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# Differential Equations Midterm 2 Study Guide Math 240

KSU

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This 18 page Study Guide was uploaded by Laurel Knust on Thursday October 13, 2016. The Study Guide belongs to Math 240 at Kansas State University taught by Andrew G. Bennett in Fall 2016. Since its upload, it has received 2 views. For similar materials see Elem Differential Equations in Mathematics at Kansas State University.

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Date Created: 10/13/16

1. Write down the general solution of each of the following ODEs: ′′ ′ (a) y + 3y + 2y = 0. ′′ ′ (b) y + 4y + 8y = 0. (c) y ▯ 4y + 4y = 0. |||||||||||||||||||||||| (a) y(t) = c e1 −t + c 2 −2t . (b) y(t) = c e −2t cos(2t) + c e −2t sin(2t) . 1 2 2t 2t (c) y(t) = c e1+ c te 2 . 2. Write down the form of the guess to use for the method of undetermined coe▯cients for the following ODE: ′′ t 2 y (t) ▯ y(t) = 2tcost + 3te ▯ 4t . You do not need to ▯nd the coe▯cients! |||||||||||||||||||||||| Y (t) = Y 1t) + Y (2) + Y (t3 where Y1(t) = A t1ost + B tsi1t + C cost 1 D sint 1 2 t t Y2(t) = A t2e + B te 2 2 Y3(t) = A t3+ B t +3C . 3 3. Solve the following initial value problem: y (t) + 3y + 2y = 4e ,t y(0) = 0, y (0) = 1 . |||||||||||||||||||||||| 1 2 y(t) = e−2t▯ e −t+ e . t 3 3 4. Write down the general solution of the following ODE: et y (t) ▯ 2y (t) + y(t) = . t + 1 |||||||||||||||||||||||| p t t t 2 t y(t) = c 1 + c te2▯ e ln t + 1 + te arctant . 5. Find the general solution of the following ODE: y (t) + 4y (t) + 8y (t) + 8y(t) = 0 . |||||||||||||||||||||||| p p y(t) = c e−2t+ c e −tcos( 3t) + c e−t sin( 3t) . 1 2 3 6. Mass, damping constant, and spring constant respectively are: 2 m = 2kg, c = 4kg/s, k = 20kg/s . Initial position and velocity respectively are: x(0) = 3cm, x (0) = 1cm. (a) Write down the di▯erential equation which governs the system. |||||||||||||||||||||||| 2x + 4x + 20x = 0 . |||||||||||||||||||||||| (b) Write down the initial energy of the system. |||||||||||||||||||||||| 2 kgcm E(0) = 91 2 . s |||||||||||||||||||||||| (c) Show, by using the di▯erential equation, that the energy is mono- tone nonincreasing. |||||||||||||||||||||||| 0 = 2x + 4x + 20x multiplying through by x gives ′ 0 = 2x x + 4(x ) + 20xx . ′ By rearranging I get: ′ ′′ ′ ′ 2 2x x + 20xx = ▯4(x ) ▯ 0. On the other hand, the energy of this system as a function of time is given by: ′ 2 2 E(t) = (x ) + 10x , and by using the chain rule we can calculate the derivative with respect to t: E (t) = 2x x + 20xx , ′ which allows us to conclude by combining our equations that ′ E (t) ▯ 0. 7. Express cos(2t)3sin(2t) in the form Acos(2t ▯ ϕ). |||||||||||||||||||||||| p ( ) cos(2t) +3sin(2t) = 2cos 2t ▯ . 3 8. Short answers (a) What does it mean if the Wronskian of two functions is zero? |||||||||||||||||||||||| If the wronskian of two solutions is zero, then the functions are linearly dependent, and therefore one of them is a multiple of the other. |||||||||||||||||||||||| (b) Suppose L is a linear di▯erential operator, and both y and y 1 2 solve L[y] = 0. What is L[12y ▯17y ]? 