Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

RolyPoly

Solve by undetermined coefficients/variation of parameters: \(x^3y'' + x^2y' -4xy = 1\) , x>0

  • one year ago
  • one year ago

  • This Question is Closed
  1. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    \[x^3y'' + x^2y' -4xy = 1\]\[x^2y'' + xy' -4y = \frac{1}{x}\]\[x^2y_c'' + xy_c' -4y_c =0\]\[\lambda (\lambda-1)+\lambda - 4 =0\]\[\lambda = \pm2\]\[y_c = c_1x^2+c_2x^{-2}\]

    • one year ago
  2. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    hi @RolyPoly are you in DE

    • one year ago
  3. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    \[u_1'(x^2)+u_2'(x^{-2})=0\]\[u_1'(2x)+u_2'(\frac{-2}{x^3})=\frac{1}{x}\] \[u_1'(x^2)+u_2'(x^{-2})=0\]\[u_1'(2x^2)+u_2'(\frac{-2}{x^2})=1\] \[4x^2u_1'=1\]\[u_1'= \frac{1}{4x^2}\]\[u_1 = -\frac{1}{4x}\] I'm doing something wrong, probably.. @mark_o. Yes.

    • one year ago
  4. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    are you familiar with cauchy -euler equation? that is the problem you are having there

    • one year ago
  5. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    are your class in that chapter already?

    • one year ago
  6. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    chapter in cauchy -euler equation ?

    • one year ago
  7. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes. I'm having trouble with finding the particular solutions.

    • one year ago
  8. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    The Cauchy-Euler part is done, I think.

    • one year ago
  9. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    I divided both sides by x...

    • one year ago
  10. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    And isn't it a second order DE only?

    • one year ago
  11. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Wwhat is it??

    • one year ago
  12. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    thats the formula for getting the characteristic equation

    • one year ago
  13. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Why doesn't my method work?

    • one year ago
  14. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    hold on lets double check it by using this formula

    • one year ago
  15. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    And for Cauchy-Euler equation, isn't it the power of x and the no. of times of derivatives of y the same? (eg: x^2y'')

    • one year ago
  16. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    it may not be the same so lets try doing the the formula one im giving you

    • one year ago
  17. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    But this is not a 3rd order DE!

    • one year ago
  18. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    im sorry i mess up,, yes its a 2nd order

    • one year ago
  19. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    So will that m^3 thing work here?

    • one year ago
  20. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    no,we have to use the (am^2+(b-a)m+c)x^m=0

    • one year ago
  21. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Wait.. Please refer to my first comment here. The Cauchy-Euler thing is done, I think.

    • one year ago
  22. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i think your ok

    • one year ago
  23. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    The problem is how to find the particular solution.

    • one year ago
  24. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    ok take x=e^t

    • one year ago
  25. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Why?

    • one year ago
  26. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    so that we can reduced i/x into function of t

    • one year ago
  27. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Not i/x but 1/x

    • one year ago
  28. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    yes 1/x

    • one year ago
  29. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    and it's y=...(in terms of x) How come we have a t?!

    • one year ago
  30. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    let x=e^t then 1/x==1/e^t=e^-t

    • one year ago
  31. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    temporarily so that we can get Yp

    • one year ago
  32. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    How come...

    • one year ago
  33. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    assume Yp=Ae^-t

    • one year ago
  34. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    It's not the form of yp

    • one year ago
  35. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    Y=ae^-t Y'=-Ae^t Y''=Ae^t now sub this to your equation

    • one year ago
  36. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    The yp is -1/(3x)...

    • one year ago
  37. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm try reading this first http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation

    • one year ago
  38. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    I think you need to know this is not a Cauchy-Euler equation.

    • one year ago
  39. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    http://www.sosmath.com/diffeq/second/euler/euler.html

    • one year ago
  40. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Since the right side \(\ne\) 0

    • one year ago
  41. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    eventhough its not zero you can solve it using cauchy - euler eq

    • one year ago
  42. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    this may be a good one to read http://www.sosmath.com/diffeq/second/euler/euler.html

    • one year ago
  43. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    But Cauchy-Euler is over after we get the yc, isn't it?

    • one year ago
  44. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    not until you convert everything to x=e^t

    • one year ago
  45. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    try to convert the whole thing to x=e^t

    • one year ago
  46. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    so let a=1, b=i,c=-4 then Y''(t)+0Y'(t)-4Y=e^-t that will be your equation

    • one year ago
  47. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm sorry, I must go now. Can we continue tomorrow?

