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Math 244 Jeopardy

Math 244 Jeopardy. Potluck. First Order Equations. Second Order Equations. Higher Order Misc. Matrix Fun. 100. 100. 100. 100. 100. 200. 200. 200. 200. 200. 300. 300. 300. 300. 300. 400. 400. 400. 400. 400. 500. 500. 500. 500. 500. Potluck: 100.

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Math 244 Jeopardy

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  1. Math 244 Jeopardy

  2. Potluck First OrderEquations Second OrderEquations HigherOrderMisc. MatrixFun 100 100 100 100 100 200 200 200 200 200 300 300 300 300 300 400 400 400 400 400 500 500 500 500 500

  3. Potluck: 100 Solve the differential equation

  4. Potluck: 100 So (1+y3)dy = 3x4+sin(x) Integrating both sides… Or in implicit form…

  5. Potluck: 200 A population of mice follows the differential equation dp/dt = p/4-200. If there are initially 400 mice, when will they become extinct?

  6. Potluck: 200 We have dp/dt=p/4-200 = (p-800)/4Rearranging… dp/(p-800) = dt/4.Integrating both sides, ln(p-800) = t/4 +Cso p-800 = Cet/4, and p=Cet/4+800.Using the initial condition p(0)=400, p=-400et/4+800. The population is extinct when p(t)=0so we solve 0= -400et/4+8002= et/4, so the mice are extinct when t=4 ln(2).

  7. Potluck: 300 Draw a direction field for dy/dx=y/2-1 What is the behavior as ?

  8. Potluck: 300 If y(0)>2, y  ∞If y(0)=2, y 2If y(0)<2, y -∞ y=2 is an equilibrium solution As

  9. Potluck: 400 A radioactive material decays at a rate proportional to the amount currently present. If there is 100 grams of the material decays to 75 grams in 3 days, find an expression for the amount of material at any time t.

  10. Potluck: 400 dQ/dt = -rQ dQ/Q = -r dtln(Q) = -rt+C Q=Ce-rt Initially 100 g, so Q= 100e-rt Also, 75=100e-r*3, so ¾= e-r*3, or r=-ln( ¾ )/3 So, Q(t)= 100eln(3/4)t/3

  11. Potluck: 500 A college student invests $100 dollars per year at an annual rate of return of 3%. Assume investments are made and compounded continuously. Determine the sum S(t) accumulated at any time t. How much money will be available in 30 years?

  12. Potluck: 500 dS/dt = 0.03S+100so dS/(S+100/0.03) = 0.03dtln(S+3333.333) = 0.03t+CS=Ce0.03t-3333.333Initially $0, so S=3333.33(e0.03t-1) In 30 years, there will be $3333.33(e0.03*30-1) = $4865.34

  13. First Order Equations: 100 Is this an exact equation? Why or why not? (6y2-x2+3)y'+ (3x2-2xy+2) =0

  14. First Order Equations: 100 (6y2-x2+3)y'+ (3x2-2xy+2) =0 (6y2-x2+3)dy/dx+ (3x2-2xy+2) =0 (6y2-x2+3)dy+ (3x2-2xy+2)dx =0M= 3x2-2xy+2 N= 6y2-x2+3 My= -2x Nx= -2x So, yes!

  15. First Order Equations: 200 y(5/2)=7/2 is guaranteed to have a solution. Find an interval where the equation

  16. First Order Equations: 200 Can be rewritten as… Can’t take ln of non-positive numbers, so (t-2)>0, or t>2can’t divide by 0 so t-30, or t 3 Finally, can’t take square root of negatives so 16-t20 or -4  t  4. This gives us either (2,3) or (3,4]. Since the initial value is at t=5/2, we choose the interval (2,3).

  17. First Order Equations: 300 Consider the equation dy/dt=y4-y2. Find and classify all equilibrium solutions.

  18. First Order Equations: 300 y=-1 is stable y=0 is semistable y=1 is unstable dy/dt=y4-y2= y2(y2-1) = y2(y+1)(y-1) So equilibrium solutions are y=0, y=1, y=-1

  19. First Order Equations: 400 Use an integrating factor to solve: y'-3y=t3e3t y(0)=18

  20. First Order Equations: 400 So, e-3ty=t4/4+18or y= e3tt4/4+18e3t y'e-3t-3e-3ty=t3 d/dt(e-3ty)=t^3, so e-3ty=t4/4+C Using y(0)=18, we get C=18 y'-3y=t3e3t has integrating factor

  21. First Order Equations: 500 Solve the equation y dx + (2xy-e-2y)dy=0

  22. First Order Equations: 500 Or =e2y/y y dx + (2xy-e-2y)dy=0 is not exact since My=1 but Nx=2y, so find integrating factor: Then, M=e2y and N=2xe2y-1/y, so My=Nx=2e2y Now, we must find  so that x= e2y and y = 2xe2y-1/y Which gives  = xe2y-ln(y)So the solution is xe2y-ln(y)=C

  23. Second Order Equations: 100 Find the general solution of the equation: y''+8y'+15y=0 And describe its behavior as t ∞

  24. Second Order Equations: 100 y''+8y'+15y=0 has characteristic equation r2+8r+15 = (r+3)(r+5)=0so we have roots, r=-3 and r=-5 Thus, the general solution is y=C1e-3t+C2e-5t Since enegative 0 as t ∞, we get that y 0, no matter what C1 and C2 are.

