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anonymous

  • 5 years ago

anyone a pro with differential equations? It has been too long and I need some help

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  1. anonymous
    • 5 years ago
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    I feel ok with them, not pro but hopefully I can help

  2. anonymous
    • 5 years ago
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    \[y'''+y=0 \] where y(o)=0 , y'(0) =1, y''(0)= 0 thank you please help

  3. anonymous
    • 5 years ago
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    sorry I cannot help, havent done 3rd order ones.

  4. anonymous
    • 5 years ago
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    Ok how about find all solutions to y'-2y=1 thank you for trying

  5. anonymous
    • 5 years ago
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    you can multiply with an integrating factor, do you remember that?

  6. anonymous
    • 5 years ago
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    I will get a pen and paper

  7. anonymous
    • 5 years ago
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    thank you I know its alot of work

  8. anonymous
    • 5 years ago
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    ok I solved it

  9. anonymous
    • 5 years ago
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    so do you know the technique of integrating factor?

  10. anonymous
    • 5 years ago
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    you r the best lets see if I get it :)

  11. anonymous
    • 5 years ago
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    you want to make a full derivative out of the left hand side

  12. anonymous
    • 5 years ago
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    for this you need to have something like dy/dx+ some y=whatever

  13. anonymous
    • 5 years ago
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    my computer is slowing down sorry ...yes I remeber that

  14. anonymous
    • 5 years ago
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    no problem, what is the integrating factor here?

  15. anonymous
    • 5 years ago
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    ok

  16. anonymous
    • 5 years ago
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    after we integrate x dx on the other side

  17. anonymous
    • 5 years ago
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    it is \[e ^{\int\limits_{}^{} the multiplier of y}dx\]

  18. anonymous
    • 5 years ago
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    that is -2 here

  19. anonymous
    • 5 years ago
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    the integral of -2 is -2x (you dont need the constant here) so multiply through by \[e ^{-2}\]

  20. anonymous
    • 5 years ago
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    -2x

  21. anonymous
    • 5 years ago
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    Im sorry the computer wont let me type

  22. anonymous
    • 5 years ago
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    It keeps stopping but I am here reading...Yes I remeber now the -2x

  23. anonymous
    • 5 years ago
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    the result is \[e ^{-2x}\] dy/dx -2 e^{-2x} y=e ^{-2x}

  24. anonymous
    • 5 years ago
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    do you recognise the full derivative now?

  25. anonymous
    • 5 years ago
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    yes it is coming back to me

  26. anonymous
    • 5 years ago
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    what about when we have a trig function like y''+4y=cos x

  27. anonymous
    • 5 years ago
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    this is harder but my favourite :)

  28. anonymous
    • 5 years ago
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    better your fav than mine :) :) i have not done these in 17 years it is so hard to remember you r a big help

  29. anonymous
    • 5 years ago
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    first you have to think of the homogeneous equation (meaning y''+4y=0) you have to find the general solution for that, for this we use the auxiliary equation. that just says that ay''+by'+cy=0 and write a quadratic equation for it,\[a \lambda ^{2}+b \lambda +c=0\]

  30. anonymous
    • 5 years ago
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    where lambda=e^x

  31. anonymous
    • 5 years ago
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    or in the form of e^x

  32. anonymous
    • 5 years ago
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    here you will have to solve y^2+4=0

  33. anonymous
    • 5 years ago
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    do you know how to?

  34. anonymous
    • 5 years ago
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    I will be back in 5 min

  35. anonymous
    • 5 years ago
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    i have no idea Im trying to follow along with you and my book but so confused I need to watch a tutorial i think and i will be bac

  36. anonymous
    • 5 years ago
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    you can only solve this with complex numbers

  37. anonymous
    • 5 years ago
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    thank ufor your help

  38. anonymous
    • 5 years ago
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    you had enough? How is it that after 17 years you are doing this again?

  39. anonymous
    • 5 years ago
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    http://www.analyzemath.com/calculus/Differential_Equations/second_order.html This page explains it quite well

  40. anonymous
    • 5 years ago
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    i am a teacher but needed 2 credits of advanced math i truly didnt want to take up your time

  41. anonymous
    • 5 years ago
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    I will have an exam about this in 2weeks so it is not a waste of time for me

  42. anonymous
    • 5 years ago
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    I need alot of explanation and my computer is not cooperating ...to be honest this work is due tomorrow and my brain is blank we can continue if u want to help me..

  43. anonymous
    • 5 years ago
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    well it is up to you, I am free now for an hour

  44. anonymous
    • 5 years ago
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    Consider the equation y' + (cos x) y= e^-sinx find the solution that satisfies \[\phi\](\[\pi\] ) = pi

  45. anonymous
    • 5 years ago
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    what is the end bit? that isnt clear

  46. anonymous
    • 5 years ago
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    I guess you have to use the integrating factor here as well. it will be e^integral(cosx) so e^sinx.

  47. anonymous
    • 5 years ago
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    that way you get \[(e ^{sinx}y)'=1\]

  48. anonymous
    • 5 years ago
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    so the LHS= x+C y=(x+C)/e^sinx

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