#$&* course MTH 279 Query 21 Differential Equations*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I feel pretty confident about most of my answer, but I’m not entirely sure about my initial conditions. The book said to assume that y(0) = 0 and y’(0) = 0. And since y(t) is measured from the equilibrium position, which would be 0, y(0) = 0. I assumed that since the force hadn’t initially given the mass a velocity, that y’(0) = 0. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: The motion of a mass is governed by the equation m y '' + 2 gamma y ' + omega_0^2 y = F(t), with m = 2 kg, gamma = 8 kg / s and k = 80 N / m and F(t) = 20 N * e^(- t s^-1). Solve the equation for the function y(t). What is the long-term behavior of this system? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: λ^2 + 2γ + ω^2 = 0 λ = -4±2i y_c = (A+B)e^(-4t)cos(2t) + (B-A)e^(-4t)sin(2t) y_p = Ce^(-t) y_p’ = -Ce^(-t) y_p” = Ce^(-t) 26Ce^(-t) = 20e^(-t) C = 10/13 y_p = 10/13*e^(-t) y(t) = (A+B)e^(-4t)cos(2t) + (B-A)e^(-4t)sin(2t) + 10/13*e^(-t) This is as far as we can go without some initial conditions. As t approaches infinity, the solution goes to 0, asymptotically. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: Solve the equation y '' + 2 delta y ' + omega_0^2 y = F cos( omega_1 * t), y(0) = 0, y ' (0) = 0. Give an outline of your work. A very similar problem was set up and partially solved in class on 110309, and your text gives the solution but not the steps. Find the limiting function as omega_1 approaches omega_0, and discuss what this means in terms of a real system. Find the limiting function as delta approaches 0, and discuss what this means in terms of a real system. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We start out by solving the homogeneous form of the DE. This leads to y_c = (A+B)e^(-δt)*cos(sqrt(ω_0^2-δ^2)t) + (B-A)e^(-δt)*sin(sqrt(ω_0^2-δ^2)t) We take the first derivative of y_c, and then impose the initial conditions on both, to discover that A and B are both 0, leading to y_c = 0. Using the Variation of Parameters method, we find that y_p should take a form of y_p = u_1(t)*e^(-δt)*cos(sqrt(ω_0^2-δ^2)t) + u_2(t)*e^(-δt)*sin(sqrt(ω_0^2-δ^2)t) Take the first and second derivatives, and plug them back in, and solve for u_1 and u_2. This leads to the solution the book gives. The limiting function as omega_1 approaches omega_0 is Fsin(ω_0*t)/(2δω_0) - Fe^(-δt)sin(sqrt(ω_0^2-δ^2)t)/(2δsqrt(ω_0^2-δ^2)). In a real system, this means that as the driving force approaches the natural frequency of the system, the system approaches a resonant frequency system. The limiting function as delta approaches 0 is Fcos(ω_0*t)/(ω_1^2-ω_0^2) - Fcos(ω_1*t)/(ω_1^2 - ω_0^2). In a real system, this means that as the damping force approaches 0, the system behaves more and more as a resonant system. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: An LC circuit with L = 1 Henry and C = 4 microFarads is driven by voltage V_S(t) = 10 t e^(-t). Write and solve the differential equation for the system. Interpret your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: LQ” + RQ’ + Q/C = V_S(t) R = 0 Ω Lλ^2 + 1/C = 0 λ = ±i*sqrt(1/(LC)) = ±1/2i Q_c = Ae^(-1/2*it) + Be^(1/2*it) = (A+B)cos(t/2) + (B-A)sin(t/2) Q_p = Cte^(-t) + De^(-t) Q_p’ = -Cte^(-t) + Ce^(-t) - De^(-t) Q_p” = Cte^(-t) - 2Ce^(-t) + De^(-t) Plugging in: 5/4*Cte^(-t) - 2Ce^(-t) + 5/4*De^(-t) = 10te^(-t) -2C+5D/4D = 0 C = 5D/8 5(5D/8)/4 = 10 D = 64/5 C = 8 Q_p = 8te^(-t) + 64e^(-t)/5 Q_g = (A+B)cos(t/2) + (B-A)sin(t/2) + 8te^(-t) + 64e^(-t)/5 This is as far as we can go without initial conditions. This is the amount of charge, Q, in the system at time t, due to a variable voltage source. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!