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anonymous

  • one year ago

Solve the equation for y in terms of x: 3y + 2y^2 = -e^(-x) - e^(x) + 7

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  1. dinamix
    • one year ago
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    its not differential equation ?

  2. anonymous
    • one year ago
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    Ill post the entire question and maybe that'll help

  3. anonymous
    • one year ago
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    Find the solution to the initial value problem in explicit form \[y' = \frac{e^{-x} - e^{x}}{3 + 4y}\]

  4. anonymous
    • one year ago
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    y(0) = 1

  5. anonymous
    • one year ago
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    @dan815 ?

  6. freckles
    • one year ago
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    y'*y doesn't equal y^2

  7. freckles
    • one year ago
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    \[(3+4y) dy=(e^{-x}-e^{x} )dx\] equation is seperable integrate both sides

  8. freckles
    • one year ago
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    oh wait are you saying you think you found the solution to that?

  9. anonymous
    • one year ago
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    yes i believe so and now i need it in explicit form

  10. anonymous
    • one year ago
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    I believe i have found the correct value for the constant c given the initial value as well

  11. freckles
    • one year ago
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    your solution is correct

  12. freckles
    • one year ago
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    don't believe you can find explicit form

  13. anonymous
    • one year ago
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    the answer in the back of the book has done it some how.

  14. freckles
    • one year ago
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    hmmm well you have a quadratic in terms of y

  15. freckles
    • one year ago
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    \[2y^2-3y+e^{-x}+e^{x}-7=0 \\ y=\frac{-(-3) \pm \sqrt{(-3)^2-4(2)(e^{-x}+e^{x}-7)}}{2(2)}\]

  16. freckles
    • one year ago
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    though the implicit form looks much better to me

  17. anonymous
    • one year ago
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    that looks like the answer, how'd you do that?!

  18. freckles
    • one year ago
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    it is a quadratic in terms of y

  19. freckles
    • one year ago
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    all i did was use the quadratic formula

  20. freckles
    • one year ago
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    a=2 b=-3 c=e^(-x)+e^x-7

  21. anonymous
    • one year ago
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    ohhhhh

  22. anonymous
    • one year ago
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    i gotcha now

  23. anonymous
    • one year ago
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    Thank you!!!!!!

  24. freckles
    • one year ago
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    np

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