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Dave321

Solve dy/dx=(-5x+y)^2 -4 I got 1/6ln((u-3)/(u+3))=x using a substitution of u=(-5x+y). Is that right?

  • 2 years ago
  • 2 years ago

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  1. bward10
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    the problem isnt stated properly... it looks wrong though. some clarification would help.

    • 2 years ago
  2. Dave321
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    What do you mean?

    • 2 years ago
  3. Dave321
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    \[dy/dx=(-5x+y)^2-4\] Is that better?

    • 2 years ago
  4. abtrehearn
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    That u-substitution works for this differential equation. The substitution reduces it to the variables-separable equation \[du/(u^{2}-9) + dx = 0.\]

    • 2 years ago
  5. Dave321
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    But then you still have to integrate it, right?

    • 2 years ago
  6. Dave321
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    Did you get the same answer I got?

    • 2 years ago
  7. abtrehearn
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    Computing ...

    • 2 years ago
  8. abtrehearn
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    (1/6)(ln|u - 3| - ln|u + 3| = -x + ln(C)/6 as an intermediate solution. Once that is taken back to original variables, we're finished.

    • 2 years ago
  9. abtrehearn
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    My final solution is 5x - y + 3 = C e^(6x)(5x - y - 3) in implicit form.

    • 2 years ago
  10. abtrehearn
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    Its explicit form is \[y = 5 x + 3(1 + C e^{6x})/(-1 + C e^{6x}).\]

    • 2 years ago
  11. abtrehearn
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    For increasingly large values of x, all solution curves more closely approximate the linear function 5x + 3, so the line y = 5x + 3 is the asymtote of all solution curves.

    • 2 years ago
  12. abtrehearn
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    That's a good problem, Dave :^)

    • 2 years ago
  13. Dave321
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    I'll tell my professor;-) Thanks!

    • 2 years ago
  14. abtrehearn
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    Major ten-four, good buddy :^)

    • 2 years ago
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