    • one year ago
  48. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    m^2_4=0 m=+2.-2 Yc(t)=C1e^-2t +C2e^2 next finding Yp

    • one year ago
  49. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    that was m^2-4=0 m=-2,+2

    • one year ago
  50. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    yes ok we can continue tomorrow... :D

    • one year ago
  51. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks.

    • one year ago
  52. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    ok YW good luck now and have fun :D

    • one year ago
  53. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    @sirm3d can you give us some input here?

    • one year ago
  54. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    :D

    • one year ago
  55. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    .......................... :D

    • one year ago
  56. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    by variation of parameters, let \(y=v\cdot\frac{1}{x^2}\)

    • one year ago
  57. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    its been awhile since i took DE, i need to brush up again lol

    • one year ago
  58. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    yesss variation of parameter too...but i think euler is much easier

    • one year ago
  59. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Looks like I'm not doing this right... \[y=\frac{v}{x^2}\]\[y'=-\frac{2v}{x^3}\]\[y''=-\frac{2v'}{x^3}+\frac{6v}{x^4}\] \[x^3(-\frac{2v'}{x^3}+\frac{6v}{x^4}) +x^2(-\frac{2v}{x^3}) - 4x(\frac{v}{x^2})=1\]\[-2v'+\frac{6v}{x} -\frac{2v}{x} - \frac{4v}{x}=1\]\[v'=-\frac{1}{2}\]\[v=-\frac{1}{2}x\] \[y_p=-\frac{1}{2x}\]Hmm... Where did I make mistakes?

    • one year ago
  60. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    differentiation of product yields \[\large y'=\frac{v'}{x^2}-2\frac{v}{x^3}\]

    • one year ago
  61. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    But shouldn't we ''kill'' the last term to avoid troubles when differentiate y' again?

    • one year ago
  62. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Sorry, I meant the first term :\

    • one year ago
  63. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large y''=\frac{v''}{x^2}-2\frac{v'}{x^3}-2\frac{v'}{x^3}+6\frac{v}{x^4}\]

    • one year ago
  64. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Hmm.. Would you mind explaining why you did so? It's, somehow, different from what I've learnt in the lesson..

    • one year ago
  65. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    derivative of a product \(v\) and \(1/x^2\)

    • one year ago
  66. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    I... mean why you did so...

    • one year ago
  67. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    When way I've learnt is to let \(u_1y_1\), \(u_2y_2\)be the particular solution.. But somehow.. you're doing something different (in this question and in the previous one). I really want to know why you did that and how you came up with this y=v/x^2 ..

    • one year ago
  68. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    \[x^3y''+x^2y'-4xy=x^3\left( \frac{ v'' }{ x^2 }-4\frac{ v' }{ x^3 } +6\frac{ v }{ x^4 }\right)+x^2\left( \frac{ v' }{ x^2 }-\frac{ 2v }{ x^3 }\right)-4x \left( \frac{ v }{ x^2 } \right)\]\[=xv''-4v'+6\frac{ v }{ x }+v'-2\frac{ v }{ x }-4\frac{ v }{ x }\]\[=xv''-3v'=1\]

    • one year ago
  69. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    the idea behind the variation of parameter is to use one complementary function that is a solution of the homogeneous DE. In this problem, you produced one complementary function\(1/x^2\) using cauchy-euler method.

    • one year ago
  70. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    the variation of parameter introduces a function \(v\) to solve THE particular solution using one complementary solution \(1/x^2\) by assuming that \(y_p=v(1/x^2)\) is the particular solution of the DE, where v is to be determined.

    • one year ago
  71. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large xv''-3v'=1\]let \(w=v'\) and \(w' = v''\)\[\large xw'-3w=1\]\[\large w'-\frac{3}{x}w=\frac{1}{x}\]

    • one year ago
  72. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large w(\frac{ 1 }{ x^3 })=\int\limits \frac{1}{x^4} dx=-(1/3)x^{-3}\]\[\large v'=w=(-1/3)\]\[\large v=(-1/3)x\]\[\large y_p=v(\frac{ 1 }{ x^2 })=-\frac{ 1 }{ 3x }\]

    • one year ago
  73. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    check: \[\large x^3y''+x^2y'-4xy=x^3\left( -\frac{ 2 }{ 3x^3 } \right)+x^2\left( \frac{ 1 }{ 3x^2 } \right)-4x \left( -\frac{ 1 }{ 3x } \right)\]\[\large -\frac{ 2 }{ 3 }+\frac{ 1 }{ 3 }+\frac{ 4 }{ 3 }=1\] and so it's solved.