  25. Second Order Equations: 200 Compute the Wronskian of two solutions to the equation ty''+y'+2y=42sin(t). (Hint: you don’t need to solve the equation to answer the question!)

  26. Second Order Equations: 200 In general form, the equation is: y''+y'/t+2y/t=42sin(t)/t By Abel’s Theorem, the Wronskian is

  27. Second Order Equations: 300 Solve the equation x2y''-3xy'+13y=0

  28. Second Order Equations: 300 This is an Euler equation, so guess y=xr, to get xr(r2-4r+13)=0, which has roots r=2±3i. This gives the general solution Y=|x|2(C1cos(3ln|x|)+ C2sin(3ln|x|))

  29. Second Order Equations: 400 Find the general solution of the equation: y''-4y'+4y=0, y(0)=2, y'(0)=1 And describe its behavior as t ∞

  30. Second Order Equations: 400 y''-4y'+4y=0 has characteristic equation r2-4r+4 = (r-2)(r-2)=0 So it has repeated roots r=2,2, and the general solution is y=C1e2t+C2te2t Using y(0)=2, we get that C1=2 Using y'(0)=1, we get that C2=-3 So the solution is y=(2-3t) e2t, which goes to - ∞ as t ∞.

  31. Second Order Equations: 500 A mass weighing 32 lb stretches a spring 384/5 inches. Suppose the mass is pulled down an additional 12 inches and then released, and there is a damping constant of 4 lb-sec/ft. Find an equation to describe the position of the mass at any time t. (You may assume that g=32ft/s2.)

  32. Second Order Equations: 500 W=mg, so m=1 slug F=kx, so k=5lb/ft So we have u''+4u'+5u=0, which has roots r=-2±i, and gives the general solution u=Ae-2tcos(t)+ Be-2tsin(t). Using u(0)=1 and u'(0)=0, we get u=e-2tcos(t)+ 2e-2tsin(t).

  33. Higher Order Misc.: 100 What order is the equation ? Is it linear or non-linear?

  34. Higher Order Misc.: 100 The highest derivative in this equation is the 3rd derivative, so this is a 3rd order equation. The y4 means that the equation is non-linear.

  35. Higher Order Misc.: 200 For what values of r is y=ert a solution of

  36. Higher Order Misc.: 200 If y=ert, then y'=rert, y''=r2ert, y'''=r3ert. Plugging into the differential equation, we get ert(r3+10r2+21r)=0, which has solutions r=0, r=-3, r=-7.

  37. Higher Order Misc.: 300 Transform the differential equation into a system of first order equations. y(7)+ y(5)+ y(3)+ y=0.

  38. Higher Order Misc.: 300 Let x1=y, x2=y(1), x3=y(2), x4=y(3), x5=y(4), x6=y(5), x7=y(6) x7'= y(7)=- y(5)- y(3)- y=- x6 – x4 – x1 So the system is: x1'= x2, x2'= x3, x3'= x4, x4'= x5, x5'= x6, x6'= x6, x7'=- x6 – x4 – x1.

  39. Higher Order Misc.: 400 Find the general solution to the equation y(7)-6y(6)+20y(5)-56y(4)+112y(3)-160y(2)+192y'-128y=0 Hint: r7-6r6+20r5-56r4+112r3-160r2+192r-128= (r-2)3(r2 +4) 2

  40. Higher Order Misc.: 400 (r-2)3(r2 +4) 2 has roots r=2,2,2,2i,2i,-2i,-2i So the general solution is y=Ae2t+Bte2t+Ct2e2t+Dsin(2t)+ Ecos(2t)+Ftsin(2t)+Gtcos(2t).

  41. Higher Order Misc.: 500 Guess the form of a particular solution of the equation: y''+3y'=2t4+t2e-3t+sin(3t).

  42. Higher Order Misc.: 500 The homogeneous part has solution y=C1+C2e-3t Now, we guess the particular solution Y=t(At4+Bt3+Ct2+Dt+E)+ t(Ft2+Gt+H) e-3t+Isin(3t)+Jcos(3t)

  43. Matrix Fun: 100 Compute the Wronskian of y1=e2ty2=sin(t) y3=cos(t)

  44. Matrix Fun: 100

  45. Matrix Fun: 200 Let…. Compute xTy and (x,y).

  46. Matrix Fun: 200 xTy = (1-2i)(4+i)+i(3-i)+2(5i) = 7+6i (x,y) = =(1-2i)(4-i)+i(3+i)+2(-5i)= 1-16i

  47. Matrix Fun: 300 Compute the inverse of:

  48. Matrix Fun: 300 Begin with the matrix: And do row operations to get…

  49. Matrix Fun: 400 Compute the Eigenvalues and Eigenvectors of the matrix:

  50. Matrix Fun: 400 =-(-1)(-4)(-7) Eigenvalues are=1,4,7. Now plug each  into A- I, and solve (A- I)x=0 to get A-I=

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