The general solution to a certain ODE is \(y=ax^2+bx+c\) (a, b,c are in real)
Find the differential equation.
Please, help

- Loser66

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- dan815

|dw:1436378469720:dw|

- Loser66

dan, what is your avatar?? it looks like a barbecue hahaha

- dan815

a parabola is the solution to a constant 2nd derivative, u can see it for those like projectile ball throwing up questions

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## More answers

- Empty

What is an ODE? A relationship between a function and its derivatives.
Just take the derivative of y a few times and see if you can relate it.

- Loser66

@dan815 I don't think it is ok
@Empty Yes, I did
but go nowhere
y = ax^2 +bx+c
y'= 2ax +b
y"= 2a
then??

- ganeshie8

are you given any initial conditions ?

- Empty

Yeah, y''=2a is your differential equation.

- Loser66

@ganeshie8 Yes, but the question for using initial conditions come after we get the equation.

- Empty

or even y'''=0

- ganeshie8

presumably a,b,c are arbitrary constants, so they should not be there in the ode

- ganeshie8

y''' = 0 looks good enough right ?

- Loser66

@ganeshie8 the next question is : If y" (0) = 2A (big A)
y'(0) = B
and y(0) = C (big C)
for the differential equation in part b) , find a, b, c

- Loser66

So that I don't think we can use them in part b.

- ganeshie8

Ohk so y'''=0 is not what you looking for ?

- Loser66

nope, I am looking for something like Ay" + By'+ Cy =0

- Loser66

where y = ax^2 +bx +c is the general solution of it

- ganeshie8

\[y = ax^2+bx+c\tag{1}\]
\[y(0)=C \implies C = a*0^2+b*0+c \implies c=C\tag{2}\]
are we good here ?

- Empty

y'''=0 is correct then, I don't really see the issue here.

- ganeshie8

me neither

- Loser66

@ganeshie8 Again, we are not allowed to use y''(0) = A, y'(0) = B, y(0) =C for part b) (the part we have to find out the differential equation)

- Empty

a,b, and c are all arbitrary real numbers.

- ganeshie8

we're done with finding the ode, it is y''' = 0

- Loser66

@Empty the reason we cannot use y"' =0 is

- Loser66

At the specific step, we have y^(m) =0. And it is not valid
for example, if the solution is \(ax^{m-1}+...whatever\)
we have \(y^{(m)}=0\) always, right?
Hence, what is the goal of finding a differential equation which solution is y = ....?

- ganeshie8

3 arbitrary constants require the ode to be of 3rd order, yes ?

- Loser66

Continue, please.

- dan815

|dw:1436379643346:dw|

- Loser66

My thought: if y"'=0, I can say y =0 is a solution of it, why y = ax^2 +bx+c?

- Empty

that is a particular solution with a=0, b=0, and c=0. No prblem!

- ganeshie8

y = 0 is a solution when a=b=c=0
you get to choose the initial conditions

- dan815

none of spotted that mistake!

- dan815

you

- Loser66

Hey, barbecue, use English, pleae

- ganeshie8

y''' = 0 models the path of a projectile under constant acceleration
do not belittle it just because it looks innocent ;p

- Empty

I was going to say that example haha

- Loser66

Ok, Let me ask my professor what he does really want.Because, if y"' =0, how can I solve part c) where they gave me the initial condition y"(a) = 2A....

- Loser66

sorry y"(0) = 2A....

- ganeshie8

Easy, use the solution y=ax^2+bx+c and get 3 equations based on the 3 given initial conditions. solve the system

- Loser66

I got a = A, b = B, c = C

- ganeshie8

I see why you were having second thoughts..

- Loser66

but what is the differential equation. hihihihi...

- Empty

y'''=0

- Empty

;)

- Loser66

hahahah... lol, it seems all of us are stubborn.

- ganeshie8

to my knowledge
y''' = 0 and a=A, b=B, c=C
looks good even though it leaves me wondering why ur prof wants you spend time on such trivial prob

- Empty

\(y'''=0\)
integrate
\(y'' = k_1\)
integrate
\(y'=k_1x + k_2\)
integrate
\(y=\frac{k_1}{2}x^2+k_2x+k_3\)
call \(\frac{k_1}{2}=a\), \(k_2\)=b, \(k_3=c\).
y=ax^2+bx+c
these are all arbitrary numbers because this is the general solution. :)

- dan815

it feels weird to say y'''=0 becasuse the same solution can also be worked down to y'''' =0 and y'''''''''''''''''=0

- dan815

it doesnt doesnt give enough information

- dan815

y'''=0 can also arise from y''=0
0 is just weird

- ganeshie8

3 arbitrary constants cannot make a general solution to a 4th order ode

- dan815

im not saying its that order ODE just saiyng y'''=0 lacks info imo

- Empty

y''''=0 has the general solution of
\(y=ax^3+bx^2+cx+d\)

- ganeshie8

what info is it lacking

- Loser66

I think it is y" +xy' -2y = 2a-bx-c

- Empty

So although you can have a=0 to get a specific case where it has the same answer as y'''=0, this is saying that a _particular solution_ of y''''=0 is the same as the _general solution_ of y'''=0. So not quite the same thing danny boy!

- dan815

well on 2nd thought

- Loser66

I derive from y" = 2a
y' = 2ax +b
y = ax^2 +bx +c

- dan815

i was looking at working backwards lets say suppse u are given y=ax^2+bx+c
okay then differentiative that will give us its higher order repersentation

- dan815

y=ax^2+bx+c
y'=2ax+b
y''=2a
y'''=0
but
from here on
y''''=0
y''''''''''''''''''=0

- dan815

you are saying that solution you have is general but that is just existing because u chose to work the other way around, you are following some set steps but that doesnt have to be the steps you follow for a solution

- Empty

Yes but the y'''''''''''''''''''''''''''''''''''''''''''''''''''''=0 has a different general solution which is not the same as what the question is asking.

- dan815

so i feel like y''''=0 just lacks info

- ganeshie8

Ahh right, y'''' = 0 tells us that the jerk is constant, but tells nothing about the acceleration

- ganeshie8

so we do lose info when we differentiate more than necessary

- Empty

Yeah this solution I think is supposed to not be thought about too much, I feel like the question is so easy that we're over thinking this lol

- dan815

lol we are just trolling lets face it

- dan815

ive been trolling since 2014

- ganeshie8

been wondering what is it on ur profile pic

- dan815

ribs lol

- dan815

its a secret experiment to see if this pic will increase the average weight of OS over time

- ganeshie8

lol just so it increases ur appetite is it

- ganeshie8

i mean indirectly ur weight

- dan815

working in the advertising industry must be fun, they get to do experiments on people all the time

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