- anonymous

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

- schrodinger

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- anonymous

xy'+ 2y = 0
x = 2y/y'
\[\int\limits x = \int\limits \frac{ 2y }{ y' }\]
not sure what to do after this part

- zzr0ck3r

This looks like pug and play, did they give you some functions to try?

- zzr0ck3r

plug*

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## More answers

- UsukiDoll

I think this equation is separable.
|dw:1439004269797:dw|

- UsukiDoll

wait... wow I'm reading this guy wrong... this is plug and chug -_-!
any functions given?

- 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

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

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

- anonymous

im not entirely sure what you mean but these are the answer choices
e2x
x2
xâˆ’2
None of these

- zzr0ck3r

Determine could mean solve. Since the soltution is unique

- zzr0ck3r

ahh lol

- UsukiDoll

so I should continue?

- zzr0ck3r

I doubt they have gotten this far. This seems like a first day question

- zzr0ck3r

But they should know calc so ... lol

- 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

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

I need further clarification . That's the problem. This question feels like a double edge sword.

- anonymous

umm okay what do you want to know

- anonymous

im in calc 1... this is just the supplemental topics part of the course

- 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

-_- 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

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

x^-2 is their , but can we just ignor C?

- 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

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

thank you

- 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

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

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

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

so it was just verification... did I overkill the problem? ^__^

- whpalmer4

Probably, but they get a preview of material later in the course, right? :-)

- 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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