Question

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

Answer 1

Should I just get rid of the C?

Answer 2

and if so. idk really know how tbh

Answer 3

@ganeshie8

Answer 4

@idku

Answer 5

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

Answer 6

hahahahahah

Answer 7

well thanks anyhow

Answer 8

@arthurpariz

Answer 9

@Loser66 will you be my savior?

Answer 10

thanks a lot!!!!

Answer 11

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

Answer 12

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
?

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

hmmm that's all I have...

Answer 19

@freckles

Answer 20

idk more than you guys about this question

Answer 21

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

Answer 22

I believe so

Answer 23

as long as the C is gone

Answer 24

@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.

Answer 25

thanks a lot guys!

Answer 26

can I ask another question then?

Answer 27

close this and open a new one, please.

Answer 28

Roger @Loser66 !

Answer 29

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

Answer 30

Oh so no conditions allowed you are saying @Loser66

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

I am with you now :)

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

meant to end that with a ?

Answer 43

this should work
\[(y')^3 = 27y^2\]

Answer 44

I like that requires no conditions

Answer 45

but I still think my way is valid

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

thanks a lot!