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Study23
Solve the differentiable equation: dy/dx = y^2 ?
You mean separable, right? :D
and lol, @openstudy11 It is unfortunately not that simple :D
@Study23 Shall we? :)
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?
dy = y^2 dx; I was doing this but I got confused because you couldn't integrate this could you?
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?
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
oh... \[\huge \int \frac{1}{y^2}dy \ = \int dx\] Are you sure? Maybe this ought to refresh you \[\huge \int y^{-2}dy \ = \int dx\]
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 \(n\ne -1\)
Sorry, my bad \[\huge \int ax^n = \frac{ax^{n+1}}{n+1}\]
So, I get \(\ \Huge -y^-1 + C_1 = x \) ?
That should be \(\ y^{-1} \)
Mother of sizes... o.O LOL \[\huge -y^{-1}+C = x\]
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 \[\huge y = -\frac1{x+C_0}\]
But on the whole, good job :)
So, my textbook has this interns of Y= ...
Waaaay ahead of you ^
It just needs some algebraic manipulation is all.
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?
I don't know what exam that is... better consult someone more familiar with it :)
dy/dx=y^2 dy/y^2=dx then integrate both the side -1/y=x+c