2 |||||||||| 0. |||||||||| (c) Suppose L is a linear di▯erential operator, and both y and y1 2 solve L[y] = t e cos(3t). What is L[12y ▯17y ]? 2 2 t |||||||||| 5t e cos(3t). |||||||||| (d) Solve the initial value problem: ′′′ ′′ 2 ′ p ′′ ′ y + πy ▯ e y + 17y = 0, y (0) = y (0) = y(0). |||||||||| y(t) ▯ 0. |||||||||| (e) Consider the ODE: ay + by + cy = 0. Assume that the roots of the characteristic polynomial are λ ▯ µi with the correspond- ing complex solutions e (▯+▯i)tand e (▯−▯i). How does this lead to ▯t ▯t having the real solutions e cos(µt) and e sin(µt)? |||||||||||||||||||||||| We start with the complex solutions: (▯+i▯)t ▯t ~1(t) = e = e (cos(µt) + isin(µt)) (▯−i▯)t ▯t ~2(t) = e = e (cos(µt) ▯ isin(µt)) . Now, by the principle of superposition, we can conclude that all of the functions on each line here are also solutions. ▯t ~1(t) + ~2(t) = 2e cos(µt) 1 (~1(t) + ~2(t)) = e cos(µt) 2 ▯t ~1(t) ▯ ~2(t) = 2ie sin(µt) 1 (~1(t) ▯ ~2(t)) = e sin(µt) . 2i 1. Write down the general solution of each of the following ODEs: ′′ ′ (a) y ▯ 11y + 30y = 0: y(t) = c e + c e 6t : 0 1 ′′ ′ (b) y + 3y + 10y = 0: ( p ) ( p ) − t 31 − t 31 y(t) = c0e 2 cos t + c1e 2 sin t : 2 2 (c) y + 10y + 25y = 0: y(t) = c0e −5t+ c 1e −5t: 2. Write down the form of the guess to use for the method of undetermined coe▯cients for the following ODE: ′′ ′ −t 3 y (t) + 3y (t) + 2y(t) = ▯6tcos4t + 5te ▯ 7t : You do not need to ▯nd the coe▯cients! Y 1t) = A tc1s(4t) + B tsin14t) + C cos(4t)1+ D sin(4t) 1 2 −t −t Y 2t) = A t 2 + B t2 3 2 Y 3t) = A t 3 B t + 3 t + D 3 3 Y (t) = Y 1t) + Y (2) + Y (t3 : 3. Solve the following initial value problem: y (t) + 3y (t) + 2y(t) = 4e ; t y(0) = 0; y (0) = 1 : 1 −2t −t 2 t y(t) = e ▯ e + e : 3 3 4. Suppose that you are given the following initial value problem: ′′ ′ 4 2 7t ′ y (t) + 3(y (t)) (y(t)) = 5e ; y(7) = 1; y (7) = 2 : Do the following: (a) Express the problem as an equivalent ▯rst order system. ′ 7t 4 2 z = 5e ▯ 3z y ′ y = z y(7) = 1 ; z(7) = 2 : (b) Set up a standard Euler iteration to approximate the solution of this problem with a stepsize of .01. Be sure to express y and n+1 zn+1 explicitly in terms of n;y ;nand z ; nnd give the values of y 0 and z 0 7n 4 2 zn+1 = z n :01(5e ▯ 3z n n yn+1 = y n :01z n tn= 7 + :01n y0= 1 ; z = 0 : (c) What should you use to approximate y(7:75)? y(7:75) ▯ y 75 : 5. Write down the general solution of the following ODE for t > 0: 6y + 12y + 6y = 6e −tlnt : (Hint: Integrals involving natural logarithms can frequently be con- quered using integration by parts, making sure to di▯erentiate the natural logarithm.) 1 3 y(t) = c0e−t + c1te −t+ t e2 −t lnt ▯ t e2 −t : 2 4 6. Find the general solution of the following ODE: ′′′ ′′ ′ y (t) ▯ 4y (t) + 9y (t) ▯ 6y(t) = 0 : ( p ) ( p ) 3 15 3 15 y(t) = c0e + c1e 2 cos t + c2e 2tsin t : 2 2 7. Mass, damping constant, and spring constant respectively are: 2 m = 1kg; c = 2kg=s; k = 10kg=s : Initial position and velocity respectively are: x(0) = 2cm; x (0) = 4cm=s: (a) Write down the di▯erential equation which governs the system. |||||||||||||||||||||||| ′′ ′ 1x + 