anonymous
  • anonymous
Can someone help a novice out?! Differential Equation SOS! Find the DE knowing the final answer y=(x-c)^3 Thanks a lot!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Should I just get rid of the C?
anonymous
  • anonymous
and if so. idk really know how tbh
anonymous
  • anonymous
@ganeshie8

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anonymous
  • anonymous
@idku
anonymous
  • anonymous
but the C remains! shouldn't I come to the point of which I don't have constants?
anonymous
  • anonymous
hahahahahah
anonymous
  • anonymous
well thanks anyhow
anonymous
  • anonymous
@arthurpariz
anonymous
  • anonymous
@Loser66 will you be my savior?
anonymous
  • anonymous
thanks a lot!!!!
freckles
  • freckles
You can find y' by differentiating the y=(x-c)^3.
freckles
  • 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 ?
mathmate
  • mathmate
y=(x-c)^3 y'=3(y-c)^2 y"=6(y-c) so y=(y"/6)^3 or (y")^3-216y=0
mathmate
  • mathmate
oh, sorry, I didn't see the condition.
freckles
  • freckles
I don't see how these rules in which we find a differential equation posted in the original post. @Loser66
freckles
  • freckles
I don't see that we aren't allowed to use initial conditions or find a non-linear diff equation.
freckles
  • freckles
it would be nice though if @Hipocampus can settle what we are talking about though
anonymous
  • anonymous
hmmm that's all I have...
anonymous
  • anonymous
@freckles
anonymous
  • anonymous
idk more than you guys about this question
freckles
  • freckles
So the solution can be non-linear or can have conditions ?
anonymous
  • anonymous
I believe so
anonymous
  • anonymous
as long as the C is gone
Loser66
  • 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.
anonymous
  • anonymous
thanks a lot guys!
anonymous
  • anonymous
can I ask another question then?
Loser66
  • Loser66
close this and open a new one, please.
anonymous
  • anonymous
Roger @Loser66 !
Loser66
  • 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
freckles
  • freckles
Oh so no conditions allowed you are saying @Loser66
Loser66
  • Loser66
@freckles I didn't say "no conditions allowed", just don't use the un-given conditions.
freckles
  • 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
freckles
  • freckles
you get to make up the differential equation, why can't you make up the conditions that it is subject to?
freckles
  • freckles
and I'm not really making it up I'm based my condition off what the solution should be.
Loser66
  • Loser66
Anyway!! I think if we make up a condition and have a Unique solution for that condition, we are ok :)
Loser66
  • Loser66
I am with you now :)
ganeshie8
  • ganeshie8
aren't we looking for the differential equation of order 1, whose general solution is the set of standard cubics translated horizontally ?
freckles
  • freckles
I don't know it wasn't specify that had to have order 1
ganeshie8
  • ganeshie8
the order has to be 1 because there is only 1 arbitrary constant in the given general solution
ganeshie8
  • ganeshie8
number of arbitrary constants and the order of de must agree, right
freckles
  • 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
freckles
  • freckles
meant to end that with a ?
ganeshie8
  • ganeshie8
this should work \[(y')^3 = 27y^2\]
freckles
  • freckles
I like that requires no conditions
freckles
  • freckles
but I still think my way is valid
freckles
  • freckles
only because there was not much put into the "rules" of finding the differential equation
ganeshie8
  • 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
anonymous
  • anonymous
you could just do: $$y=(x-c)^3\\y^{1/3}=x-c\\\frac13 y^{2/3}y'=1\\y'=3y^{-2/3}$$
anonymous
  • anonymous
the solutions are the one-parameter family of monic cubics up to horizontal translation \(y=(x-c)^3\)
anonymous
  • anonymous
oops, that should read \(\frac13 y^{-2/3}y'=1\implies y'=3y^{2/3}\) but same point
anonymous
  • anonymous
thanks a lot!

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