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Astrophysics

  • one year ago

@ganeshie8

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  1. Astrophysics
    • one year ago
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    How can I tell where this solution is valid, the \[(2x-y)dx+(2y-x)dy = 0\] solving this exact equation I got \[\psi (x,y) = x^2-xy+y^2-7\] but I'm not sure how to tell where this is valid

  2. ganeshie8
    • one year ago
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    what are the initial conditions ?

  3. Astrophysics
    • one year ago
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    Oh oops! y(1) = 3

  4. Mimi_x3
    • one year ago
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    are you taking a DE COURSE???

  5. Mimi_x3
    • one year ago
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    study with me pls <3

  6. ganeshie8
    • one year ago
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    so the solution curve is \(x^2-xy+y^2-7=0\)

  7. Astrophysics
    • one year ago
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    Haha yeah I am

  8. Astrophysics
    • one year ago
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    Yes

  9. Astrophysics
    • one year ago
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    the back of the book says \[|x|<\sqrt{\frac{ 28 }{ 3 }}\]

  10. Mimi_x3
    • one year ago
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    STUDY WITH ME IM LIKE 5 weeks behind LOL in class we're like up to PDE's and im still stuggling with fourier :/

  11. ganeshie8
    • one year ago
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    |dw:1443339556634:dw|

  12. ganeshie8
    • one year ago
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    Notice that the solution curve is trapped between those vertical tangents we may use that observation to find the `x` bounds

  13. Astrophysics
    • one year ago
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    Yup I see, but I'm not sure how to do it algebraically

  14. ganeshie8
    • one year ago
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    As a start, implicitly differentiate the equation of solution curve and find an expression for \(y'\)

  15. Astrophysics
    • one year ago
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    kk

  16. ganeshie8
    • one year ago
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    alternatively you may think of the solution curve as a quadratic in \(y\), then use the discriminant to see for what values of \(x\), the value of \(y\) is real

  17. Astrophysics
    • one year ago
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    \[y' = \frac{ 2x-y }{ x-2y }\]

  18. Astrophysics
    • one year ago
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    Ooh so how would you set that up using quadratic formula

  19. ganeshie8
    • one year ago
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    in above expression for derivative, \(x=2y\) makes the derivative go wild, so vertical tangents must occur when \(x=2y\) using that eliminate \(y\) in the solution curve and solve \(x\)

  20. Astrophysics
    • one year ago
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    That works!

  21. ganeshie8
    • one year ago
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    If that derivative thing doesn't strike immediately, you may simply use the disrcriminat : \(\color{red}{y^2}-\color{red}{y}x+x^2-7=0\) this is a quadratic in \(\color{red}{y}\) : \(a=1\) \(b=-x\) \(c=x^2-7\) Discriminant must be nonnegative for \(y\) to be real. so \(x^2-4(1)(x^2-7)\ge 0\), we can solve \(x\)

  22. Astrophysics
    • one year ago
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    Ha, I like the derivative thing, so this is finding the same thing as interval of definition?

  23. ganeshie8
    • one year ago
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    Yes this is like finding the largest interval in which the solution curve is defined

  24. Astrophysics
    • one year ago
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    So how come it doesn't say \[-\sqrt{\frac{ 28 }{ 3 }} < x < \sqrt{\frac{ 28 }{ 3 }}\]

  25. Astrophysics
    • one year ago
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    < = to I believe it should be

  26. ganeshie8
    • one year ago
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    dy/dx is not defined at either ends, so we must exclude them

  27. Astrophysics
    • one year ago
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    Oh right!

  28. Astrophysics
    • one year ago
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    Haha

  29. ganeshie8
    • one year ago
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    the derivative thingy is easy because the derivative is readily available for us : \((2x-y)dx+(2y-x)dy = 0 \implies \dfrac{dy}{dx}=\dfrac{y-2x}{2y-x}\) Clearly you don't wanto plug \(2y=x\) in above equation

  30. Astrophysics
    • one year ago
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    Yup, I really like this method

  31. Astrophysics
    • one year ago
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    Solving O.D.E's isn't too bad, it's this interval stuff that can be confusing as it requires you to remember stuff from calc 1 >.<

  32. Astrophysics
    • one year ago
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    Thanks again

  33. ganeshie8
    • one year ago
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    the interval stuff is really not that easy, there are several things that you need to consider when working the domain of the solution

  34. ganeshie8
    • one year ago
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    this pdf talks about "why" domain is important and "how" to find the domain http://apcentral.collegeboard.com/apc/members/repository/ap07_calculus_DE_domain_fin.pdf

  35. Astrophysics
    • one year ago
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    Awesome, thanks by the way...what exactly is the difference from the problems that say "determine the interval in which the solution is defined" and from this where it asks you for "where solution is valid"

  36. ganeshie8
    • one year ago
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    both are same

  37. Astrophysics
    • one year ago
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    Ah ok haha, thanks xD

  38. ganeshie8
    • one year ago
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    except for philosophical differences, both refer to the same thing

  39. Astrophysics
    • one year ago
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    Ok cool, yeah I guess it's just they word each question differently, so I thought there must be something different about them even though they sound the same

  40. ganeshie8
    • one year ago
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    1) "determine the interval in which the solution is defined" 2) "where solution is valid" first one is in ur control, "You" can define the interval where your solution would be valid while modeling the differential equaiton based on physical constraints. for the second one, you need to work out the domain by looking at the solution but there is not much difference between them, at least i never worried about the differences..

