## anonymous one year ago Can someone help a novice out?! Differential Equation SOS! Find the DE knowing the final answer y=(x-c)^3 Thanks a lot!!!

1. anonymous

Should I just get rid of the C?

2. anonymous

and if so. idk really know how tbh

3. anonymous

@ganeshie8

4. anonymous

@idku

5. anonymous

but the C remains! shouldn't I come to the point of which I don't have constants?

6. anonymous

hahahahahah

7. anonymous

well thanks anyhow

8. anonymous

@arthurpariz

9. anonymous

@Loser66 will you be my savior?

10. anonymous

thanks a lot!!!!

11. freckles

You can find y' by differentiating the y=(x-c)^3.

12. freckles

why can't the differential equation be y'=3(x-c)^2 with condition .. let me figure it out... y=(x-c)^3+C y(0)=(0-c)^3+C y(0)=-c^3+C y(0)+c^3=C So why can't the differential equation be y'=3(x-c)^2 with condition y(0)=-c^3 ?

13. mathmate

y=(x-c)^3 y'=3(y-c)^2 y"=6(y-c) so y=(y"/6)^3 or (y")^3-216y=0

14. mathmate

oh, sorry, I didn't see the condition.

15. freckles

I don't see how these rules in which we find a differential equation posted in the original post. @Loser66

16. freckles

I don't see that we aren't allowed to use initial conditions or find a non-linear diff equation.

17. freckles

it would be nice though if @Hipocampus can settle what we are talking about though

18. anonymous

hmmm that's all I have...

19. anonymous

@freckles

20. anonymous

21. freckles

So the solution can be non-linear or can have conditions ?

22. anonymous

I believe so

23. anonymous

as long as the C is gone

24. Loser66

@freckles You see, @mathmate derives to another way, we have another ODE. I have a bunch of ODE satisfy the given information. YOu add more condition to get another one. That is the reason why it is invalid!! We know that the solution of ODE is UNIQUE for a specific condition. Unfortunately, we didn't have it.

25. anonymous

thanks a lot guys!

26. anonymous

can I ask another question then?

27. Loser66

close this and open a new one, please.

28. anonymous

Roger @Loser66 !

29. Loser66

@ganeshie8 @dan815 @oldrin.bataku @SithsAndGiggles @Hipocampus That is the reason why I don't want you to mix this question with the other one

30. freckles

Oh so no conditions allowed you are saying @Loser66

31. Loser66

@freckles I didn't say "no conditions allowed", just don't use the un-given conditions.

32. freckles

you want to find a differential equation subject to no conditions is what you are looking for the rest are valid though if you provide conditions in which I have

33. freckles

you get to make up the differential equation, why can't you make up the conditions that it is subject to?

34. freckles

and I'm not really making it up I'm based my condition off what the solution should be.

35. Loser66

Anyway!! I think if we make up a condition and have a Unique solution for that condition, we are ok :)

36. Loser66

I am with you now :)

37. ganeshie8

aren't we looking for the differential equation of order 1, whose general solution is the set of standard cubics translated horizontally ?

38. freckles

I don't know it wasn't specify that had to have order 1

39. ganeshie8

the order has to be 1 because there is only 1 arbitrary constant in the given general solution

40. ganeshie8

number of arbitrary constants and the order of de must agree, right

41. freckles

well I was thinking that this would be okay: $y'=3(x-c)^2 \text{ with condition } y(0)=-c^3$ but are you saying we are suppose to be treating the c from y=(x-c)^3 has like the constant of integration

42. freckles

meant to end that with a ?

43. ganeshie8

this should work $(y')^3 = 27y^2$

44. freckles

I like that requires no conditions

45. freckles

but I still think my way is valid

46. freckles

only because there was not much put into the "rules" of finding the differential equation

47. ganeshie8

Haha true, they should have asked it less ambiguously: eliminate the arbitrary constant to get the differential equation that represents the family of standard cubics translated horizontally

48. anonymous

you could just do: $$y=(x-c)^3\\y^{1/3}=x-c\\\frac13 y^{2/3}y'=1\\y'=3y^{-2/3}$$

49. anonymous

the solutions are the one-parameter family of monic cubics up to horizontal translation $$y=(x-c)^3$$

50. anonymous

oops, that should read $$\frac13 y^{-2/3}y'=1\implies y'=3y^{2/3}$$ but same point

51. anonymous

thanks a lot!