Math Assignment 6

Transcription

1 Math Assignment 6 Dylan Zwick Fall 2013 Section 3.7-1, 5, 10, 17, 19 Section 3.8-1, 3, 5, 8, 13 Section 4.1-1, 2, 13, 15, 22 Section 4.2-1, 10, 19, 28 1

2 Section Electrical Circuits This problem deals with the RL circuit pictured below. It is a se ries circuit containing an inductor with an inductance of L henries, a resistor with a resistance of R ohms, and a source of electromotive force (emf), but no capacitor. In this case the equation governing our system is the first-order equation LI + RI = E(t). C Suppose that L = 511, R = 25Q, and the source E of emf is a battery supplying 1OOl to the circuit. Suppose also that the switch has been in position 1 for a long time, so that a steady current of 4A is flowing in the circuit. At time t = 0, the switch is thrown to position 2, so thatl(0)=r4ande=ofort>0. FindI(t). Solution - The ODE that describes this system is: = 0. We can rewrite this as: I + SI = 0. 2

3 This is a first-order linear ODE, and the corresponding integrating factor is ρ(t) = e R 5dt = e 5t. Multiplying both sides of the ODE by this integrating factor we get: e 5t I + 5e 5t I = 0 d dt ( e 5t I ) = 0. Integrating both sides of the above equation we get: e 5t I = C I(t) = Ce 5t. If we plug in the initial condition I(0) = 4 = C we get the final solution: I(t) = 4e 5t. 3

4 In the circuit from Problem 3.7.1, with the switch in position 1, suppose that E(t) = 100e 10t cos 60t, R = 20, L = 2, and I(0) = 0. Find I(t). Solution - The differential equation governing this system will be: 2I + 20I = 100e 10t cos (60t). If we divide both sides by 2 we get the ODE: I + 10I = 50e 10t cos (60t). The integrating factor for this first-order linear ODE will be ρ(t) = e R 10dt = e 10t. Multiplying both sides of our equation by this integrating factor we get: d ( e 10t I ) = 50 cos(60t). dt Integrating both sides of this equation gives us: e 10t I = 5 sin (60t) + C 6 I(t) = 5 6 e 10t sin (60t) + Ce 10t. If we plug in the initial condition I(0) = 0 we get C = 0, and our solution is: I(t) = 5 6 e 10t sin (60t). 4

5 This problem deals with an RC circuit pictured below, containing a resistor (R ohms), a capacitor (C farads), a switch, a source of emf, but no inductor. This system is governed by the linear first-order differential equation dq 1 R-- + = E(t). for the charge Q Q (t). = Q(t) on the capacitor at time t. Note that 1(t) C Suppose an emf of voltage E(t) = B0 cos wt is applied to the RC circuit at time t = 0 (with the switch closed), and Q(0) = 0. Substitute Q5(t) = A cos wt + B sin wt in the differential equation to show that the steady periodic charge on the capacitor is E0C Q5 (t) = p cos (wt 3) where /3 = tan (wrc). 5

6 Solution - Our steady periodic charge will be of the form: Q sp = A cos (ωt) + B sin (ωt). Its derivative will is: Q sp = ωa sin(ωt) + ωb cos (ωt). If we plug these values into the ODE: RQ sp + 1 C Q sp = E 0 cos (ωt), we get: ARω sin (ωt) + BRω cos (ωt) + A C cos (ωt) + B C sin (ωt) = E 0 cos (ωt). Grouping like terms together and multiplying both sides by C we have: (B AωRC) sin (ωt) + (A + BωRC) cos (ωt) = CE 0 cos (ωt). Equating coefficients we get B AωRC = 0, which means B = AωRC. Plugging these into the above equation gives us: A(1 + ω 2 R 2 C 2 ) cos (ωt) = CE 0 cos (ωt). So, A = E 0 C 1 + ω 2 R 2 C, and B = E 0C 2 ωr. From this we get: ω 2 R 2 C2 6

