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Denebel

What happens after this..? use separation of variables to solve the initial value problem

  • 2 years ago
  • 2 years ago

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  1. Denebel
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    • 2 years ago
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  2. Denebel
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    Forgot the parenthesis: e^( (x^2/2) + 2x + c)

    • 2 years ago
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  3. Roachie
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    sub youe initial values and solve for C y= C(e^( .5x^2 + 2x ))- 5 for x = 0, y = 1 1 = C - 5 C = 6 so y= 6e^( .5x^2 + 2x )- 5

    • 2 years ago
  4. cinar
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    ln(6)=c

    • 2 years ago
  5. Denebel
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    How did you bring C out?

    • 2 years ago
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  6. Roachie
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    e^C = C its math magic

    • 2 years ago
  7. Denebel
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    ? Huh? But for example, e^(2) does not equal to 2...

    • 2 years ago
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  8. cinar
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    |dw:1327563797536:dw|

    • 2 years ago
  9. Roachie
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    e^c = 6

    • 2 years ago
  10. cinar
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    |dw:1327563871600:dw|

    • 2 years ago
  11. Roachie
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    cinar!

    • 2 years ago
  12. cinar
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    yes!

    • 2 years ago
  13. Denebel
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    ahh I see now! Thank you very much

    • 2 years ago
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  14. Roachie
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    just try to have fun with it.

    • 2 years ago
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