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amistre64

  • 2 years ago

\[y''-4y=(x^2-3)sin(2x)\] solve the diffyQ; oy what a nightmare ...

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  1. Lukecrayonz
    • 2 years ago
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    I'd help you if I could, but when you solve this I posted one for you to help me with.

  2. amistre64
    • 2 years ago
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    \[y=y_h+y_p\] \[y_h=c_1 e^{2x}+c_2 e^{-2x}\] its that yp thats making me loose my religion :)

  3. Calculus102
    • 2 years ago
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    Wooooooow. Just wondering, where did you get this problem from?

  4. Lukecrayonz
    • 2 years ago
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    Where the hell did c come into play?

  5. amistre64
    • 2 years ago
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    i call it #14 form section 4.3 ... from HELL!!! lol

  6. Hermeezey
    • 2 years ago
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    @Calculus102 It's differential equations. It's like a class after Calc III

  7. Calculus102
    • 2 years ago
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    Calc 3??? Like multivariate calculus?

  8. Lukecrayonz
    • 2 years ago
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    I'm done here, I am zero help.

  9. Lukecrayonz
    • 2 years ago
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    I'm in PRE calculus. This is madness.

  10. amistre64
    • 2 years ago
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    \[y_p=(ax^2+bx+c)sin(2x)+(fx^2+gx+h)cos(2x)\]

  11. amistre64
    • 2 years ago
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    calciii is cake compared to having to deal with this tripe ....

  12. bahrom7893
    • 2 years ago
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    Bernoulli's method, anyone?

  13. krypton
    • 2 years ago
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    this is very simple ,chain rule,product rule and some pellet mixed together

  14. amistre64
    • 2 years ago
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    burnoodle aint gonna help on this one

  15. bahrom7893
    • 2 years ago
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    wait nevermind... bernoulli was if u had y^n lol not nth derivative.. i suck at math

  16. amistre64
    • 2 years ago
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    its keeping track of the mindnumbingness till the end ... i dont think i have a happy personality left in me after this lol

  17. bahrom7893
    • 2 years ago
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    I took diff eq's.. use the char equation? r^2-4r = 0 r(r-4)=0 r = 0, r = 4

  18. bahrom7893
    • 2 years ago
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    but what about the right? i forgot.. and i only took this like a year ago.

  19. amistre64
    • 2 years ago
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    yeah the homogenous parts a breeze done did that

  20. amistre64
    • 2 years ago
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    the right is one of those "make a good guess" type deals

  21. amistre64
    • 2 years ago
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    heres my guess \[y_p=(ax^2+bx+c)sin(2x)+(fx^2+gx+h)cos(2x)\]

  22. bahrom7893
    • 2 years ago
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    ohh one of the rules? anyway i gotta eat dinner, mom's yelling at me lol.

  23. amistre64
    • 2 years ago
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    \[y_p=(ax^2+bx+c)sin(2x)+(fx^2+gx+h)cos(2x)\] \[y' _p=(2ax+b)sin(2x)+(2ax^2+2bx+2c)cos(2x)\]\[\hspace{5em} +(2fx+g)cos(2x)+(-2fx^2-2gx-2h)sin(2x)\] \[y'' _p=(2a)sin(2x)+(4ax+2b)cos(2x)\]\[\hspace{3em}+(4ax+2b)cos(2x)+(-4ax^2-4bx-4c)sin(2x)\]\[\hspace{6em} +(2f)cos(2x)+(-4fx-2g)sin(2x)\]\[\hspace{9em}+(-4fx-2g)sin(2x)+(-4fx^2-4gx-4h)cos(2x)\]

  24. amistre64
    • 2 years ago
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    \[(-4ax^2-4bx-4c)sin(2x)\]\[(2a-4ax^2-4bx-4c-4fx-2g-4fx-2g)sin(2x)\] \[(-4fx^2-4gx-4h)cos(2x)\]\[(4ax+2b+4qx+2b+2f-4fx^2-4gx-4h)cos(2x)\] \[(-8ax^2-8bx-8c+2a-8fx-4)sin(2x)=(x^2-3)sin(2x)\]\[a=-\frac{1}{8}\] \[(-8bx-8c-\frac{1}{4}-8fx-4)=(-3)\] ugh ... messed it up again \[(-8fx^2-8gx-8h+4b+8ax+4b+2f)cos(2x)=0cos(2x)\]

  25. krypton
    • 2 years ago
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    LOL...THIS CAN BE pelletY

  26. bahrom7893
    • 2 years ago
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    satellite, where does your math knowledge come from?

