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lilMissMindset

  • 2 years ago

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

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  1. lilMissMindset
    • 2 years ago
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    @AnimalAin this.

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

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

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

  5. AnimalAin
    • 2 years ago
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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.

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

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

  8. lilMissMindset
    • 2 years ago
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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 \[=C _{1} \cos 3x+ C _{2} \sin3x + \frac{ 1 }{ 2 } e ^{x} -18xy \]

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

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

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

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

  13. AnimalAin
    • 2 years ago
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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?

  14. lilMissMindset
    • 2 years ago
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    yeps. ok.

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

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

  17. AnimalAin
    • 2 years ago
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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.

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

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

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

  21. lilMissMindset
    • 2 years ago
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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,

  22. lilMissMindset
    • 2 years ago
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    not*

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

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

  25. lilMissMindset
    • 2 years ago
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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,

  26. lilMissMindset
    • 2 years ago
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    SORRY!

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

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

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

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

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

  32. bashpow
    • 3 months ago
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    may i ask what book the problem referred?

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