7 Q sp = If we define α = tan 1 (ωrc) then E 0 C (cos (ωt) + ωrc sin (ωt)). 1 + ω 2 R 2 C2 sin α = cos α = ωrc 1 + ω2 R 2 C 2, ω2 R 2 C 2. Using these we can rewrite Q sp as: Q sp = E 0 C (cos (α)cos (ωt) + sin (α) sin (ωt)). 1 + ω2 R 2 C2 If we use the trigonometric identity cos (θ φ) = cos (θ) cos (φ) + sin (θ) sin (φ) we can rewrite the above equation as: Q sp = E 0 C cos (ωt α). 1 + ω2 R 2 C2 This is our desired form. 7

8 For the RLC circuit pictured below find the current 1(t) using the given values of R, L, C and V(t), and the given initial values. L C(7 R = 16Q, L = 2H, C =.02F; E(t) = 100V; 1(0) = 0, Q(0) 5. Solution - The differential equation that governs this system is: i-50Q= 100. We can rewrite this as: I Q 50. The corresponding homogeneous equation is: I Q = 0. The characteristic polynomial for this linear homogeneous ODE is: r+8r+25, 8

9 which has roots: 8 ± 8 2 4(1)(25) 2 = 4 ± 3i. So, the corresponding homogeneous solution is: Q h (t) = c 1 e 4t cos (3t) + c 2 e 4t sin (3t). As for a particular solution, we guess our particular solution is of the form Q p (t) = A. Plugging this into our ODE we get: 25A = 50, so A = 2. Therefore the solution to our differential equation is: Q(t) = c 1 e 4t cos (3t) + c 2 e 4t sin (3t) + 2. Its derivative is: I(t) = 3c 1 e 4t sin (3t) 4c 1 e 4t cos (3t) + 3c 2 e 4t cos (3t) 4c 2 e 4t sin (3t). Plugging in our initial conditions gives us: Q(0) = c = 5, I(0) = 4c 1 + 3c 2 = 0. Solving this system of equations gives us c 1 = 3, c 2 = 4. So, our solution is: 9

10 Q(t) = 3e 4t cos (3t) + 4e 4t sin (3t) + 2, and I(t) = 25e 4t sin (3t). 10

11 Same instructions as Problem , but with the values: R = 60Ω, L = 2H, C =.0025F ; E(t) = 100e 10t V ; I(0) = 0, Q(0) = 1. Solution - The differential equation that governs this circuit is: 2I + 60I + 400Q = 100e 10t. We can rewrite this ODE as: I + 30I + 200Q = 50e 10t. The corresponding homogeneous equation is: I + 30I + 200Q = 0. This has characteristic polynomial: r r = (r + 20)(r + 10). The roots of this polynomial are: r = 10, 20, and so the homogeneous solution is: Q h = c 1 e 20t + c 2 e 10t. 11

12 As for a particular solution, our first guess is the particular solution will be of the form Ae 10t, but this would not be linearly independent of our homogeneous solution, so we need to use the guess Ate 10t. With this guess we get: Q p = Ate 10t, Q p = A( 10te 10t + e 10t ), Q p = A(100te 10t 20e 10t ). Plugging these into the ODE we get: 100Ate 10t 20Ae 10t 300Ate 10t + 30Ae 10t + 200Ate 10t = 10Ae 10t = 50e 10t So, our solution is: Its derivative is: A = 5. Q(t) = 5te 10t + c 1 e 20t + c 2 e 10t. Q (t) = I(t) = 50te 10t + 5e 10t 20c 1 e 20t 10c 2 e 10t. Plugging in our initial conditions we get: Q(0) = c 1 + c 2 = 1, I(0) = 5 20c 1 10c 2 = 0. Solving this system we get c 1 = 3 2, c 2 = 1, and our current will be: 2 I(t) = 10e 20t 10e 10t 50te 10t. 12

13 Section Endpoint Problems and Eigenvalues For the eigenvalue problem y + λy = 0; y (0) = 0, y(1) = 0, first determine whether λ = 0 is an eigenvalue; then find the positive eigenvalues and associated eigenfunctions. Solution - First, if λ = 0 then the solution to the differential equation y = 0 is y = Ax + B. From this we get y = A, and so if y (0) = 0 we must have A = 0. This would mean y = B, and if y(1) = 0 then B = 0. So, only the trivial solution A = B = 0 works, and therefore λ = 0 is not an eigenvalue. For λ > 0 the characteristic polynomial for our linear differential equation is: r 2 + λ = 0, which has roots r = ± λ. The corresponding solution to our ODE will be: y = A cos( λx) + B sin ( λx). with derivative y = A λ sin ( λx) + B λcos ( λx). 13