  27. bahrom7893
    • 2 years ago
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    Wait LOOOLL, u're amistre!!! Doesn't matter man, where does your knowledge come from amistre?

  28. bahrom7893
    • 2 years ago
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    You guys are epic! Like seriously! I'm very impressed, look up to you guys :))!

  29. amistre64
    • 2 years ago
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    \[(-8ax^2-8bx-8c+2a-8fx-4g)=(x^2-3)\] \[(-8a)x^2+(-8b-8f)x+(2a-8c-4g)=x^2+0x-3\] i gots no math knowledge ;)

  30. amistre64
    • 2 years ago
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    well, I know a = -1/8 regardless of what that other stuff does :)

  31. amistre64
    • 2 years ago
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    and b=-f ...

  32. amistre64
    • 2 years ago
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    i spose we gotta undo the cosines now and have some systems of equations to play with for the remaing terms

  33. amistre64
    • 2 years ago
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    (-8f) x^2+(-8g+8a)x+(-8h+8b+2f)=0x^2+0x+0 f = 0 so b = 0

  34. amistre64
    • 2 years ago
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    which means h=0 too for that constant

  35. amistre64
    • 2 years ago
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    (-8(0)) x^2+(-8g+8a)x+(-8(0)+8(0)+2(0))=0x^2+0x+0 -8g+8a = 0; and a = -1/8 -8g-1 = 0 ; g = -1/8

  36. amistre64
    • 2 years ago
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    2(-1/8)-8c-4(-1/8)=-3 -1/4 -8c +1/2=-3 -8c =-3' 1/4 c =13/32

  37. amistre64
    • 2 years ago
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    \[y''-4y=(x^2-3)sin(2x)\] \[y=c_1e^{2x}+c_2e^{-2x}-(\frac{1}{8}x^2-\frac{13}{32})sin(2x)-\frac{1}{8}xcos(2x)\]

  38. amistre64
    • 2 years ago
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    and i believe the wolf agrees .... yay!!

  39. amistre64
    • 2 years ago
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    if JamesJ had been here he could of answered it with one hand tied behind his back ;)

  40. gopeder
    • 2 years ago
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    Amistre64? is that you?

  41. gopeder
    • 2 years ago
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    remember me its gopeder :)

  42. amistre64
    • 2 years ago
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    thats what it says on my nametag :)

  43. gopeder
    • 2 years ago
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    Do you Remember me im the bear guy :)

  44. amistre64
    • 2 years ago
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    i vaguely recall a screenname like yours; at my age its all i can do to remember to wear socks

  45. gopeder
    • 2 years ago
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    xD we talked alot maybe 4 mounts ago

  46. amistre64
    • 2 years ago
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    yeah, 4 months is an eternity around here lol how you been?

  47. gopeder
    • 2 years ago
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    haha so true :) Not much Just starting Geometry.. its loads of fun

  48. amistre64
    • 2 years ago
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    if you ever meet euclid, punch him in the breadbasket for me. hes to blame for alot of that geometry stuff

  49. gopeder
    • 2 years ago
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    xD oh i will :P

  50. gopeder
    • 2 years ago
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    But at the end of the day its really cool because it relates to real world

  51. bahrom7893
    • 2 years ago
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    Amistre, sorry to bother you guys, but can you take a look at this? please? http://openstudy.com/study#/updates/4f4d9f77e4b019d0ebade348

  52. robtobey
    • 2 years ago
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    Refer to the attachment.

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  53. amistre64
    • 2 years ago
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    We got "derive" at the college that looks quite similar to that kind of thing.

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