Differential Equations: Homework 8

Size: px

Start display at page:

Download "Differential Equations: Homework 8"

Delphia Stone
5 years ago
Views:

1 Differential Equations: Homework 8 Alvin Lin January 08 - May 08

2 Section.6 Exercise Find a general solution to the differential equation using the method of variation of parameters. y + y = tan(t) r + = 0 r = 0 ± i α = 0 β = y h = c cos(t) + c sin(t) y = cos(t) y = sin(t) y p = v y + v y W [y, y ] = cos(t) sin(t) sin(t) cos(t) = cos (t) + sin (t) = tan(t) sin(t) v = dt = cos (t) dt cos(t) = sec(t) cos(t) dt = ( ) ln sec(t) + tan(t) sin(t) tan(t) cos(t) v = dt = cos(t) ln sec(t) + tan(t) cos(t) sin(t) cos(t) cos(t) sin(t) y p = ln sec(t) + tan(t) cos(t) y = c cos(t) + c sin(t)

3 Exercise Find a general solution to the differential equation using the method of variation of parameters. Exercise y + y = sec(t) r + = 0 r = 0 ± i α = 0 β = y h = c cos(t) + c sin(t) y = cos(t) y = sin(t) y p = v y + v y W [y, y ] = cos(t) sin(t) sin(t) cos(t) = cos (t) + sin (t) = sec(t) sin(t) v = dt = ln sec(t) sec(t) cos(t) v = dt = t y p = ln sec(t) cos(t) + t sin(t) y = c cos(t) + c sin(t) ln sec(t) cos(t) + t sin(t) Find a general solution for the differential equation using the method of variation of parameters. y + y + y = e t r + r + = (r + ) = 0 r = y h = c e t + c te t y = e t y = te t y p = v y + v y W [y, y ] = e t te t e t te t + e t = e t e t v = te dt = t t e t v = dt = t e t y p = t e t + t e t y = c e t + c te t + t e t 3

4 Exercise 5 Find a general solution for the differential equation using the method of variation of parameters. y (θ) + 6y(θ) = sec(θ) r + 6 = 0 r = 0 ± i α = 0 β = y h = c cos(θ) + c sin(θ) y = cos(θ) y = sin(θ) y p = v y + v y W [y, y ] = cos(θ) sin(θ) sin(θ) cos(θ) = cos (θ) + sin (θ) = sec(θ) sin(θ) ln sec θ v = dθ = 6 sec(θ) cos(θ) v = dθ = θ ln sec θ cos(θ) y p = + θ sin(θ) 6 y = c cos(θ) + c sin(θ) + θ sin(θ) ln sec θ cos(θ) 6 Exercise 7 Find a general solution to the differential equation using the method of variation of parameters. y + t + y = e t ln(t) r + t + = (r + ) = 0 r = y h = c e t + c te t y = e t y = te t y p = v y + v y W [y, y ] = e t te t e t te t + e t = e t e t ln(t)te t v = dt = t ln(t) dt = t e t t ln(t) e t ln(t)e t v = dt = ln(t) dt = t ln(t) t e t y p = t e t = t ln(t)e t t ln(t)e t 3t e t y = c e t + c te t + t ln(t)e t + t ln(t)e t t e t 3t e t

5 Exercise Find a general solution to the differential equation..9 Exercise y + y = tan(t) + e 3t r + = 0 r = 0 ± i α = 0 β = y h = c cos(t) + c sin(t) y = cos(t) y = sin(t) y p = v y + v y W [y, y ] = cos(t) sin(t) sin(t) cos(t) = cos (t) + sin (t) = v = (tan(t) + e 3t ) sin(t) dt = tan(t) sin(t) e 3t sin(t) + sin(t) dt = sin(t) ln sec(t) + tan(t) e3t (cos(t) 3 sin(t)) cos(t) 0 v = (tan(t) + e 3t ) cos(t) dt = sin(t) + e 3t sin(t) cos(t) = cos(t) + e3t (sin(t) + 3 cos(t)) 0 y p = e3t ln sec(t) + tan(t) cos(t) 0 sin(t) y = c cos(t) + c sin(t) + e3t ln sec(t) + tan(t) cos(t) 0 A -kg mass is attached to a spring with stiffness k = 50 N. The mass is displaced m to the left of the m equilibrium point and given a velocity of m to the left. Neglecting damping, find the equation of motion sec of the mass along with the amplitude, period, and frequency. How long after release does the mass pass through the equilibrium position? y + 50y = 0 y(0) = y (0) = r + 50 = 0 r = 0 ± 5i α = 0 β = 5 y(t) = c cos(5t) + c sin(5t) y (t) = 5c sin(5t) + 5c cos(5t) y(0) = = c y (0) = = 5c c = 5 y(t) = cos(5t) 5 sin(5t) 5

