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Dave321

  • 3 years ago

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?

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

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

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

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

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

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

  7. abtrehearn
    • 3 years ago
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    Computing ...

  8. abtrehearn
    • 3 years ago
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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.

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

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

  11. abtrehearn
    • 3 years ago
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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.

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

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

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

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