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mathslover

  • one year ago

Solve the given differential equation. \[\cfrac{dy}{dx} = \cfrac{(x+y)^2}{(x+2)(y-2)}\]

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  1. mathslover
    • one year ago
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    @ganeshie8 @Australopithecus @robtobey

  2. Australopithecus
    • one year ago
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    What methods have you learned? I don't think this is separable

  3. mathslover
    • one year ago
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    I tried putting y = vx and x + y = v but none of them seem to work.

  4. Australopithecus
    • one year ago
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    First bring everything to one side and make it equal to zero

  5. mathslover
    • one year ago
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    \(\cfrac{(x+2)(y-2) dy}{(x+y)^2 dx} - 1 = 0 \) What next?

  6. Australopithecus
    • one year ago
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    That is in incorrect form

  7. Australopithecus
    • one year ago
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    You want the form: \[M(x,y)dx + N(x,y)dy = 0\]

  8. Australopithecus
    • one year ago
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    You can use substitution or you can check to see if it is exact and use that method

  9. Australopithecus
    • one year ago
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    Substitution is as follows: You have two choices for substitution: u = x/y then, dx = udy + ydu Or u = y/x then, dy = xdu + udx

  10. Australopithecus
    • one year ago
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    For setting it in the right form do you understand?

  11. mathslover
    • one year ago
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    Yeah, I've studied the method of substitution. In fact, I did try substituting y = vx but that didn't seem to work.

  12. Australopithecus
    • one year ago
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    You wrote it in the wrong form so that could be your problem

  13. mathslover
    • one year ago
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    I understand that point. But, am not sure where to put that \((x+y)^2\) ... \((y-2)dy = \cfrac{dx}{x+2} \times (x+y)^2\)

  14. Australopithecus
    • one year ago
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    So, for example: \[\frac{dx}{dy} = \frac{(y + x)}{(x + 2y)^2}\] in standard form would be: \[dx(x+2y)^2 - (y+x)dy = 0\] M(x,y) = (x + 2y)^2 N(x,y) = -(y+x)

  15. Australopithecus
    • one year ago
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    Sorry this would be standard form \[(x+2y)^2dx - (y+x)dy = 0\]

  16. mathslover
    • one year ago
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    Okay, so, I have : \((y-2)(x+2) dy - (x+y)^2 dx = 0 \) This is the standard form, right?

  17. Australopithecus
    • one year ago
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    Yup :D

  18. Australopithecus
    • one year ago
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    Ok now you have to test to see if you can even use the separation method

  19. Australopithecus
    • one year ago
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    Or substitution sorry

  20. Australopithecus
    • one year ago
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    This is clearly not seperable

  21. mathslover
    • one year ago
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    Now, I will simply use substitution : y = vx \(dy = vdx + xdv \) The equation becomes: \((vx - 2)(x+2)(vdx + xdv) -(x+vx)^2 dx = 0 \) \((vx-2)(x+2)(vdx + xdv) = x^2 (v+1)^2 dx \) Should I proceed further or am I doing something wrong?

  22. Australopithecus
    • one year ago
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    Expand (y-2)(x+2)

  23. Australopithecus
    • one year ago
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    Also expand (x+y)^2

  24. Australopithecus
    • one year ago
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    MY browser is lagging like crazy one second

  25. Australopithecus
    • one year ago
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    there is a test you have to do to make sure you can even use substitution method

  26. mathslover
    • one year ago
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    Okay, it will give me : \[(y-2)(x+2) dy - (x+y)^2 dx = 0 \\ xydy + 2(y-x) dy - 4dy - x^2 dx -y^2 dx -2xy dx = 0 \\ xydy - 2y dy - 2xdy - x^2 dx - y^2 dx - 2xydx = 0 \]

  27. mathslover
    • one year ago
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    Test? Never heard of it. Can you please explain?

  28. Australopithecus
    • one year ago
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    Now factor out dy and dx

  29. mathslover
    • one year ago
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    \[(xy - 2y - 2x)dy - (x^2 + y^2 + 2xy)dx = 0 \\ (xy - 2y - 2x)dy = (x^2 + y^2 + 2xy)dx \] Seems like I'm reverting everything back to the original equation :/ :(

  30. Australopithecus
    • one year ago
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    You can only do substitution method if \[M(\lambda x, \lambda y) = \lambda^pM(x,y)\ for\ all\ \lambda,x,y\] and \[N(\lambda x, \lambda y) = \lambda^qN(x,y)\ for\ all\ \lambda,x,y\] to use separation p must equal q

  31. Australopithecus
    • one year ago
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    First to check if M(x,y) or N(x,y) is homogenous here are some examples:

  32. Australopithecus
    • one year ago
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    The degree of homogeneity must be equal for both terms N(x,y) and M(x,y) to use the substitution method

  33. Australopithecus
    • one year ago
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    hence p = q

  34. Australopithecus
    • one year ago
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    its a pretty simple check you can do in your mind almost

  35. mathslover
    • one year ago
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    I seem to get your point! Will give it a try next morning. Gotta sleep now! :) Thanks a lot for your help! I'll let you know if I get stuck anywhere.

