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# chapter1 - Chapter 1 Solutions to 1.1 1 d2y/dt2 = g dy/dt =...

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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 initial–value problem is: y(t) = gt 2 /2. The object hits the ground at the time, t 0 , when y(t 0 ) = 100. Hence 100 = gt 0 2 /2, so that t 0 = (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 initial–value problem d 2 y/dt 2 = g, y(0) = 0, dy/dt(0) = v 0 . 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 0 c 1 = v 0 . Hence the solution to the initial–value problem is y(t) = gt 2 /2 + v 0 t. We are given that y(t 0 ) = h. Hence h = gt 0 2 + v 0 t 0 . Solving for v 0 yields v 0 = (2h – gt 0 2 )/(2t 0 ). 5. y(t) = c 1 cos ϖ t + c 2 sin ϖ t dy/dt = –c 1 ϖ sin ϖ t + c 2 ϖ cos ϖ t d 2 y/dt 2 = –c 1 ϖ 2 cos ϖ t – c 2 ϖ 2 sin ϖ t = – ϖ 2 (c 1 cos ϖ t + c 2 sin ϖ t) = 2 y. Consequently, d 2 y/dt 2 + ϖ 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 ϖ t + c 2 f8e5f8e5f8e5f8e5f8e5 c 1 2 + c 2 2 sin ϖ t . We can now define an angle φ by cos φ = c 1 f8e5f8e5f8e5f8e5f8e5 c 1 2 + c 2 2 , sin φ = 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 ϖ t cos φ + sin ϖ t sin φ ) = f8e5f8e5f8e5f8e5 c 1 2 + c 2 2 cos( ϖ t – φ ). 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 = 0 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 .

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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 m–1 , 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 . 15. y 2 = mx + c dy/dx = m/(2y) . Orthogonal trajectories satisfy: dy/dx = –2y/m
Sec. 1.1 How Differential Equations Arise 3 y –1 dy/dx = –2/m ln|y| = – 2x/m + c 1 y = c 2 e –2x/m . 17. m 1 = tan(a 1 ) = tan(a 2 – a) = tan(a 2 ) – tan(a) 1 + tan(a 2 )tan(a) = m 2 – tan(a) 1 + m 2 tan(a) .

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chapter1 - Chapter 1 Solutions to 1.1 1 d2y/dt2 = g dy/dt =...

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