6 ( ) A = + ( ) = 5 00 T = π ω = π 5 f = ω π = 5 ( π ) ( ) φ = tan c 5 = tan c ( ( )) 5 y(t) = 0 sin 5t + tan = 0 ( ( )) 5 sin 5t + tan = 0 ( ) 5 5t + tan = π t = π tan ( 5 ) 5 Exercise 3 The motion of a mass-spring system with damping is governed by y (t) + by (t) + 6y(t) = 0 y(0) = y (0) = 0 Find the equation of motion and sketch its graph for b = 0, 6, 8, 0. b = 0 y (t) + 6y(t) = 0 r + 6 = 0 r = 0 ± i α = 0 β = y(t) = c cos(t) + c sin(t) y (t) = c sin(t) + c cos(t) y(0) = = c y (0) = 0 = c c = 0 y(t) = cos(t) 6

7 b = 6 y (t) + 6y (t) + 6y(t) = 0 r + 6r + 6 = 0 r = 6 ± 36 ()(6) = 3 ± 7i α = 3 β = 7 y(t) = e 3t (c cos( 7t) + c sin( 7t)) y (t) = e 3t ( 7c sin( 7t) + 7c cos( 7t)) + 3e 3t (c cos( 7t) + c sin( 7t)) y(0) = = c y (0) = 0 = 7c 3c c = 3 7 y(t) = e 3t (cos( 7t) sin( 7t)) b = 8 y (t) + 8y (t) + 6y(t) = 0 r + 8r + 6 = (r + ) = 0 r = y(t) = c e t + c te t y (t) = c e t c te t + c e t y(0) = = c y (0) = 0 = c + c c = y(t) = e t + te t b = 0 y (t) + 0y (t) + 6y(t) = 0 r + 0r + 6 = (r + 8)(r + ) = 0 r = 8 r = y(t) = c e 8t + c e t y (t) = 8c e 8t c e t y(0) = = c + c y (0) = 0 = 8c c c = 3 c = 3 y(t) = 3 e 8t + 3 e t 7

8 Exercise 7 A kg mass is attached to a spring with stiffness 6 N N sec. The damping constant for the system is. If 8 m m the mass is moved 3 m m to the left of equilibrium and given an initial leftward velocity of, determine sec the equation of motion of the mass and give its damping factor, quasiperiod, and quasifrequency. 8 y + y + 6y = 0 y(0) = 3 8 r + r + 6 = 0 r + 6r + 8 = 0 y (0) = r = 6 ± 6 ()(8) = 8 ± 8i α = 8 β = 8 y(t) = e 8t (c cos(8t) + c sin(8t)) y (t) = e 8t ( 8c sin(8t) + 8c cos(8t)) 8e 8t (c cos(8t) + c sin(8t)) y(0) = 3 = c y (0) = = 8c 8c c = y(t) = e 8t ( 3 cos(8t) sin(8t)) = 3e 8t cos(8t) e 8t sin(8t) 5 A = c + c = 6 e 6t = 5 e 8t k 6 ω = m = 0.5 T = π β = π 8 = π f = β π = π If you have any questions, comments, or concerns, please contact me at alvin@omgimanerd.tech 8

Math 308 Exam II Practice Problems

Math 308 Exam II Practice Problems Math 38 Exam II Practice Problems This review should not be used as your sole source for preparation for the exam. You should also re-work all examples given in lecture and all suggested homework problems..