    • one year ago
  74. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    But I think there are two complementary solutions? \(x^2\) and \(x^{-2}\)?!

    • one year ago
  75. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    variation of parameter allows you to choose any of the complementary solutions. i chose \(x^{-2}\) over \(x^2\) for no particular reason. why don't you try \(y=vx^2\) ? i have yet to learn your method to find why you get -1/4 when i got -1/3

    • one year ago
  76. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Wouldn't it be extremely weird?! I think I chose both the complementary functions..but you just chose one.. It's weird...

    • one year ago
  77. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Consider y''-4y'+4y = e^(2x) / x \[y_c = c_1e^{2x} +c_2xe^{2x}\] The way I find the particular solution is \[u_1'e^{2x}+u_2'(xe^{2x})=0\]\[u_1'(2e^{2x})+u_2'(2xe^{2x}+e^{2x})=\frac{e^{2x}}{x}\]and find u1 u2. Then, there are two parts in the particular solution...

    • one year ago
  78. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    is the particular solution \[\large y_p=-x e^{2x}+(\ln x) x e^{2x}\]

    • one year ago
  79. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    i got that result following your method.

    • one year ago
  80. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    if i use my method, and choose the complementary solution \(e^{2x}\)\[\large y=v e^{2x}\]\[\large y'=v'e^{2x}+2v e^{2x}\]\[\large y''=v'' e^{2x}+4v'e^{2x}+4v e^{2x}\]

    • one year ago
  81. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large y''-4y' + 4y=(v''e^{2x}+4v'e^{2x}+4v e^{2x})-4(v'e^{2x} + 2v e^{2x})+4(v e^{2x})\]\[\large =v''e^{2x}=\frac{e^{2x}}{x}\]\[\large v''=\frac{1}{x}\]

    • one year ago
  82. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large v'=\ln x\]\[\large v=x\ln x - x\]\[\large y=\left(x \ln x - x \right) e^{2x}\], no different from the result using your method.

    • one year ago
  83. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Is.. it...magic? :S Please let me try to do it using your method!!

    • one year ago
  84. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    can you provide me a link on your method?

    • one year ago
  85. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Link!? I don't have it :( Or can I just type the notes I copied in the lesson here? (scanner is not working now :( )

    • one year ago
  86. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    i found a link detailing your method. i'm reading it now.

    • one year ago
  87. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    @RolyPoly i saw some mistakes in your solution. your two equations are \[\large u_1'x^2+u_2x^{-2}=0\]\[\large u_1(2x)+u_2(-2x^{-3})=\frac{1/x}{x^2}=\frac{1}{x^3}\]

    • one year ago
  88. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Why should it be (1/x)/x^2 ??

    • one year ago
  89. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    just solve \(u_1\) and \(u_2\) and sub them in \[\large y_p= u_1x^2+u_2x^{-2}\] and you should get the same answer \(-1/(3x)\)

    • one year ago
  90. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    the second equation in the system is of the form \[\large u_1'y_1'+u_2'y_2'=\frac{r(x)}{b_nx^n}\] from the cauchy-euler \[\large b_nx^n y^{(n)}+\cdots +b_1xy'+b_0y=r(x)\]

    • one year ago
  91. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    you should get \(y_p=-1/(3x)\) this time after solving \(u_1\) and \(u_2\). this are the values i got: \[\large u_1=-\frac{1}{12x^3}\]\[\large u_2=-\frac{x}{4}\]

    • one year ago
  92. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    @RolyPoly hope i have given you peace of mind in this problem.

    • one year ago
  93. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    My... teacher... didn't .... told... us... that... the ...second equation is in the form \[ u_1'y_1'+u_2'y_2'=\frac{r(x)}{b_nx^n}\]

    • one year ago
  94. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    that may be the cause of your problems.

    • one year ago
  95. RolyPoly
    Best Response
    You've already chosen the best response.
    Medals 0

    Most probably.. That seems your method works better.. I have to try both methods to decide which to use.. Thanks!!! I hope you don't mind me tagging you if I still have problems concerning this question!

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.