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anonymous

  • one year ago

Differential Equation Problem: How Can i solve the Initial value problem. I am not sure if to use integrating factor? Equation below.

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  1. anonymous
    • one year ago
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    \[2y \prime -y^2 + 4 =0, y(0) = 0\]

  2. amistre64
    • one year ago
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    what methods are available to you?

  3. anonymous
    • one year ago
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    integrating factory and separable....

  4. amistre64
    • one year ago
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    2 dy/dx = y^2-4 looks separable to me ... not sure if it makes life easier tho

  5. anonymous
    • one year ago
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    Yep, i can work with that.

  6. anonymous
    • one year ago
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    I'll post the result soon.

  7. anonymous
    • one year ago
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    It does not. It is complex.

  8. anonymous
    • one year ago
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    |dw:1443368763254:dw|

  9. Loser66
    • one year ago
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    Use partial fraction.

  10. Loser66
    • one year ago
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    |dw:1443368964936:dw|

  11. anonymous
    • one year ago
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    I'm using that, A = -1/2 and b = 1/2

  12. anonymous
    • one year ago
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    is it safe to say....|dw:1443369358723:dw|

  13. anonymous
    • one year ago
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    but as y (0) = 0. I noticed that the equation initial condition will bring take the solution to.... e^(2c) = -1, but c does not exist since the ln(-1) is undefined.

  14. amistre64
    • one year ago
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    as per the wolf ... might help you see where to go.

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  15. amistre64
    • one year ago
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    A and B are +- 1/4 right?

  16. amistre64
    • one year ago
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    i ignored the 2 lol

  17. amistre64
    • one year ago
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    dy/(y^2-4) = dx/2 ln((y-2)/(y+2))/4 = x/2 +C ln((y-2)/(y+2)) = 2x +4C (y-2)/(y+2) = e^(2x+4C) = C e^(2x) x=0,y=0 -1 = C

  18. amistre64
    • one year ago
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    y-2 = -e^(2x) (y+2) y+ye^(2x) = -2e^(2x)+2 y(1+e^(2x)) = -2(1-e^(2x)) ...

  19. amistre64
    • one year ago
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    mighta factored the rhs off by a negative in my head

  20. anonymous
    • one year ago
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    Thanks all.

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