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Loser66

  • one year ago

The general solution to a certain ODE is \(y=ax^2+bx+c\) (a, b,c are in real) Find the differential equation. Please, help

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  1. dan815
    • one year ago
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    |dw:1436378469720:dw|

  2. Loser66
    • one year ago
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    dan, what is your avatar?? it looks like a barbecue hahaha

  3. dan815
    • one year ago
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    a parabola is the solution to a constant 2nd derivative, u can see it for those like projectile ball throwing up questions

  4. Empty
    • one year ago
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    What is an ODE? A relationship between a function and its derivatives. Just take the derivative of y a few times and see if you can relate it.

  5. Loser66
    • one year ago
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    @dan815 I don't think it is ok @Empty Yes, I did but go nowhere y = ax^2 +bx+c y'= 2ax +b y"= 2a then??

  6. ganeshie8
    • one year ago
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    are you given any initial conditions ?

  7. Empty
    • one year ago
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    Yeah, y''=2a is your differential equation.

  8. Loser66
    • one year ago
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    @ganeshie8 Yes, but the question for using initial conditions come after we get the equation.

  9. Empty
    • one year ago
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    or even y'''=0

  10. ganeshie8
    • one year ago
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    presumably a,b,c are arbitrary constants, so they should not be there in the ode

  11. ganeshie8
    • one year ago
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    y''' = 0 looks good enough right ?

  12. Loser66
    • one year ago
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    @ganeshie8 the next question is : If y" (0) = 2A (big A) y'(0) = B and y(0) = C (big C) for the differential equation in part b) , find a, b, c

  13. Loser66
    • one year ago
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    So that I don't think we can use them in part b.

  14. ganeshie8
    • one year ago
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    Ohk so y'''=0 is not what you looking for ?

  15. Loser66
    • one year ago
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    nope, I am looking for something like Ay" + By'+ Cy =0

  16. Loser66
    • one year ago
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    where y = ax^2 +bx +c is the general solution of it

  17. ganeshie8
    • one year ago
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    \[y = ax^2+bx+c\tag{1}\] \[y(0)=C \implies C = a*0^2+b*0+c \implies c=C\tag{2}\] are we good here ?

  18. Empty
    • one year ago
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    y'''=0 is correct then, I don't really see the issue here.

  19. ganeshie8
    • one year ago
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    me neither

  20. Loser66
    • one year ago
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    @ganeshie8 Again, we are not allowed to use y''(0) = A, y'(0) = B, y(0) =C for part b) (the part we have to find out the differential equation)

  21. Empty
    • one year ago
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    a,b, and c are all arbitrary real numbers.

  22. ganeshie8
    • one year ago
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    we're done with finding the ode, it is y''' = 0

  23. Loser66
    • one year ago
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    @Empty the reason we cannot use y"' =0 is

  24. Loser66
    • one year ago
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    At the specific step, we have y^(m) =0. And it is not valid for example, if the solution is \(ax^{m-1}+...whatever\) we have \(y^{(m)}=0\) always, right? Hence, what is the goal of finding a differential equation which solution is y = ....?

  25. ganeshie8
    • one year ago
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    3 arbitrary constants require the ode to be of 3rd order, yes ?

  26. Loser66
    • one year ago
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    Continue, please.

  27. dan815
    • one year ago
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    |dw:1436379643346:dw|

  28. Loser66
    • one year ago
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    My thought: if y"'=0, I can say y =0 is a solution of it, why y = ax^2 +bx+c?

  29. Empty
    • one year ago
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    that is a particular solution with a=0, b=0, and c=0. No prblem!

  30. ganeshie8
    • one year ago
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    y = 0 is a solution when a=b=c=0 you get to choose the initial conditions

  31. dan815
    • one year ago
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    none of spotted that mistake!

  32. dan815
    • one year ago
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    you

  33. Loser66
    • one year ago
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    Hey, barbecue, use English, pleae

  34. ganeshie8
    • one year ago
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    y''' = 0 models the path of a projectile under constant acceleration do not belittle it just because it looks innocent ;p

  35. Empty
    • one year ago
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    I was going to say that example haha

  36. Loser66
    • one year ago
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    Ok, Let me ask my professor what he does really want.Because, if y"' =0, how can I solve part c) where they gave me the initial condition y"(a) = 2A....

