anonymous
  • anonymous
Determine which function is a solution to the differential equation xy ' + 2y = 0.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
xy'+ 2y = 0 x = 2y/y' \[\int\limits x = \int\limits \frac{ 2y }{ y' }\] not sure what to do after this part
zzr0ck3r
  • zzr0ck3r
This looks like pug and play, did they give you some functions to try?
zzr0ck3r
  • zzr0ck3r
plug*

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UsukiDoll
  • UsukiDoll
I think this equation is separable. |dw:1439004269797:dw|
UsukiDoll
  • UsukiDoll
wait... wow I'm reading this guy wrong... this is plug and chug -_-! any functions given?
UsukiDoll
  • UsukiDoll
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.
jim_thompson5910
  • jim_thompson5910
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
UsukiDoll
  • UsukiDoll
yeah.... but the question doesn't ask to solve with separation of variables. IT's a verification problem . that's why I stopped.
anonymous
  • anonymous
im not entirely sure what you mean but these are the answer choices e2x x2 x−2 None of these
zzr0ck3r
  • zzr0ck3r
Determine could mean solve. Since the soltution is unique
zzr0ck3r
  • zzr0ck3r
ahh lol
UsukiDoll
  • UsukiDoll
so I should continue?
zzr0ck3r
  • zzr0ck3r
I doubt they have gotten this far. This seems like a first day question
zzr0ck3r
  • zzr0ck3r
But they should know calc so ... lol
UsukiDoll
  • UsukiDoll
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.
anonymous
  • anonymous
-_-
anonymous
  • anonymous
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
UsukiDoll
  • UsukiDoll
I need further clarification . That's the problem. This question feels like a double edge sword.
anonymous
  • anonymous
umm okay what do you want to know
anonymous
  • anonymous
im in calc 1... this is just the supplemental topics part of the course
UsukiDoll
  • UsukiDoll
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.
UsukiDoll
  • UsukiDoll
-_- I'm gonna experiment with something.... if I finish separation of variables. \[\Large \frac{dy}{y} = \frac{-2}{x}dx \] \[\Large \ln y = -2 lnx +C \] \[\Large e^{\ln y} = e^{ lnx^{-2} +C} \] \[\Large e^{\ln y} = e^{ lnx^{-2}} e^{C} \] \[\Large y = x^{-2} e^{C} \] \[\Large y = x^{-2}C \] \[\Large y = \frac{1}{x^2}C \]
UsukiDoll
  • UsukiDoll
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
anonymous
  • anonymous
x^-2 is their , but can we just ignor C?
UsukiDoll
  • UsukiDoll
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.
UsukiDoll
  • UsukiDoll
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.
anonymous
  • anonymous
thank you
whpalmer4
  • whpalmer4
\[y=x^{-2}\]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.
whpalmer4
  • whpalmer4
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.
UsukiDoll
  • UsukiDoll
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.
whpalmer4
  • whpalmer4
Trying to get people not to copy/paste is like trying to get water to flow uphill :-) Just gotta learn to recognize the damage...
UsukiDoll
  • UsukiDoll
so it was just verification... did I overkill the problem? ^__^
whpalmer4
  • whpalmer4
Probably, but they get a preview of material later in the course, right? :-)
UsukiDoll
  • UsukiDoll
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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