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Denebel

  • 2 years ago

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

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

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  3. Roachie
    • 2 years ago
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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

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

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

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

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

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

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

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

  11. Roachie
    • 2 years ago
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    cinar!

  12. cinar
    • 2 years ago
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    yes!

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

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

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