anonymous
  • anonymous
Solve the differentiable equation: dy/dx = y^2 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
2y
terenzreignz
  • terenzreignz
You mean separable, right? :D
terenzreignz
  • terenzreignz
and lol, @openstudy11 It is unfortunately not that simple :D

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terenzreignz
  • terenzreignz
@Study23 Shall we? :)
anonymous
  • anonymous
Yup :D
terenzreignz
  • terenzreignz
Okay, thing about separable differential equations... they are.... separable! LOL Yeah, I know, big surprise :) What I mean is that you can rearrange them so that you can separate them into expressions involving only y and involving only x. Let's have a look at this \[\huge \frac{dy}{dx} = y^2\] You can treat the dy and dx as variables like any other, when separating. Just you try multiplying both sides by dx, and what do you get?
anonymous
  • anonymous
dy = y^2 dx; I was doing this but I got confused because you couldn't integrate this could you?
terenzreignz
  • terenzreignz
Not yet. Now, you could bring all expressions involving y, to the appropriate side of the equation (namely, the side with the dy) You could do this by multiplying \(\large y^{-2}\) or \(\Large \frac1{y^2}\) on both sides, and what would you get then?
anonymous
  • anonymous
dy/y^2=dx (?)
terenzreignz
  • terenzreignz
Correct :D \[\huge \frac{dy}{y^2}=dx\] Now, slip a \(\Huge \int\) sign on both sides of the equation~ and be done with it :D
anonymous
  • anonymous
So, ln|y^2|=x ?
terenzreignz
  • terenzreignz
oh... \[\huge \int \frac{1}{y^2}dy \ = \int dx\] Are you sure? Maybe this ought to refresh you \[\huge \int y^{-2}dy \ = \int dx\]
terenzreignz
  • terenzreignz
Okay, we started with this, right \[\huge \frac{dy}{y^2}=dx\] And we simply integrated both sides \[\huge\color{blue}\int \frac{dy}{y^2}=\color{blue}{\int}dx\] And yes.... \[\huge \int ax^n=\frac{x^{n+1}}{n+1} \] when \(n\ne -1\)
terenzreignz
  • terenzreignz
Sorry, my bad \[\huge \int ax^n = \frac{ax^{n+1}}{n+1}\]
anonymous
  • anonymous
So, I get \(\ \Huge -y^-1 + C_1 = x \) ?
anonymous
  • anonymous
That should be \(\ y^{-1} \)
terenzreignz
  • terenzreignz
Mother of sizes... o.O LOL \[\huge -y^{-1}+C = x\]
terenzreignz
  • terenzreignz
If you want it as a function of y, you can bring the constant to the other side \[\huge -\frac1y=x+C_0\] and then isolate \[\huge y = -\frac1{x+C_0}\]
terenzreignz
  • terenzreignz
But on the whole, good job :)
anonymous
  • anonymous
So, my textbook has this interns of Y= ...
terenzreignz
  • terenzreignz
Waaaay ahead of you ^
anonymous
  • anonymous
In terms (autocorrect)
terenzreignz
  • terenzreignz
Already done... ^
terenzreignz
  • terenzreignz
It just needs some algebraic manipulation is all.
anonymous
  • anonymous
Oh, haha I didn't see that! :D. Thanks for all your help @terenzreignz ! I appreciate your encouragement! By the way, any tips for the AP Calc BC exam by the way?
terenzreignz
  • terenzreignz
I don't know what exam that is... better consult someone more familiar with it :)
anonymous
  • anonymous
Oh haha okay. Thanks!
anonymous
  • anonymous
dy/dx=y^2 dy/y^2=dx then integrate both the side -1/y=x+c

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