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Study23

Solve the differentiable equation: dy/dx = y^2 ?

  • one year ago
  • one year ago

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  1. openstudy11
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    2y

    • one year ago
  2. terenzreignz
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    You mean separable, right? :D

    • one year ago
  3. terenzreignz
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    and lol, @openstudy11 It is unfortunately not that simple :D

    • one year ago
  4. terenzreignz
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    @Study23 Shall we? :)

    • one year ago
  5. Study23
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    Yup :D

    • one year ago
  6. terenzreignz
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    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?

    • one year ago
  7. Study23
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    dy = y^2 dx; I was doing this but I got confused because you couldn't integrate this could you?

    • one year ago
  8. terenzreignz
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    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?

    • one year ago
  9. Study23
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    dy/y^2=dx (?)

    • one year ago
  10. terenzreignz
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    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

    • one year ago
  11. Study23
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    So, ln|y^2|=x ?

    • one year ago
  12. terenzreignz
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    oh... \[\huge \int \frac{1}{y^2}dy \ = \int dx\] Are you sure? Maybe this ought to refresh you \[\huge \int y^{-2}dy \ = \int dx\]

    • one year ago
  13. terenzreignz
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    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 \(n\ne -1\)

    • one year ago
  14. terenzreignz
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    Sorry, my bad \[\huge \int ax^n = \frac{ax^{n+1}}{n+1}\]

    • one year ago
  15. Study23
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    So, I get \(\ \Huge -y^-1 + C_1 = x \) ?

    • one year ago
  16. Study23
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    That should be \(\ y^{-1} \)

    • one year ago
  17. terenzreignz
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    Mother of sizes... o.O LOL \[\huge -y^{-1}+C = x\]

    • one year ago
  18. terenzreignz
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    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 \[\huge y = -\frac1{x+C_0}\]

    • one year ago
  19. terenzreignz
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    But on the whole, good job :)

    • one year ago
  20. Study23
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    So, my textbook has this interns of Y= ...

    • one year ago
  21. terenzreignz
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    Waaaay ahead of you ^

    • one year ago
  22. Study23
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    In terms (autocorrect)

    • one year ago
  23. terenzreignz
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    Already done... ^

    • one year ago
  24. terenzreignz
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    It just needs some algebraic manipulation is all.

    • one year ago
  25. Study23
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    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?

    • one year ago
  26. terenzreignz
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    I don't know what exam that is... better consult someone more familiar with it :)

    • one year ago
  27. Study23
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    Oh haha okay. Thanks!

    • one year ago
  28. vickky
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    dy/dx=y^2 dy/y^2=dx then integrate both the side -1/y=x+c

    • one year ago
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