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AmTran_Bus

  • one year ago

Need DE help!

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  1. AmTran_Bus
    • one year ago
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    |dw:1440809128053:dw|

  2. AmTran_Bus
    • one year ago
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    So I do not have a clue what to do about the boundary equation.

  3. AmTran_Bus
    • one year ago
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    But I can show you how I solved the original part.

  4. tkhunny
    • one year ago
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    I'd try something like \(y(x) = Ae^{2x} + Be^{-2x} + Cx + D\) Or, I might notice that \(r^{2} - 4 = 0\) has solutions r = 2 and r = -2 and then forget about the C and D terms.

  5. AmTran_Bus
    • one year ago
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    |dw:1440809556496:dw| is what I got. But I have no clue what a boundary is nor how to apply it.

  6. AmTran_Bus
    • one year ago
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    and the book says that is right.

  7. tkhunny
    • one year ago
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    ?? Substitute x = 0 \(c_{1}e^{2(0)} + c_{2}e^{-2(0)} = 2\) Now, find the first derivative and do it again. This gives two equations in two unknowns and you are nearly done.

  8. AmTran_Bus
    • one year ago
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    So you gotta take deriv. and then substitute.

  9. tkhunny
    • one year ago
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    If you want to use the second boundary condition, yes.

  10. AmTran_Bus
    • one year ago
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    Ohhhhhh. That is where I am messing up.

  11. AmTran_Bus
    • one year ago
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    Duh duh me. Thanks so much.

  12. AmTran_Bus
    • one year ago
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    So 2=c1+c2 and 2= c1-c2

  13. tkhunny
    • one year ago
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    Shedding a little light is very rewarding.

  14. AmTran_Bus
    • one year ago
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    Sure nuff. Especially since this is pchem.

  15. AmTran_Bus
    • one year ago
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    Hum. So how where does the 2e^(2x) come from at the end @tkhunny ? I know 4=2c1 and so c1=2 and c2=0 so does that 2 just apply to both the coeff. and the e^whatever?

  16. tkhunny
    • one year ago
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    "pchem"?! You have my sympathies. You CAN get through it.

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