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anonymous

  • 5 years ago

Solve the differential equation: (x * y' - 1) * ln(x) = 2 * y

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  1. anonymous
    • 5 years ago
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    it's already been differentiated?

  2. anonymous
    • 5 years ago
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    i dont understand what you mean.

  3. anonymous
    • 5 years ago
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    nvm...do you know implicit differentiation?

  4. anonymous
    • 5 years ago
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    yeah

  5. anonymous
    • 5 years ago
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    ok that makes it easier...I'll calc it in a second

  6. nowhereman
    • 5 years ago
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    substituting \[x := e^t\] seems to make it a lot easier

  7. anonymous
    • 5 years ago
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    the derivative of ln(x) is 1/x, but we have to do product rule

  8. anonymous
    • 5 years ago
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    sorry man not sure on this one

  9. anonymous
    • 5 years ago
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    and gotta get back studying for lin alg final -___-

  10. anonymous
    • 5 years ago
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    try www.wolframalpha.com for an answer

  11. anonymous
    • 5 years ago
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    oh ok thanks for your help :) @ nowhereman: if i do that i get y'*(e^t)*t=2y

  12. nowhereman
    • 5 years ago
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    I got \[(\frac{dy}{dt} - 1)t = 2y\]

  13. nowhereman
    • 5 years ago
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    assuming the function is analytic, that can be solved with power series

  14. anonymous
    • 5 years ago
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    would i be able to have \[(dy/dt) - (2/t)y=1\] and then do an integrating factor:\[e^(intergral of (-2/t))\]?

  15. nowhereman
    • 5 years ago
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    no, I don't see how that would work.

  16. anonymous
    • 5 years ago
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    ... i dont understand what i should do then

  17. anonymous
    • 5 years ago
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    umm and i dont get how you got the dt part

  18. nowhereman
    • 5 years ago
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    I used the chain rule \[\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}\]

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