  41. Astrophysics
    • one year ago
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    Ok that makes sense, like I had a separable equation problem earlier and it asked for where it was defined \[\frac{ dy }{ dx } = \frac{ arcsinx }{ y^2(1-x^2)^{1/2} }\] I sort of just looked at the derivative here and got -1<x<1, but I guess I should be paying attention to vertical tangents as well?

  42. ganeshie8
    • one year ago
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    what are the initial conditions ?

  43. Astrophysics
    • one year ago
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    y(0)=1

  44. ganeshie8
    • one year ago
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    whats the solution ?

  45. IrishBoy123
    • one year ago
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    .

  46. ganeshie8
    • one year ago
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    you must consider the DE, the initial conditions and the solution while working the domain

  47. ganeshie8
    • one year ago
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    All 3 things can change the domain

  48. Astrophysics
    • one year ago
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    Ok yeah that's what I did, that was sort of a bad example

  49. Astrophysics
    • one year ago
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    I didn't include all the info

  50. Astrophysics
    • one year ago
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    The solution was \[y(x) = \left[ \frac{ 3 }{ 2 }\arcsin^2x \right]^{1/3}+1\]

  51. Astrophysics
    • one year ago
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    \[y^2dy = \frac{ arcsinx }{ (1-x^2) }dx\] y(0)=1 so not too hard of a problem

  52. ganeshie8
    • one year ago
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    -1<x<1 looks good to me why do you think it is wrong ?

  53. Astrophysics
    • one year ago
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    Oh no I didn't think it was wrong it made sense to me, but I didn't really consider the vertical tangent

  54. Astrophysics
    • one year ago
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    Ok nvm, I think I have got a decent idea of this stuff now hehe, thanks!

  55. ganeshie8
    • one year ago
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    try finding the domain of this IVP \[\dfrac{dy}{dx}=\dfrac{2}{\sqrt{y}};~~y(0) = 9\]

  56. Astrophysics
    • one year ago
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    Ok give me a sec

  57. Astrophysics
    • one year ago
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    Omg C seems ugly X_X

  58. Astrophysics
    • one year ago
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    nvm its 18?

  59. ganeshie8
    • one year ago
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    you may simply use wolfram to find the solution http://www.wolframalpha.com/input/?i=solve+dy%2Fdx%3D2%2Fsqrt%28y%29%2C+y%280%29%3D9

  60. Astrophysics
    • one year ago
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    Ok haha, \[y=3^{2/3}(x+9)^{2/3}\] so the domain then is x>-9

  61. Astrophysics
    • one year ago
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    \[x \ge -9\]

  62. ganeshie8
    • one year ago
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    why

  63. ganeshie8
    • one year ago
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    what happens if i plugin x = -10

  64. Astrophysics
    • one year ago
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    you get -3^(2/3)

  65. Astrophysics
    • one year ago
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    So that works

  66. ganeshie8
    • one year ago
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    so nothing wrong with x<-9 ?

  67. Astrophysics
    • one year ago
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    Nope, so all real numbers then?

  68. ganeshie8
    • one year ago
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    Look at the graph of solution again : |dw:1443344660260:dw|

  69. ganeshie8
    • one year ago
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    then look at the differential equation again

  70. ganeshie8
    • one year ago
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    the graph is not matching with the given differential equation, why ?

  71. ganeshie8
    • one year ago
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    |dw:1443344789571:dw|

  72. Astrophysics
    • one year ago
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    Nooo, because we then get undefined

  73. ganeshie8
    • one year ago
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    right, the slope of tangent to solution curve, y, is negative for y < -9, yes ?

  74. Astrophysics
    • one year ago
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    Yup

  75. ganeshie8
    • one year ago
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    |dw:1443345018801:dw|

  76. Astrophysics
    • one year ago
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    But see that's where I was thinking, when we look at the derivative itself we would have to consider the vertical tangets

  77. Astrophysics
    • one year ago
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    tangets*

  78. ganeshie8
    • one year ago
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    its not primarily about vertical tangents here its much more deeper than that, the slope field of the given differential equation cannot have negative slope segments

  79. ganeshie8
    • one year ago
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    so we must exclude x<-9 from the domain of the solutions

  80. Astrophysics
    • one year ago
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    Ah!

  81. ganeshie8
    • one year ago
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    as you can see, finding domain is not so straightforward and definitely not easy

  82. Astrophysics
    • one year ago
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    Yeah, it's pretty confusing haha

  83. Astrophysics
    • one year ago
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    So we're sort of restricting it?

  84. ganeshie8
    • one year ago
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    since the derivative is always "positive", x <= -9 makes no sense, so the domain is x > -9

  85. Astrophysics
    • one year ago
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    Ok I see

  86. Astrophysics
    • one year ago
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    Oh haha I'm reading the link, that's where you got the problem

  87. Astrophysics
    • one year ago
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    This stuff is confusing haha it's like I get it and at the same time I don't

  88. ganeshie8
    • one year ago
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    domains for differential equations are not same as that of algebraic equations nothing to feel bad about, it takes a while to get used to...

  89. Astrophysics
    • one year ago
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    I guess, I'll have to do lots of practice

  90. Astrophysics
    • one year ago
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    Well thanks a lot ganeshie it's nearly 4 am, been doing this stuff nearly all day haha, time for sleep, take care, bye!

  91. ganeshie8
    • one year ago
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    tc, have good sleep :)

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