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Should I just get rid of the C?

and if so. idk really know how tbh

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

hahahahahah

well thanks anyhow

thanks a lot!!!!

You can find y' by differentiating the y=(x-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 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

thanks a lot guys!

can I ask another question then?

close this and open a new one, please.

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

I am with you now :)

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

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

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

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

thanks a lot!