Laplace Transform Problems

Transcription

1 AP Calculus BC Name: Laplace Transformation Day 3 2 January 206 Laplace Transform Problems Example problems using the Laplace Transform.. Solve the differential equation y! y = e t, with the initial value conditions y(0) = 0. Ans. y(t) = te t Solution: sy(s) 0 Y(s) = s (s )Y = s The solution is Y(s) = (s ) 2 = A (s ) + B (s ) 2 = 0 (s ) + (s ) 2 = Y(s ) = L { et f (t)} = e t t. 2. Solve the differential equation 0. i " + i " + 00i(t) = 40000cos( 400t) with the initial value conditions i(0) = 0, i "(0) = 0. Ans. i(t) = e 0t e 00t cos400t sin400t This problem describes the current in an RLC circuit with resistance R = Ohms, capacitance C = /00 Farad, and inductance L = /0 Henry, with an imposed sinusoidal voltage. Special thanks to Mr. Olsen and Calculus (4 th edition) by Deborah Hughes-Hallet, et. al. for most of these problems.

2 Laplace Transformation Day 3 Application Problems page 2 Solution: 0.s 2 s I(s) +si(s) +00I(s) = s (s 2 +0s +000)I(s) = I(s) = (s +0)(s +00) s Now s (s +0)(s +00)( ) = A s +0 + B s Cs + D, giving s = A(s +00)( ) + B(s +0)( ) + (Cs + D)(s +0)(s +00) Now, for s = 0, we get = 90( )A, or A = For s = 00, we have = 90( )B, or B = By looking at s 3 terms, we get 0 = A + B + C, or C = By looking at s 2 terms, we get Then 0 =00A +0B +0C + D, or D = I(s) = s s s Taking the inverse transform, we get the solution: i(t) = e 0t e 00t cos400t sin400t

3 Laplace Transformation Day 3 Application Problems page 3 3. Solve the differential equation y " + 4 y " +3y =45cos2t, with y(0) =0, and y "(0) =4 Ans. y(t) = e 2t cos3t + 9cos2t + 8sin2t. Solution: Taking the Laplace Transform gives (s 2 Y(s) 0s 4) + 4(sY(s) 0) +3Y(s) = 45s (s +3)Y(s) =0s s Y(s) = 0s + 54 s +3 + The last term can be written 45s (s +3)() 45s (s +3)() = As + B s +3 + Cs + D 45s = (As + B)() + (Cs + D)(s +3) 45s = s 3 (A + C) + s 2 (B + 4C + D) + s(4a +3C + 4D) + (4B +3D) From this we get # C = A % D = 4 A B $ % 7A 4B =45 &% 52A 9B = 0 This yields A = 9; B = 52; C = 9; D =6. Thus (0s 9s) + (54 52) Y(s) = + s +3 Y(s) = 9s +6 (s + 2) (s + 2) s 2 s s where in the last line we have completed the square in the first term. Taking the inverse transform, we obtain the answer above.

4 Laplace Transformation Day 3 Application Problems page 4 Practice Laplace Transform problems 5. dy dt = t 2 + y(t) F(s) = 2 s 3 (s ) + f (0) s = 2 s 2 s 2 2 s f (0) 3 s y(t) = 2 2t t 2 + (2 + f (0))e t 6. Find the Laplace Transform for f (t) = te t L{ te t } = (this leads to L 2 { teat} = (s ) (s a) 2 ) 7. Solve y!! + 3 y! + 2y = where y = 0 and y! = 0 for t = 0. F(s) = s(s 2 + 3s + 2) = 2 s s s + 2 y(t) = 2 e t + 2 e 2t 8. Solve y!! + 4y = 5e t +6t where y = 4 and y! = 7 for t = 0. (This might take another piece of paper.) F(s) = 4s (s )() + 6 s 2 () = 3s s s s + 4 s 2 y(t) = 3cos(2t)+ sin(2t)+ e t + 4t

5 Laplace Transformation Day 3 Application Problems page 5 Application problems You may solve these using any method but the last problem requires the Laplace Transform. 9. Dead leaves accumulate on the ground in a forest at a rate of 3 grams per square centimeter per year. At the same time, these leaves decompose at a continuous rate of 75% per year. Write a differential equation for the total quantity of dead leaves (per square centimeter) at time t. Sketch a solution showing that the quantity of dead leaves tends toward an equilibrium level. What is that equilibrium level? dl dt = L = 3 (L 4) which can be 4 solved with separation of variables to get L(t) = 4 + Ae 3 4 t. The equilibrium level is L = 4 because that makes the derivative equal zero (no change). 0. As you know, when a course ends, students start to forget the material they have learned. One model (called the Ebbinghaus model) assumes that the rate at which a student forgets material is proportional to the difference between material currently remembered and some positive constant, a. a. Let y = f(t) be the fraction of the original material remembered t weeks after the course has ended. Set up a differential equation for y. Your equation will contain two constants; the constant a is less than y for all t, and the constant b is the constant of proportionality. b. Solve the differential equation, using the initial condition y =.00 at t = 0. c. Describe the practical meaning (in terms of the amount remembered) of the constants in the solution y = f(t).. An object of mass m is thrown vertically upward from the surface of the earth with initial velocity v 0. We will calculate the value of v 0, called the escape velocity, with which the object can escape the pull of the gravity and never return to earth. Since the object is moving far from the surface of the earth, we must take into account the variation of gravity with altitude. If the acceleration due to gravity at sea level is g, and R is the radius of the earth, the gravitational force, F, on the object of mass m at an altitude h above the surface of the earth is given by F = mgr2 (R + h) 2 a. Suppose v is the velocity of the object (measured upward) at time t. Using Newton s Law of Motion one can show that dv dt = gr2 (R + h). 2 This equation can be written with h instead of t as the independent variable using the chain rule dv dt = dv dh dh dv. Hence show that v dt dh = gr2 (R + h) 2 b. Solve the differential equation in part a. c. Find the escape velocity, the smallest value of v 0 such that v is never zero.

6 Laplace Transformation Day 3 Application Problems page 6 2. Use the Laplace Transform on this problem. In a nuclear reactor, about 6. percent of fission reactions produce the isotope Iodine-35. I-35 then decays to Xenon-35 with a half-life of 6.7 hours, and Xe-35 decays to Cesium-35 with a half-life of 9.2 hours. a. When the reactor is shut off, I 35 is no longer produced. The amount of I 35 present is then subject to the differential equation di dt = a I. Here I(t) is the amount of I 35 present at t seconds and a = t I = ln(2) 6.7 hr = sec. Show that this equation has the solution where I 0 is equal to I(0). I(t) = I 0 exp( a t), () b. The amount of Xe 35 present at time t seconds after shutdown is subject to the differential equation dx = b X + a I (2) dt where I(t) is the amount of I 35 present from part a), and b = t X = ln(2) 9.2 hr = sec. Substitute the expression () for I(t) from part a) into the differential equation (2) and show that the Laplace transform of the result is X(s) = " s + b X 0 + a I % 0 # $ s + a& ' (3) c. Take the inverse Laplace transform of (3) and show that the solution is [ ] (4) X(t) = X 0 exp( b t)+ a I 0 exp( a t) exp( b t) b a d. To get a sense of how the amount of Xe 35 varies with time, graph X(t) using X 0 = 0 and I 0 = in equation (4). Set the t-scale to [0, 60 hours] with marks every 0 hrs, and the y- scale to [0, 0.5], with marks every 0..