14 So, y (0) = B λ, and therefore if y (0) = 0 then we must have B = 0, as λ > 0. So, our solution must be of the form: y = A cos ( λx). If we plug in y(1) = 0 we get: y(1) = A cos ( λ) = 0. If A 0 we must have cos( λ) = 0, which is true only if λ = π + nπ. So, the eigenvalues are: 2 λ n = ( ( )) 2 1 π 2 + n, with n N, and corresponding eigenfunctions (( π ) ) y n = cos 2 + nπ x. 14

15 3.8.3 Same instructions as Problem 3.8.1, but for the eigenvalue problem: y + λy = 0; y( π) = 0, y(π) = 0. Solution - If λ = 0 then, as in Problem 3.8.1, our solution will be of the form: y = Ax + B. This means y(π) = Aπ + B = 0, and y( π) = Aπ + B = 0. Adding these two equations we get 2B = 0, which means B = 0. If B = 0 then Aπ = 0, which means A = 0. So, the only solution is the trivial solution A = B = 0, and therefore λ = 0 is not an eigenvalue. Now if λ > 0 then again just as in Problem we ll have a solution of the form: y(x) = A cos ( λx) + B sin ( λx). If we plug in our endpoint values we get: y(π) = A cos ( λπ) + B sin ( λπ) = 0, y( π) = A cos ( λπ) + B sin ( λπ) = A cos ( λπ) B sin ( λπ) = 0, where in the second line above we use that cos is an even function, while sin is odd. If we add these two equations together we get: 2A cos ( λπ) = 0. 15

16 This is true if either A = 0 or ( ) 1 λ = 2 + n. If ( ) 1 λ = 2 + n then (( ) ) 1 y(π) = B sin 2 + n π = 0. (( ) ) 1 As sin 2 + n π = ±1 we must have B = 0. On the other hand, if A = 0 above then we have: y(π) = B sin ( λπ). If B 0 then we must have λ = n. Combining our two results we get that the possible eigenvalues are: λ n = n2 4, for n N, and n > 0, with corresponding eigenfunctions: { ( cos n y n (x) = x) n odd 2 sin ( n x) n even 2 16

17 3.8.5 Same instructions as Problem 3.8.1, but for the eigenvalue problem: y + λy = 0; y( 2) = 0, y (2) = 0. Solution - If λ = 0 then, just as in Problem 3.8.1, the solution to the ODE will be: y(x) = Ax + B, y (x) = A. If we plug in our endpoint conditions we get y( 2) = 2A + B = 0 and y (2) = A = 0. These equations are satisfied if and only if A = B = 0, which is the trivial solution. So, λ = 0 is not an eigenvalue. If λ > 0 then, just as in Problem 3.8.1, the solution to the ODE will be of the form: y(x) = A cos ( λx) + B sin ( λx), with y (x) = A λsin ( λx) + B λ cos ( λx). Plugging in the endpoint conditions, and using that cos is even and sin is odd, we get: y( 2) = A cos ( 2 λ) + B sin ( 2 λ) = A cos (2 λ) B sin (2 λ) = 0, y (2) = A λ sin (2 λ) + B λcos (2 λ) = 0. If we divide both sides of the second equality by λ we get 17

18 A sin (2 λ) + B cos (2 λ) = 0. From these equations we get: A cos (2 λ) = B sin (2 λ) A B = tan (2 λ), B cos (2 λ) = A sin (2 λ) B A = tan (2 λ). So, A B = B A A2 = B 2. So, either A = B or A = B. If A = B then tan (2 λ) = 1, which means 2 λ = π 4 therefore + nπ, and (( ) n λ = π). 8 If A = B then tan(2 λ) = 1, which means 2 λ = 3π 4 therefore + nπ, and (( ) n λ = π). 8 So, the eigenvalues are: 18

