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

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

Mathematics
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Should I just get rid of the C?
and if so. idk really know how tbh

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

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