Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

RolyPoly

  • 2 years ago

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

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

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

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

    hi @RolyPoly are you in DE

  3. RolyPoly
    • 2 years ago
    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.

  4. mark_o.
    • 2 years ago
    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

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

    are your class in that chapter already?

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

    chapter in cauchy -euler equation ?

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

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

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

    The Cauchy-Euler part is done, I think.

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

    I divided both sides by x...

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

    And isn't it a second order DE only?

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

    Wwhat is it??

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

    thats the formula for getting the characteristic equation

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

    Why doesn't my method work?

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

    hold on lets double check it by using this formula

  15. RolyPoly
    • 2 years ago
    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'')

  16. mark_o.
    • 2 years ago
    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

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

    But this is not a 3rd order DE!

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

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

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

    So will that m^3 thing work here?

  20. mark_o.
    • 2 years ago
    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

  21. RolyPoly
    • 2 years ago
    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.

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

    yes i think your ok

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

    The problem is how to find the particular solution.

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

    ok take x=e^t

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

    Why?

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

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

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

    Not i/x but 1/x

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

    yes 1/x

  29. RolyPoly
    • 2 years ago
    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?!

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

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

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

    temporarily so that we can get Yp

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

    How come...

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

    assume Yp=Ae^-t

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

    It's not the form of yp

  35. mark_o.
    • 2 years ago
    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

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

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

  37. mark_o.
    • 2 years ago
    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

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

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

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

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

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

    Since the right side \(\ne\) 0

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

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

  42. mark_o.
    • 2 years ago
    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

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

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

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

    not until you convert everything to x=e^t

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

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

  46. mark_o.
    • 2 years ago
    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

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

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

  48. mark_o.
    • 2 years ago
    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

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

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

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

    yes ok we can continue tomorrow... :D

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

    Thanks.

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

    ok YW good luck now and have fun :D

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

    @sirm3d can you give us some input here?

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

    :D

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

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

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

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

  57. mark_o.
    • 2 years ago
    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

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

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

  59. RolyPoly
    • 2 years ago
    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?

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

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

  61. RolyPoly
    • 2 years ago
    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?

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

    Sorry, I meant the first term :\

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

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

  64. RolyPoly
    • 2 years ago
    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..

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

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

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

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

  67. RolyPoly
    • 2 years ago
    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 ..

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

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

  69. sirm3d
    • 2 years ago
    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.

  70. sirm3d
    • 2 years ago
    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.

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

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

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

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

  73. sirm3d
    • 2 years ago
    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.

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

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

  75. sirm3d
    • 2 years ago
    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

  76. RolyPoly
    • 2 years ago
    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...

  77. RolyPoly
    • 2 years ago
    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...

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

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

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

    i got that result following your method.

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

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

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

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

  82. sirm3d
    • 2 years ago
    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.

  83. RolyPoly
    • 2 years ago
    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!!

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

    can you provide me a link on your method?

  85. RolyPoly
    • 2 years ago
    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 :( )

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

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

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

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

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

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

  89. sirm3d
    • 2 years ago
    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 \(-1/(3x)\)

  90. sirm3d
    • 2 years ago
    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 \[\large b_nx^n y^{(n)}+\cdots +b_1xy'+b_0y=r(x)\]

  91. sirm3d
    • 2 years ago
    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: \[\large u_1=-\frac{1}{12x^3}\]\[\large u_2=-\frac{x}{4}\]

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

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

  93. RolyPoly
    • 2 years ago
    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 \[ u_1'y_1'+u_2'y_2'=\frac{r(x)}{b_nx^n}\]

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

    that may be the cause of your problems.

  95. RolyPoly
    • 2 years ago
    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!

  96. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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.