Question

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

Answer 1

2y

Answer 2

You mean separable, right? :D

Answer 3

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

Answer 4

@Study23 Shall we? :)

Answer 5

Yup :D

Answer 6

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?

Answer 7

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

Answer 8

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?

Answer 9

dy/y^2=dx (?)

Answer 10

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

Answer 11

So, ln|y^2|=x ?

Answer 12

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

Answer 13

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 \(n\ne -1\)

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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 \[\huge y = -\frac1{x+C_0}\]

Answer 19

But on the whole, good job :)

Answer 20

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

Answer 21

Waaaay ahead of you ^

Answer 22

In terms (autocorrect)

Answer 23

Already done... ^

Answer 24

It just needs some algebraic manipulation is all.

Answer 25

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?

Answer 26

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

Answer 27

Oh haha okay. Thanks!

Answer 28

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