  37. Loser66
    • one year ago
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    sorry y"(0) = 2A....

  38. ganeshie8
    • one year ago
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    Easy, use the solution y=ax^2+bx+c and get 3 equations based on the 3 given initial conditions. solve the system

  39. Loser66
    • one year ago
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    I got a = A, b = B, c = C

  40. ganeshie8
    • one year ago
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    I see why you were having second thoughts..

  41. Loser66
    • one year ago
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    but what is the differential equation. hihihihi...

  42. Empty
    • one year ago
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    y'''=0

  43. Empty
    • one year ago
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    ;)

  44. Loser66
    • one year ago
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    hahahah... lol, it seems all of us are stubborn.

  45. ganeshie8
    • one year ago
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    to my knowledge y''' = 0 and a=A, b=B, c=C looks good even though it leaves me wondering why ur prof wants you spend time on such trivial prob

  46. Empty
    • one year ago
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    \(y'''=0\) integrate \(y'' = k_1\) integrate \(y'=k_1x + k_2\) integrate \(y=\frac{k_1}{2}x^2+k_2x+k_3\) call \(\frac{k_1}{2}=a\), \(k_2\)=b, \(k_3=c\). y=ax^2+bx+c these are all arbitrary numbers because this is the general solution. :)

  47. dan815
    • one year ago
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    it feels weird to say y'''=0 becasuse the same solution can also be worked down to y'''' =0 and y'''''''''''''''''=0

  48. dan815
    • one year ago
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    it doesnt doesnt give enough information

  49. dan815
    • one year ago
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    y'''=0 can also arise from y''=0 0 is just weird

  50. ganeshie8
    • one year ago
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    3 arbitrary constants cannot make a general solution to a 4th order ode

  51. dan815
    • one year ago
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    im not saying its that order ODE just saiyng y'''=0 lacks info imo

  52. Empty
    • one year ago
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    y''''=0 has the general solution of \(y=ax^3+bx^2+cx+d\)

  53. ganeshie8
    • one year ago
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    what info is it lacking

  54. Loser66
    • one year ago
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    I think it is y" +xy' -2y = 2a-bx-c

  55. Empty
    • one year ago
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    So although you can have a=0 to get a specific case where it has the same answer as y'''=0, this is saying that a _particular solution_ of y''''=0 is the same as the _general solution_ of y'''=0. So not quite the same thing danny boy!

  56. dan815
    • one year ago
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    well on 2nd thought

  57. Loser66
    • one year ago
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    I derive from y" = 2a y' = 2ax +b y = ax^2 +bx +c

  58. dan815
    • one year ago
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    i was looking at working backwards lets say suppse u are given y=ax^2+bx+c okay then differentiative that will give us its higher order repersentation

  59. dan815
    • one year ago
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    y=ax^2+bx+c y'=2ax+b y''=2a y'''=0 but from here on y''''=0 y''''''''''''''''''=0

  60. dan815
    • one year ago
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    you are saying that solution you have is general but that is just existing because u chose to work the other way around, you are following some set steps but that doesnt have to be the steps you follow for a solution

  61. Empty
    • one year ago
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    Yes but the y'''''''''''''''''''''''''''''''''''''''''''''''''''''=0 has a different general solution which is not the same as what the question is asking.

  62. dan815
    • one year ago
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    so i feel like y''''=0 just lacks info

  63. ganeshie8
    • one year ago
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    Ahh right, y'''' = 0 tells us that the jerk is constant, but tells nothing about the acceleration

  64. ganeshie8
    • one year ago
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    so we do lose info when we differentiate more than necessary

  65. Empty
    • one year ago
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    Yeah this solution I think is supposed to not be thought about too much, I feel like the question is so easy that we're over thinking this lol

  66. dan815
    • one year ago
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    lol we are just trolling lets face it

  67. dan815
    • one year ago
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    ive been trolling since 2014

  68. ganeshie8
    • one year ago
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    been wondering what is it on ur profile pic

  69. dan815
    • one year ago
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    ribs lol

  70. dan815
    • one year ago
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    its a secret experiment to see if this pic will increase the average weight of OS over time

  71. ganeshie8
    • one year ago
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    lol just so it increases ur appetite is it

  72. ganeshie8
    • one year ago
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    i mean indirectly ur weight

  73. dan815
    • one year ago
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    working in the advertising industry must be fun, they get to do experiments on people all the time

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