## Study23 Group Title Solve the differentiable equation: dy/dx = y^2 ? one year ago one year ago

1. openstudy11 Group Title

2y

2. terenzreignz Group Title

You mean separable, right? :D

3. terenzreignz Group Title

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

4. terenzreignz Group Title

@Study23 Shall we? :)

5. Study23 Group Title

Yup :D

6. terenzreignz Group Title

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 Group Title

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

8. terenzreignz Group Title

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 Group Title

dy/y^2=dx (?)

10. terenzreignz Group Title

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 Group Title

So, ln|y^2|=x ?

12. terenzreignz Group Title

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 Group Title

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 Group Title

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

15. Study23 Group Title

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

16. Study23 Group Title

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

17. terenzreignz Group Title

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

18. terenzreignz Group Title

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 Group Title

But on the whole, good job :)

20. Study23 Group Title

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

21. terenzreignz Group Title

22. Study23 Group Title

In terms (autocorrect)

23. terenzreignz Group Title

24. terenzreignz Group Title

It just needs some algebraic manipulation is all.

25. Study23 Group Title

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 Group Title

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

27. Study23 Group Title

Oh haha okay. Thanks!

28. vickky Group Title

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