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

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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.

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.