19 (( ) ) n λ n = π 8 with n N and n > 0, with corresponding eigenfunctions: { (( cos 1+2n ) ) (( y n = 8 πx + sin 1+2n ) ) 8 πx n even cos (( ) ) (( 1+2n 8 πx sin 1+2n ) ) 8 πx n odd 19

20 Consider the eigenvalue problem y + λy = 0; y(0) = 0 y(1) = y (1) (not a typo).; all its eigenvalues are nonnegative. (a) Show that λ = 0 is an eigenvalue with associated eigenfunction y 0 (x) = x. (b) Show that the remaining eigenfunctions are given by y n (x) = sin β n x, where β n is the nth positive root of the equation tanz = z. Draw a sketch showing these roots. Deduce from this sketch that β n (2n + 1)π/2 when n is large. Solution - (a) - If λ = 0 then the solution to the ODE will be of the form: y(x) = Ax + B, with y (x) = A. So, y(0) = B = 0, and y(1) = A = y (1). So, any function of the form y(x) = Ax will work, and our eigenfunction for λ = 0 is: y 0 = x. (b) - For λ > 0 the solutions will all be of the form: y(x) = A cos(λx) + B sin (λx). If we plug in y(0) = A = 0 we get the solutions are of the form: 20

21 y(x) = Bsin(\/ 5x), with = Bvcos (\/I). If we plug in the other endpoint values we get: y(l) = Bsin(v) = BVcos(v ) = y (l). If B 0 then we must have: tan(/)= So, v X works if it s a root of the equation tan z = z, and if 8, is the iith such root, then the associated eigenfunction is: sin(/ 3x). A sketch intersect: of z and tan z are below. The roots are where they L As ii gets large it occurs at approximately /277 + in J7F. 2 21

22 Consider the eigenvalue problem y + 2y + λy = 0; y(0) = y(1) = 0. (a) Show that λ = 1 is not an eigenvalue. (b) Show that there is no eigenvalue λ such that λ < 1. (c) Show that the nth positive eigenvalue is λ n = n 2 π 2 + 1, with associated eigenfunction y n (x) = e x sin (nπx). Solution - (a) - If λ = 1 then the characteristic polynomial is: r 2 + 2r + 1 = (r + 1) 2, which has roots r = 1, 1. So, 1 is a root with multiplicity 2. The corresponding solution to the ODE will be: y(x) = Ae x + Bxe x. If we plug in the endpoint values we get: y(0) = A = 0, y(1) = Ae 1 + Be 1 = Be 1 = 0. From these we see the only solution is the trivial solution A = B = 0, so λ = 1 is not an eigenvalue. 22

23 (b) - If λ < 1 then the characteristic polynomial will be: r 2 + 2r + λ, which has roots r = 2 ± 2 2 4(1)λ 2 = 1 ± 1 λ. If λ < 1 then 1 λ will be real, and the solution to our ODE will be of the form: y(x) = Ae ( 1+ 1 λ)x + Be ( 1 1 λ)x. Plugging in our endpoint values we get: y(0) = A + B = 0, y(1) = Ae 1+ 1 λ + Be 1 1 λ = 0. From these we get, after a little algebra: A(1 e 2 1 λ ) = 0. If λ < 1 then e 2 1 λ < 1, and therefore 1 e 2 1 λ > 0. So, for the above equality to be true we must have A = 0, which means B = 0, and so the only solution is the trivial solution A = B = 0. Therefore, no value λ < 1 is an eigenvalue. (c) - If λ > 1 then again using the roots from the quadratic equation in part (b) we get that our solutions will be of the form: 23

24 y(x) = Ae x cos ( λ 1x) + Be x sin ( λ 1x). If we plug in the endpoint values we get: y(0) = A = 0, and so y(x) = Be x sin ( λ 1x). If we plug in our other endpoint value we get: y(1) = Be 1 sin ( λ 1) = 0. If B 0 then we must have sin ( λ 1) = 0, which is only possible if λ 1 = nπ, λ n = n 2 π So, the eigenvalues are given above, and the corresponding eigenfunctions are: y n = e x sin (nπx), for n N, n > 0. 24

