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

- anonymous

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

- katieb

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- anonymous

2y

- terenzreignz

You mean separable, right? :D

- terenzreignz

and lol, @openstudy11 It is unfortunately not that simple :D

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## More answers

- terenzreignz

@Study23 Shall we? :)

- anonymous

Yup :D

- 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

dy = y^2 dx; I was doing this but I got confused because you couldn't integrate this could you?

- 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

dy/y^2=dx (?)

- 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

So, ln|y^2|=x ?

- 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

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

Sorry, my bad
\[\huge \int ax^n = \frac{ax^{n+1}}{n+1}\]

- anonymous

So, I get \(\ \Huge -y^-1 + C_1 = x \) ?

- anonymous

That should be \(\ y^{-1} \)

- terenzreignz

Mother of sizes... o.O
LOL
\[\huge -y^{-1}+C = x\]

- 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

But on the whole, good job :)

- anonymous

So, my textbook has this interns of Y= ...

- terenzreignz

Waaaay ahead of you ^

- anonymous

In terms (autocorrect)

- terenzreignz

Already done... ^

- terenzreignz

It just needs some algebraic manipulation is all.

- 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

I don't know what exam that is... better consult someone more familiar with it :)

- anonymous

Oh haha okay. Thanks!

- anonymous

dy/dx=y^2
dy/y^2=dx
then integrate both the side
-1/y=x+c

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