2x + 10x = 0 : |||||||||||||||||||||||| (b) Write down the initial energy of the system. |||||||||||||||||||||||| kg cm 2 E(0) = 28 : s2 |||||||||||||||||||||||| (c) Show, by using the di▯erential equation, that the energy is mono- tone nonincreasing. |||||||||||||||||||||||| 0 = x + 2x + 10x multiplying through by x gives ′ 0 = x x + 2(x ) + 10xx : ′ By rearranging I get: x x + 10xx = ▯2(x ) ▯ 0:2 On the other hand, the energy of this system as a function of time is given by: 1 E(t) = (x ) + 5x ;2 2 and by using the chain rule we can calculate the derivative with respect to t: E (t) = x x + 10xx ; ′ which allows us to conclude by combining our equations that E (t) ▯ 0: 8. (a) Match up the following \factored" di▯erential equations to their \expanded" counterparts. (\D" stands for di▯erentiation with re- spect to x:) Factored: i. (D + 5)(D + 5)y = 0: | D ii. (D + 5)(D + x)y = 0: | C iii. (D + x)(D + 5)y = 0: | A iv. (D + x)(D + x)y = 0: | B Expanded : ′′ ′ A) y + (x + 5)y + 5xy = 0: |||| iii ′′ ′ 2 B) y + 2xy + (1 + x )y = 0: |||{ iv ′′ ′ C) y + (x + 5)y + (1 + 5x)y = 0: |{ ii D) y + 10y + 25y = 0: ||||||{ i (b) Expand the folowing into a third order di▯erential operator: L[y] = (D + 2x)(D + 3x)(D + 4x)y : ′′′ ′′ 2 ′ 3 L[y] = y + 9xy + (11 + 26x )y + (32x + 24x )y : 9. Short answers (a) What does it mean if the Wronskian of two functions is zero? |||||||||||||||||||||||| If the wronskian of two solutions is zero, then the functions are linearly dependent, and therefore one of them is a multiple of the other. |||||||||||||||||||||||| (b) Suppose L is a linear di▯erential operator, and both y and y1 2 solve L[y] = 0: What is L[6y ▯15y ]?2 |||||||||| 0: |||||||||| 3 4t (c) Suppose L is a linear di▯erential operator, L[y ] =12t e ; and L[y 2 = ▯3cos(5t): Then what is L[6y ▯ 51 ]? 2 ||||||||| 12t e + 15cos(5t): ||||||||| (d) Consider the ODE: ay + by + cy = 0: Assume that the roots of the characteristic polynomial are ▯ ▯ ▯i with the correspond- ing complex solutions e (▯+▯i)tand e (▯−▯i): How does this lead to ▯t ▯t having the real solutions e cos(▯t) and e sin(▯t)? |||||||||||||||||||||||| We start with the complex solutions: ~1(t) = e(▯+i▯)t= e (cos(▯t) + isin(▯t)) ~2(t) = e(▯−i▯)t= e (cos(▯t) ▯ isin(▯t)) : Now, by the principle of superposition, we can conclude that all of the functions on each line here are also solutions. ~ (t) + ~ (t) = 2e cos(▯t) 1 2 1 ▯t (~1(t) + ~2(t)) = e cos(▯t) 2 ~ (t) ▯ ~ (t) = 2ie sin(▯t) 1 2 1 ▯t (~1(t) ▯ ~2(t)) = e sin(▯t) : 2i 10. For the initial value problem y + 49y = F(t); y(0) = A; y (0) = 0; answer the following questions: (a) Find a bounded function F(t) which leads to an unbounded solution. F(t) = cos(7t) is one example which produces resonance. (b) Assuming that we take A = 1; then ▯nd a function F(t) satisfying jF(t)j ▯ 1 for all t; which leads to a solution that has a maximum value which is greater than 100; but which remains bounded. F(t) = cos(7:0000000001t) : (c) If cos(5t) is the solution to the initial value problem above, then determine A and F(t): A = 1 and F(t) = 24cos(5t) :

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