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Loser66
 one year ago
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
 one year ago
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

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.1dan, what is your avatar?? it looks like a barbecue hahaha

dan815
 one year ago
Best ResponseYou've already chosen the best response.1a parabola is the solution to a constant 2nd derivative, u can see it for those like projectile ball throwing up questions

Empty
 one year ago
Best ResponseYou've already chosen the best response.2What 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
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.1are you given any initial conditions ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, y''=2a is your differential equation.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 Yes, but the question for using initial conditions come after we get the equation.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1presumably a,b,c are arbitrary constants, so they should not be there in the ode

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1y''' = 0 looks good enough right ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.1So that I don't think we can use them in part b.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ohk so y'''=0 is not what you looking for ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1nope, I am looking for something like Ay" + By'+ Cy =0

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1where y = ax^2 +bx +c is the general solution of it

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[y = ax^2+bx+c\tag{1}\] \[y(0)=C \implies C = a*0^2+b*0+c \implies c=C\tag{2}\] are we good here ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2y'''=0 is correct then, I don't really see the issue here.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.2a,b, and c are all arbitrary real numbers.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1we're done with finding the ode, it is y''' = 0

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@Empty the reason we cannot use y"' =0 is

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1At the specific step, we have y^(m) =0. And it is not valid for example, if the solution is \(ax^{m1}+...whatever\) we have \(y^{(m)}=0\) always, right? Hence, what is the goal of finding a differential equation which solution is y = ....?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.13 arbitrary constants require the ode to be of 3rd order, yes ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1My thought: if y"'=0, I can say y =0 is a solution of it, why y = ax^2 +bx+c?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2that is a particular solution with a=0, b=0, and c=0. No prblem!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1y = 0 is a solution when a=b=c=0 you get to choose the initial conditions

dan815
 one year ago
Best ResponseYou've already chosen the best response.1none of spotted that mistake!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Hey, barbecue, use English, pleae

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1y''' = 0 models the path of a projectile under constant acceleration do not belittle it just because it looks innocent ;p

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I was going to say that example haha

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Ok, 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....

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Easy, use the solution y=ax^2+bx+c and get 3 equations based on the 3 given initial conditions. solve the system

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I got a = A, b = B, c = C

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I see why you were having second thoughts..

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1but what is the differential equation. hihihihi...

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1hahahah... lol, it seems all of us are stubborn.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1to 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
 one year ago
Best ResponseYou've already chosen the best response.2\(y'''=0\) 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
 one year ago
Best ResponseYou've already chosen the best response.1it feels weird to say y'''=0 becasuse the same solution can also be worked down to y'''' =0 and y'''''''''''''''''=0

dan815
 one year ago
Best ResponseYou've already chosen the best response.1it doesnt doesnt give enough information

dan815
 one year ago
Best ResponseYou've already chosen the best response.1y'''=0 can also arise from y''=0 0 is just weird

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.13 arbitrary constants cannot make a general solution to a 4th order ode

dan815
 one year ago
Best ResponseYou've already chosen the best response.1im not saying its that order ODE just saiyng y'''=0 lacks info imo

Empty
 one year ago
Best ResponseYou've already chosen the best response.2y''''=0 has the general solution of \(y=ax^3+bx^2+cx+d\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1what info is it lacking

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I think it is y" +xy' 2y = 2abxc

Empty
 one year ago
Best ResponseYou've already chosen the best response.2So 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!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I derive from y" = 2a y' = 2ax +b y = ax^2 +bx +c

dan815
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.1y=ax^2+bx+c y'=2ax+b y''=2a y'''=0 but from here on y''''=0 y''''''''''''''''''=0

dan815
 one year ago
Best ResponseYou've already chosen the best response.1you 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
 one year ago
Best ResponseYou've already chosen the best response.2Yes but the y'''''''''''''''''''''''''''''''''''''''''''''''''''''=0 has a different general solution which is not the same as what the question is asking.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1so i feel like y''''=0 just lacks info

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ahh right, y'''' = 0 tells us that the jerk is constant, but tells nothing about the acceleration

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so we do lose info when we differentiate more than necessary

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.1lol we are just trolling lets face it

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ive been trolling since 2014

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1been wondering what is it on ur profile pic

dan815
 one year ago
Best ResponseYou've already chosen the best response.1its a secret experiment to see if this pic will increase the average weight of OS over time

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1lol just so it increases ur appetite is it

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1i mean indirectly ur weight

dan815
 one year ago
Best ResponseYou've already chosen the best response.1working in the advertising industry must be fun, they get to do experiments on people all the time
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