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lilMissMindset Group Title

(D^2 + 9)y =5e^x - 162x i need to check if my answer is right.

  • 2 years ago
  • 2 years ago

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  1. lilMissMindset Group Title
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    @AnimalAin this.

    • 2 years ago
  2. partyrainbow276 Group Title
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    Awesome! u did it write

    • 2 years ago
  3. lilMissMindset Group Title
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    this is my answer: 2Ae^x + B + Cx = 5e^x -162x

    • 2 years ago
  4. partyrainbow276 Group Title
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    lets see...... AWESOME AWESOME :) thats right good job

    • 2 years ago
  5. AnimalAin Group Title
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    It's been a minute since I've done these, but I think the way is to consider the homogeneous solution by using an auxilliary (sp?) equation, then fit a particular solution that will match the RHS.

    • 2 years ago
  6. partyrainbow276 Group Title
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    lilmissmindset is right thats what i got for my answer

    • 2 years ago
  7. AnimalAin Group Title
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    That's the problem we did before.....

    • 2 years ago
  8. lilMissMindset Group Title
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    yes yes, this is the first question i asked. it's just that the book says the general solution for this is \[=C _{1} \cos 3x+ C _{2} \sin3x + \frac{ 1 }{ 2 } e ^{x} -18xy \]

    • 2 years ago
  9. lilMissMindset Group Title
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    there's no y after -18x SORRY

    • 2 years ago
  10. AnimalAin Group Title
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    Lets look at this in more detail.

    • 2 years ago
  11. lilMissMindset Group Title
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    and our answer is 5/2 e^x -162x

    • 2 years ago
  12. lilMissMindset Group Title
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    i thought, maybe there's something wrong with me, getting the solution after all

    • 2 years ago
  13. AnimalAin Group Title
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    \[(D^2 + 9)y =5e^x - 162x\]Consider\[(D^2 + 9)y =0\]Aux equation\[r^2+9=0 \implies r= \pm 3i \implies y _{h}=C _{1} \cos 3x +C_2 \sin3x\]OK so far?

    • 2 years ago
  14. lilMissMindset Group Title
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    yeps. ok.

    • 2 years ago
  15. lilMissMindset Group Title
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    ok with that part,

    • 2 years ago
  16. AnimalAin Group Title
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    Then we find the particular solution. Consider the first part of the RHS by itself, so\[(D^2 + 9)y_p =5e^x \implies y_p=\frac{e^x}{2}\]

    • 2 years ago
  17. AnimalAin Group Title
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    This is true since the derivative of the e^x term is always itself. That means the second derivative is the same as the original function. Added to the nine copies of the original function, we get 10ysubp=5e^x.

    • 2 years ago
  18. lilMissMindset Group Title
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    owwww. i understand,

    • 2 years ago
  19. AnimalAin Group Title
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    Now consider the polynomial part. The second derivative of a linear function is zero, so we get\[ 9y_{p2} =- 162x \implies y_{p2} =-19x\]

    • 2 years ago
  20. AnimalAin Group Title
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    So we get something like\[y=y_h+y_p=C_1\cos 3x+C_2\sin3x+\frac{e^x}{2}-19x\]

    • 2 years ago
  21. lilMissMindset Group Title
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    i forgot to multiply \[y _{p}\] with 9 when i substitute, right? that's why i did ot get the answer,

    • 2 years ago
  22. lilMissMindset Group Title
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    not*

    • 2 years ago
  23. AnimalAin Group Title
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    Actually, my division is horrible today.... should be 18 in the last term; exactly as your text had said.

    • 2 years ago
  24. AnimalAin Group Title
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    I hope you've got this; you're wearing me out....LOL

    • 2 years ago
  25. lilMissMindset Group Title
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    actually, it's the same for the two of us. LOL hey, thanks a lot. you've been helping me large. Sorry for the stress,

    • 2 years ago
  26. lilMissMindset Group Title
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    SORRY!

    • 2 years ago
  27. lilMissMindset Group Title
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    i get it, i'm just confused on some parts.

    • 2 years ago
  28. AnimalAin Group Title
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    No stress. It's fun working with students that work hard and get it. Do math every day.

    • 2 years ago
  29. lilMissMindset Group Title
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    i'll practice harder, :)

    • 2 years ago
  30. AnimalAin Group Title
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    I'm sure you'll get it.

    • 2 years ago
  31. lilMissMindset Group Title
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    i hope i get it the soonest possible, my exam is tomorrow. O.O

    • 2 years ago
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