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Denebel Group Title

use separation of variables to solve the initial value problem. dy/dx = (y+2)(x+2) and y=1 when x=0

  • 2 years ago
  • 2 years ago

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  1. Denebel Group Title
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    *oops I meant (y+5)

    • 2 years ago
  2. Denebel Group Title
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    dy/dx= (y+5)(x+2) and y=1 when x=0

    • 2 years ago
  3. Roachie Group Title
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    dy/(y+5) = (x+2) dx integrate both sides

    • 2 years ago
  4. Denebel Group Title
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    I did, but I don't know what to do after.. I did this? |dw:1327548662810:dw|

    • 2 years ago
  5. Denebel Group Title
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    Can you show me your steps?

    • 2 years ago
  6. Roachie Group Title
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    ok, take your ln ( y+5) = 1/2x^2+2x + C and raise expotentiate to get rid of the ln y+5 = Ce^(1/2x^2+2x ) then y = Ce^(1/2x^2+2x ) - 5

    • 2 years ago
  7. Roachie Group Title
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    y=6e^(.5x^2+2x) + 5

    • 2 years ago
  8. Denebel Group Title
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    Why can't I plug in the numbers before putting in e?

    • 2 years ago
  9. Roachie Group Title
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    Because this is how I was taught :) lemme try it that way

    • 2 years ago
  10. Roachie Group Title
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    that gives ln(6) = C... 1.7917594692280550008124773583807022727229906921830047058553

    • 2 years ago
  11. Denebel Group Title
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    Oh ok. thanks :) How did you get your C to become a coefficient?

    • 2 years ago
  12. Denebel Group Title
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    Referring to this: y+5 = Ce^(1/2x^2+2x )

    • 2 years ago
  13. Denebel Group Title
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    Is this a rule?

    • 2 years ago
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