## anonymous one year ago Determine which function is a solution to the differential equation xy ' + 2y = 0.

1. anonymous

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

2. zzr0ck3r

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

3. zzr0ck3r

plug*

4. UsukiDoll

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

5. UsukiDoll

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

6. 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.

7. 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

8. UsukiDoll

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

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

10. zzr0ck3r

Determine could mean solve. Since the soltution is unique

11. zzr0ck3r

ahh lol

12. UsukiDoll

so I should continue?

13. zzr0ck3r

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

14. zzr0ck3r

But they should know calc so ... lol

15. 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.

16. anonymous

-_-

17. 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

18. UsukiDoll

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

19. anonymous

umm okay what do you want to know

20. anonymous

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

21. 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.

22. UsukiDoll

-_- I'm gonna experiment with something.... if I finish separation of variables. $\Large \frac{dy}{y} = \frac{-2}{x}dx$ $\Large \ln y = -2 lnx +C$ $\Large e^{\ln y} = e^{ lnx^{-2} +C}$ $\Large e^{\ln y} = e^{ lnx^{-2}} e^{C}$ $\Large y = x^{-2} e^{C}$ $\Large y = x^{-2}C$ $\Large y = \frac{1}{x^2}C$

23. 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

24. anonymous

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

25. 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.

26. 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.

27. anonymous

thank you

28. 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.

29. 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.

30. 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.

31. 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...

32. UsukiDoll

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

33. whpalmer4

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

34. UsukiDoll

yeah XD. This equation is also a first order linear ODE. in the form of $\large \frac{dy}{dx} + p(x)y = q(x)$