## Study23 2 years ago Solve the differentiable equation: dy/dx = y^2 ?

1. openstudy11

2y

2. terenzreignz

You mean separable, right? :D

3. terenzreignz

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

4. terenzreignz

@Study23 Shall we? :)

5. Study23

Yup :D

6. 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?

7. Study23

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

8. 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?

9. Study23

dy/y^2=dx (?)

10. 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

11. Study23

So, ln|y^2|=x ?

12. terenzreignz

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

13. 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 $$n\ne -1$$

14. terenzreignz

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

15. Study23

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

16. Study23

That should be $$\ y^{-1}$$

17. terenzreignz

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

18. 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 $\huge y = -\frac1{x+C_0}$

19. terenzreignz

But on the whole, good job :)

20. Study23

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

21. terenzreignz

Waaaay ahead of you ^

22. Study23

In terms (autocorrect)

23. terenzreignz

24. terenzreignz

It just needs some algebraic manipulation is all.

25. Study23

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?

26. terenzreignz

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

27. Study23

Oh haha okay. Thanks!

28. vickky

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