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anonymous

  • one year ago

Determine which function is a solution to the differential equation xy ' + 2y = 0.

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  1. anonymous
    • one year ago
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    xy'+ 2y = 0 x = 2y/y' \[\int\limits x = \int\limits \frac{ 2y }{ y' }\] not sure what to do after this part

  2. zzr0ck3r
    • one year ago
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    This looks like pug and play, did they give you some functions to try?

  3. zzr0ck3r
    • one year ago
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    plug*

  4. UsukiDoll
    • one year ago
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    I think this equation is separable. |dw:1439004269797:dw|

  5. UsukiDoll
    • one year ago
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    wait... wow I'm reading this guy wrong... this is plug and chug -_-! any functions given?

  6. UsukiDoll
    • one year ago
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    which function is a solution? meaning there's gotta be a y and then take the derivative of that y and then plug it into the equation to see if left hand side = right hand side.

  7. jim_thompson5910
    • one year ago
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    UsukiDoll you're on the right track \[\Large x \frac{dy}{dx} = -2y\] \[\Large \frac{dy}{y} = \frac{-2}{x}dx\] then integrate both sides

  8. UsukiDoll
    • one year ago
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    yeah.... but the question doesn't ask to solve with separation of variables. IT's a verification problem . that's why I stopped.

  9. anonymous
    • one year ago
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    im not entirely sure what you mean but these are the answer choices e2x x2 x−2 None of these

  10. zzr0ck3r
    • one year ago
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    Determine could mean solve. Since the soltution is unique

  11. zzr0ck3r
    • one year ago
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    ahh lol

  12. UsukiDoll
    • one year ago
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    so I should continue?

  13. zzr0ck3r
    • one year ago
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    I doubt they have gotten this far. This seems like a first day question

  14. zzr0ck3r
    • one year ago
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    But they should know calc so ... lol

  15. UsukiDoll
    • one year ago
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    that's what I mean. When I first had ODEs I was asked this determine which function is a solution to the differential equation. So I had to get my y = ? , take the derivative of y, and plug it into the equation.

  16. anonymous
    • one year ago
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    -_-

  17. anonymous
    • one year ago
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    i think i just have to find y i was told to isolate each variable on one side and integrate both sides and find y

  18. UsukiDoll
    • one year ago
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    I need further clarification . That's the problem. This question feels like a double edge sword.

  19. anonymous
    • one year ago
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    umm okay what do you want to know

  20. anonymous
    • one year ago
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    im in calc 1... this is just the supplemental topics part of the course

  21. UsukiDoll
    • one year ago
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    like it's separable I'll admit that.. but there's also those you're given y = something and then verify that when you take the derivative of y = something and plug it back into the equation, the result should come true. ahhhhhhhhhhhhhhhhhhhh.......... I see... so it's elementary type of ODE ok... I had some ODEs in calc 2 years ago.

  22. UsukiDoll
    • one year ago
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    -_- I'm gonna experiment with something.... if I finish separation of variables. \[\Large \frac{dy}{y} = \frac{-2}{x}dx \] \[\Large \ln y = -2 lnx +C \] \[\Large e^{\ln y} = e^{ lnx^{-2} +C} \] \[\Large e^{\ln y} = e^{ lnx^{-2}} e^{C} \] \[\Large y = x^{-2} e^{C} \] \[\Large y = x^{-2}C \] \[\Large y = \frac{1}{x^2}C \]

  23. UsukiDoll
    • one year ago
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    hmmm then maybe take product rule \[\Large y = x^{-2}C \] \[\Large y' = x^{-2}(0) +C(-2)x^{-3}\] \[\Large y' = -2Cx^{-3}\] \[\large x \frac{-2C}{x^3}+\frac{2C}{x^2} = 0\] \[\large \frac{-2Cx}{x^3}+\frac{2C}{x^2} = 0\] \[\large \frac{-2C}{x^2}+\frac{2C}{x^2} = 0\] \[\large 0 = 0\] yay.... but I'm not sure if it's related to the answer choices. I see x^2 but it's missing something

  24. anonymous
    • one year ago
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    x^-2 is their , but can we just ignor C?

  25. UsukiDoll
    • one year ago
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    like if one of the answer choices was \[\LARGE y = x^{-2} \] that would work \[\LARGE y' = -2x^{-3} \] \[\LARGE xy'+2y=0\] \[\LARGE (\frac{-2x}{x^3})+\frac{2}{x^2}=0\] \[\LARGE (\frac{-2}{x^2})+\frac{2}{x^2}=0\] \[\LARGE 0=0\] Yeah we can ignore C. C just means constant.

  26. UsukiDoll
    • one year ago
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    so taking separation of variables is necessary . It's just that whoever wrote this question needs to be more specific as I had questions that had y = ? given to me, take derivatives, and just verify. I was trying to do that on here, but it didn't work, so I used separation of variables and we got our y.

  27. anonymous
    • one year ago
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    thank you

  28. whpalmer4
    • one year ago
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    \[y=x^{-2}\]is one of the answer choices, it just got mangled by copy and paste losing the formatting. All this problem asks is that you take each of the 3 candidate answer choices that specify a function, and plug them in one at a time in the differential equation. Discard all that do not make a true number sentence, and if one works, that's your answer, otherwise "none of these" is your answer.

  29. whpalmer4
    • one year ago
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    For example, if \(y=3x^2\) was among the choices, you find \[y' = 6x\] substitute into the differential equation to be verified: \[xy' + 2y=0\]\[x(6x) + 2(3x^2) = 0\]\[6x^2 + 6x^2 = 0\]That's not true for any non-zero value of \(x\), so that would not be a good choice.

  30. UsukiDoll
    • one year ago
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    mhm... I think taking screenshots of the problem is better than copy and paste as some of the formatting may not carry over and will lead to some confusion or problems.

  31. whpalmer4
    • one year ago
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    Trying to get people not to copy/paste is like trying to get water to flow uphill :-) Just gotta learn to recognize the damage...

  32. UsukiDoll
    • one year ago
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    so it was just verification... did I overkill the problem? ^__^

  33. whpalmer4
    • one year ago
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    Probably, but they get a preview of material later in the course, right? :-)

  34. UsukiDoll
    • one year ago
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    yeah XD. This equation is also a first order linear ODE. in the form of \[\large \frac{dy}{dx} + p(x)y = q(x) \]

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