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anonymous

  • 5 years ago

solve the initial value problem dy/dx=2yx-x^2y and determine where the solution attains its minimal value

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  1. anonymous
    • 5 years ago
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    dy = (2yx - x^2y) dx dy = y(2x - x^2) dx dy / y = (2x - x^2) dx ln(y) = x^2 - x^3 / 3 y = e^((x^2 - x^3) / 3) + C

  2. anonymous
    • 5 years ago
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    but how to get the value of c

  3. anonymous
    • 5 years ago
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    C is a constant - it can be anything.

  4. anonymous
    • 5 years ago
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    Because if you differentiate a constant, you get 0

  5. amistre64
    • 5 years ago
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    the value of "C" will depend on the point that is provided in the initial question. Otherwise, "C" is any constant....

  6. anonymous
    • 5 years ago
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    but i got to solve it by separable equation

  7. anonymous
    • 5 years ago
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    yes y(0)=3

  8. anonymous
    • 5 years ago
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    Right, right - plug in that initial to find C.

  9. anonymous
    • 5 years ago
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    i will type the question again

  10. amistre64
    • 5 years ago
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    there are an infinite number of "graphs" that have the same derivitive. They all differ by the value of their constant (C). the only way to determine the value of (C) is to have information regarding some point onthe graph that contains (C) to anchor it in...

  11. anonymous
    • 5 years ago
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    I saw the question. You need your initial value (x = 0 y = __)

  12. anonymous
    • 5 years ago
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    solve the initial value problem dy/dx=2y^2+xy^2, y(0)=1 and determine where the solution attain its minimal value i am sending the same example and solution i got to do the same with the above one

  13. anonymous
    • 5 years ago
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    im sending the solution now

  14. anonymous
    • 5 years ago
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    solution; separating variables gives; y^-2dy=(2+x)dx

  15. anonymous
    • 5 years ago
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    Get out of here if you're not willing to listen to us.

  16. anonymous
    • 5 years ago
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    -y^-1=2x+x^2/2+C

  17. anonymous
    • 5 years ago
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    OK GO ON

  18. anonymous
    • 5 years ago
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    so explain me plz

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