  36. Australopithecus
    • one year ago
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    Wait lets at least check to see

  37. Australopithecus
    • one year ago
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    because you might have to use another method

  38. mathslover
    • one year ago
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    Yeah sure!

  39. Australopithecus
    • one year ago
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    Is M(x,y) homogenous?

  40. Australopithecus
    • one year ago
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    and to what degree

  41. Australopithecus
    • one year ago
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    Sorry I know you are probably tired just I want to save you time

  42. Australopithecus
    • one year ago
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    If you are lost I dont mind giving you another example

  43. mathslover
    • one year ago
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    It doesn't seem to be homogeneous :( because of that constant there

  44. Australopithecus
    • one year ago
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    So you have to use a different method

  45. mathslover
    • one year ago
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    Oh! :( Any ideas for which method to use?

  46. Australopithecus
    • one year ago
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    Yup

  47. Australopithecus
    • one year ago
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    You can check to see if it is exact

  48. Australopithecus
    • one year ago
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    Do you want to sleep we can work on this later if you want?

  49. Australopithecus
    • one year ago
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    its actually a pretty long method

  50. Australopithecus
    • one year ago
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    kind of

  51. mathslover
    • one year ago
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    Oh! Well, I will love to give it a try now. I can control my sleep for studies!

  52. Australopithecus
    • one year ago
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    Alright, first step you need to test to see if the DE is exact or not

  53. Australopithecus
    • one year ago
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    A DE is said to be exact when, \[M_y(x,y) = N_x(x,y)\] You know how to take partial derivatives right?

  54. mathslover
    • one year ago
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    Yeah!

  55. Australopithecus
    • one year ago
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    Also I would put it in another form but I am not sure how to use write daran in latex

  56. Australopithecus
    • one year ago
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    So first step take both partial derivatives and check to see if they are equal

  57. Australopithecus
    • one year ago
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    if they are not this question is even longer ha

  58. mathslover
    • one year ago
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    \(\delta\) You mean this?

  59. Australopithecus
    • one year ago
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    that is lower case delta

  60. mathslover
    • one year ago
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    \(\Delta\) .. ?

  61. Australopithecus
    • one year ago
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    |dw:1432921439849:dw|

  62. Australopithecus
    • one year ago
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    anyways we are getting off topic

  63. Australopithecus
    • one year ago
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    Have you taken the partial derivatives? You should always take both of them because you will always need them to solve this kind of problem

  64. mathslover
    • one year ago
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    I will do that part. What I'm thinking right now is that this question appeared in my book before the topic of "Exact DE" ... and it wil surely involve the method of solving H.D.E

  65. Australopithecus
    • one year ago
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    I Dont know the HDE method

  66. mathslover
    • one year ago
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    I meant, solving Homogeneous Differential Equations

  67. Australopithecus
    • one year ago
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    oh im out of my mind lol

  68. Australopithecus
    • one year ago
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    anyways this is the I would use to solve this problem

  69. Australopithecus
    • one year ago
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    it should work

  70. Australopithecus
    • one year ago
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    I mean unless you make a mistake with algebra substitution method wont work with this

  71. mathslover
    • one year ago
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    Well, I guess, I might have made a mistake then. I will see it tomorrow morning and will let you know about the progress, if you don't mind. As I'm getting a bit out of mind too :(

  72. Australopithecus
    • one year ago
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    That's fine just message me or tag me or something I can keep going over this with you, This is essentially a really brief note on how to solve this problem IF it is exact

  73. Australopithecus
    • one year ago
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    If you need help later have a nice sleep

  74. mathslover
    • one year ago
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    :) Thanks a lot for that, @Australopithecus . And same to you (whenever you schedule to sleep :D )

  75. Australopithecus
    • one year ago
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    It is never

  76. mathslover
    • one year ago
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    Sleep is important ! Have one now, please! :D

  77. mathslover
    • one year ago
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    Well, I got it! We had to substitute x + 2 = u and y -2 = v and then work out with y = kx

  78. anonymous
    • one year ago
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    The ODE as given isn't homogeneous. For \(\dfrac{dy}{dx}=f(x,y)\) to be considered homogeneous, \(f\) has to satisfy \(f(tx,ty)=t^kf(x,y)\) for some real \(k\). \[\begin{align*}f(x,y)&= \frac{(x+y)^2}{(x+2)(y-2)}\\\\ f(tx,ty)&= \frac{(tx+ty)^2}{(tx+2)(ty-2)}\\ &= t^2\frac{(x+y)^2}{(tx+2)(ty-2)}\\ &\neq t^kf(x,y) \end{align*}\] which explains why you were running into trouble with the substitution \(y=vx\). As for your substitution, I was actually in the process of suggesting that! Nice work.

  79. mathslover
    • one year ago
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    Yeah! :) I don't know how did it click just when I woke up ! :D Indeed, I solved it in 5 minutes :P As compared to hour of work last night. And thanks for that explanation, @SithsAndGiggles ! :)

  80. anonymous
    • one year ago
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    Sometimes the hardest thing is remembering that \(2-2=0\) :P

  81. mathslover
    • one year ago
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    Haha! Agreed. This is what "Maths" is...! Happens in programming too, when the error is ";" .. and we keep finding it throughout the code :D It's fun...

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