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anonymous

  • 5 years ago

Solve the differential equation. xy' - (x + 1)y = 0

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  1. anonymous
    • 5 years ago
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    x(dy/dx) - (x+1)y = 0 x(dy/dx) = (x+1)y dy/dx = [ (x+1)/x ] y 1/y (dy) = (x+1)/x dx ::: integrate can yu go from there?

  2. anonymous
    • 5 years ago
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    somehow i got (x-1)/x

  3. anonymous
    • 5 years ago
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    as a final answeR?

  4. anonymous
    • 5 years ago
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    noo instead of (x+1)/x dx

  5. anonymous
    • 5 years ago
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    my final answer was y = -1/2Ce^x(x)^-2

  6. anonymous
    • 5 years ago
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    i have a feeling that's not right

  7. anonymous
    • 5 years ago
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    hmm not what i got, wait i'm gonna type what i have

  8. anonymous
    • 5 years ago
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    ok now i'm up to ln(y) = x + ln(x) + C.. not sure what to do next

  9. anonymous
    • 5 years ago
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    \[xy' - (x + 1)y = 0\]\[x{dy \over dx} - (x+1)y = 0\]\[x{dy \over dx} = (x+1)y\]\[{dy \over dx} = {(x+1) \over x} (y)\]\[{1 \over y} dy = {(x+1) \over x} dx\] Integrate:\[\ln|y|= \ln|x|+x +C\]\[e^{lny} = e^{lnx+x+C}=e^{lnx}*e^{x}*e^{C}\]\[y = A*x*e^{x}\] \[y = Axe^{x}\] that's what i have

  10. anonymous
    • 5 years ago
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    is that right?

  11. anonymous
    • 5 years ago
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    I think soo.. what is A?

  12. anonymous
    • 5 years ago
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    e^C = constant a constant which i called A, you can called it whatever..

  13. anonymous
    • 5 years ago
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    but did yu understand everything?

  14. anonymous
    • 5 years ago
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    ohh gotcha.. i understand it.. thank you.. can you help me with other problems too?

  15. anonymous
    • 5 years ago
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    good. i sure can, just post them and i'll try

  16. anonymous
    • 5 years ago
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    \[\int\limits_{1/2}^{2} \] [(e^1/x) / (x^2)] dx

  17. anonymous
    • 5 years ago
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    \[\int\limits_{2}^{2}{{e^{1} \over x} \over x^{2}} dx\] is that what yu tried to write

  18. anonymous
    • 5 years ago
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    \[{e^{1/x} \over x^{2}}\]

  19. anonymous
    • 5 years ago
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    that didn't come out on my computer - maybe bc i'm using a mac. but its the integration bounded from 1/2 to 2

  20. anonymous
    • 5 years ago
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    and the function that you just wrote

  21. anonymous
    • 5 years ago
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    ok

  22. anonymous
    • 5 years ago
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    do a product rule between e^(1/x) and x^(-2)

  23. anonymous
    • 5 years ago
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    can you do that ?

  24. anonymous
    • 5 years ago
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    i'm not really sure.. the integration of e^(1/x) is just e^(1/x)?

  25. anonymous
    • 5 years ago
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    -e^(1/x)*x^-1??

  26. anonymous
    • 5 years ago
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    the final answer comes up to be e^2 - sqrt(e) i think

  27. anonymous
    • 5 years ago
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    how did you get that?

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