see that flow is to the right in the right half plane and...
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527solns6 - 642:527 SOLUTIONS: ASSIGNMENT 6 FALL 2007 NOTE:...

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