25 Section First-Order Systems and Applications Transform the given differential equation into an equivalent system of first-order differential equations. x + 3x + 7x = t 2. Solution - If we define x = x 1 then define: x 1 = x 2, x 2 = t2 3x 2 7x 1. So, the system is: x 1 = x 2 x 2 = 7x 1 3x 2 + t 2. 25

26 Transform the given differential equation into an equivalent system of first-order differential equations. x (4) + 6x 3x + x = cos 3t. Solution - Define x = x 1. Then the equivalent system is: x 1 = x 2 x 2 = x 3 x 3 = x 4 x 4 = 6x 3 + 3x 2 x 1 + cos (3t). 26

27 Find the particular solution to the system of differential equations below. Use a computer system or graphing calculator to construct a direction field and typical solution curves for the given system. x = 2y, y = 2x; x(0) = 1, y(0) = 0. Solution - If we differentiate y = 2x, we get y = 2x = 4y. So, we have the differential equation: y + 4y = 0. The solution to this ODE is: y(t) = A cos (2t) + B sin (2t). Now, x(t) = 1 2 y = 1 ( 2A sin (2t) + 2B cos (2t)) = A sin (2t) + B cos (2t). 2 If we plug in x(0) = B = 1 and y(0) = A = 0 we get: x(t) = cos (2t) y(t) = sin (2t). 27

28 More room, if necessary, for Problem Pirec lvi V 4 / 7 o( Ndfe; 5hoI) be CI(cIJ 40 A 1 I bei 19

29 Find the general solution to the system of differential equations below. Use a computer system or graphing calculator to construct a direction field and typical solution curves for the given system. x = 1 2 y, y = 8x. Solution - If we differentiate y = 8x we get y = 8x = 4y. So, our ODE is: y + 4y = 0. The solution to this ODE is: y(t) = A cos (2t) + B sin (2t). The function x(t) is: x(t) = 1 8 y (t) = A 4 sin (2t) + B 4 cos (2t). So, the general solution to this system of ODEs is: x(t) = A 4 sin (2t) + B 4 cos (2t) y(t) = A cos (2t) + B sin (2t). 29

30 Pirecio More room, if necessary, for Problem

31 (a) - Beginning with the general solution of the system from Problem 13, calculate x 2 + y 2 to show that the trajectories are circles. (b) - Show similarly that the trajectories of the system from Problem 15 are ellipses of the form 16x 2 + y 2 = C 2. (a) - The general solution to the system of ODEs from Problem is: From these we get: x(t) = A sin (2t) + B cos (2t) y(t) = A cos (2t) + B sin (2t). x(t) 2 + y(t) 2 = ( A sin (2t) + B cos (2t)) 2 + (A cos (2t) + B sin (2t)) 2 = A 2 sin 2 (2t) 2AB sin (2t) cos (2t) + B 2 cos 2 (2t) + A 2 cos 2 (2t) + 2AB sin (2t) cos (2t) + B 2 sin 2 (2t) So, circles. = A 2 + B 2. (b) - The general solution to the system of ODEs from Problem is: x(t) = A 4 sin (2t) + B 4 cos (2t) y(t) = A cos (2t) + B sin (2t). So, 16x(t) 2 = A 2 sin 2 (2t) 2AB sin (2t) cos (2t) + B 2 cos 2 (2t), y(t) 2 = A 2 cos 2 (2t) + 2AB sin (2t) cos (2t) + B 2 sin 2 (2t). Combining these we get 16x(t) 2 + y(t) 2 = A 2 + B 2 = C 2. So, ellipses. 31

32 Section The Method of Elimination Find a general solution to the linear system below. Use a computer system or graphing calculator to construct a direction field and typical solution curves for the system. x = x + 3y y = 2y Solution - The differential equation y = 2y has the solution y(t) = Ae 2t. So, x = x + 3Ae 2t x + x = 3Ae 2t. This is a first-order linear ODE. Its integrating factor is: ρ(t) = e R 1dt = e t. Multiplying both sides by this integrating factor our linear ODE becomes: d ( e t x ) = 3Ae 3t. dt 32

33 33 The direction field looks kind of like this: (I)4(i)2t+B(i)t We can write this in vector form as: j(t) = Ae 2t. x(t) = Ae 2t + Be_t, So, the general solution to this system is: x Ae 2 + Be. etx = Ae 3t + B Integrating both sides we get:

34 Find a particular solution to the given system of differential equations that satisfies the given initial conditions. x + 2y = 4x + 5y, 2x y = 3x; x(0) = 1, y(0) = 1. Solution - If we add 2 times the second equation to the first we get: 5x = 10x + 5y. If we subtract 2 times the first equation from the second we get: 5y = 5x 10y 5y = 5x + 10y. Differentiating 5x = 10x+5y and plugging in 5y = 5x+10y we get: 5x = 10x + 5y = 10x + (5x + 10y) 5x = 10x + (5x + 10x 20x) 5x = 20x 15x x = 4x 3x. The linear homogeneous differential equation x 4x + 3x = 0 has characteristic equation: r 2 4r + 3 = (r 3)(r 1). 34

35 So, the roots are r = 3, 1, and the general solution to the ODE is: x(t) = c 1 e 3t + c 2 e t. From the equation 5y = 5x + 10y we get y = x + 2y, and therefore: y 2y = c 1 e 3t + c 2 e t. If we multiply both sides by the integrating factor e 2t we get: d ( e 2t y ) = c 1 e t + c 2 e t. dt Integrating both sides we get: e 2t y = c 1 e t c 2 e t + C, and so: y(t) = c 1 e 3t c 2 e t + Ce 2t. Plugging this into any of the equations in our system gives us C = 0. So, y(t) = c 1 e 3t c 2 e t. We can write this solution in matrix form as: x(t) = c 1 ( 1 1 ) e 3t + c 2 ( 1 1 ) e t. 35

36 If we plug in x(0) = ( 1 1 ) ( 1 = c 1 1 ) + c 2 ( 1 1 ) we can see immediately that c 1 = 0 and c 2 = 1. So, the solution to our initial value problem is: x(t) = ( 1 1 ) e t. 36

37 Find a general solution to the given system of differential equations. x = 4x 2y, y = 4x + 4y 2z, z = 4y + 4z. Solution - If we differentiate the first equation we get: x = 4x 2y = 4x 2( 4x + 4y 2z) x = 4x + 8x 8y + 4z. Differentiating again we get: x (3) = 4x + 8x 8y + 4z = 4x + 8x 8y + 4( 4y + 4z) x (3) = 4x + 8x 8y 16y + 16z x (3) = 4x + 8x 8y 16y + 8( y 4x + 4y) x (3) = 4x + 8x 16y + 16y 32x x (3) = 4x + 8x 8(4x x ) + 8(4x x ) 32x x (3) = 12x 32x x (3) 12x + 32x = 0. The characteristic equation for this ODE is r 3 12r r = r(r 8)(r 4). So, 37

38 x(t) = c 1 + c 2 e 8t + c 3 e 4t. From this we get: x (t) = 8c 2 e 8t + 4c 3 e 4t and y(t) = 2x 1 2 x = 2c 1 2c 2 e 8t. Finally, z(t) = 2x (t) + 2y(t) 1 2 y (t) = 2c 1 + 2c 2 e 8t 2c 3 e 4t. So, x(t) = c 1 + c 2 e 8t + c 3 e 4t, y(t) = 2c 1 2c 2 e 8t, z(t) = 2c 1 + 2c 2 e 8t 2c 3 e 4t. 38

39 For the system below first calculate the operational determinant to determine how many arbitrary constants should appear in a general solution. Then attempt to solve the system explicitly so as to find such a general solution. (D 2 + D)x + D 2 y = 2e t (D 2 1)x + (D 2 D)y = 0 Solution - The operational determinant of the system above is: (D 2 +D)(D 2 D) D 2 (D 2 1) = D 4 D 3 +D 3 D 2 D 4 +D 2 = 0. So, there are 0(!) arbitrary constants. How is this possible? Well, if we subtract the second relation from the first we get: (D + 1)x + Dy = 2e t Dy = 2e t (D + 1)x D 2 y = 2e t (D 2 + D)x (D 2 + D)x + D 2 y = 2e t. However, this cannot be, as our first relation above is: (D 2 + D)x + D 2 y = 2e t, and 2e t 2e t. So, there is no solution to the system. 39