chapter1 - 1 Chapter 1 Solutions to 1.1 1. d 2 y/dt 2 = g...

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Unformatted text preview: 1 Chapter 1 Solutions to 1.1 1. d 2 y/dt 2 = g dy/dt = gt + c 1 y(t) = gt 2 /2 + c 1 t + c 2 . Now impose the initial conditions. y(0) = 0 c 2 = 0. dy/dt(0) = 0 c 1 = 0. Hence the solution to the initialvalue problem is: y(t) = gt 2 /2. The object hits the ground at the time, t , when y(t ) = 100. Hence 100 = gt 2 /2, so that t = (200/g) 1/2 4.52 s, where we have taken g = 9.8 ms 2 . 3. If y(t) denotes the displacement of the object from its initial position at time t, the motion of the object can be described by the initialvalue problem d 2 y/dt 2 = g, y(0) = 0, dy/dt(0) = v . We first integrate the differential equation: d 2 y/dt 2 = g dy/dt = gt + c 1 y(t) = gt 2 /2 + c 1 t + c 2 . Now impose the initial conditions. y(0) = 0 c 2 = 0. dy/dt(0) = v c 1 = v . Hence the solution to the initialvalue problem is y(t) = gt 2 /2 + v t. We are given that y(t ) = h. Hence h = gt 2 + v t . Solving for v yields v = (2h gt 2 )/(2t ). 5. y(t) = c 1 cos t + c 2 sin t dy/dt = c 1 sin t + c 2 cos t d 2 y/dt 2 = c 1 2 cos t c 2 2 sin t = 2 (c 1 cos t + c 2 sin t) = - 2 y. Consequently, d 2 y/dt 2 + 2 y = 0. To determine the amplitude of the motion we write the solution to the differential equation in the equivalent form y(t) = f8e5f8e5f8e5f8e5f8e5 c 1 2 + c 2 2 c 1 f8e5f8e5f8e5f8e5 c 1 2 + c 2 2 cos t + c 2 f8e5f8e5f8e5f8e5f8e5 c 1 2 + c 2 2 sin t . We can now define an angle by cos = c 1 f8e5f8e5f8e5f8e5f8e5 c 1 2 + c 2 2 , sin = c 2 f8e5f8e5f8e5f8e5 c 1 2 + c 2 2 . Then the expression for the solution to the differential equation is y(t) = f8e5f8e5f8e5f8e5f8e5 c 1 2 + c 2 2 (cos t cos + sin t sin ) = f8e5f8e5f8e5f8e5 c 1 2 + c 2 2 cos( t ). Consequently the motion corresponds to an oscillation with amplitude A = f8e5f8e5f8e5f8e5f8e5 c 1 2 + c 2 2 . 7. Given family of curves satisfies: x 2 + 4y 2 = c 1 2x + 8y dy/dx = dy/dx = x/(4y) . Orthogonal trajectories satisfy: y 1 dy/dx = 4/x d(ln |y|)/dx = 4/x ln |y| = 4ln |x| + c 2 y = kx 4 , where k = e c 2 . 2 Ch. 1 First-Order Differential Equations 9. Given family of curves satisfies: y = cx 2 c = yx 2 . Hence dy/dx = 2cx = 2(yx 2 )x = 2y/x. Orthogonal trajectories satisfy: dy/dx = x/(2y) 2ydy/dx = x d(y 2 )/dx = x x 2 + 2y 2 = k. 11. Given family of curves satisfies: y = ce x dy/dx = ce x = y. Orthogonal trajectories satisfy: dy/dx = 1/y ydy/dx = 1 y 2 /2 = x + c 1 y 2 = 2x + c 2 . 13. y = cx m dy/dx = cmx m1 , but c = y/x m so dy/dx = my/x. Orthogonal trajectories satisfy: dy/dx = x/(my) ydy/dx = x/m d(y 2 /2)/dx = x/m y 2 = 1 m x 2 + c 1 ....
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This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue University-West Lafayette.

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chapter1 - 1 Chapter 1 Solutions to 1.1 1. d 2 y/dt 2 = g...

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