Loser66
  • Loser66
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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dan815
  • dan815
|dw:1436378469720:dw|
Loser66
  • Loser66
dan, what is your avatar?? it looks like a barbecue hahaha
dan815
  • 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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Empty
  • 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
  • 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
  • ganeshie8
are you given any initial conditions ?
Empty
  • Empty
Yeah, y''=2a is your differential equation.
Loser66
  • Loser66
@ganeshie8 Yes, but the question for using initial conditions come after we get the equation.
Empty
  • Empty
or even y'''=0
ganeshie8
  • ganeshie8
presumably a,b,c are arbitrary constants, so they should not be there in the ode
ganeshie8
  • ganeshie8
y''' = 0 looks good enough right ?
Loser66
  • 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
  • Loser66
So that I don't think we can use them in part b.
ganeshie8
  • ganeshie8
Ohk so y'''=0 is not what you looking for ?
Loser66
  • Loser66
nope, I am looking for something like Ay" + By'+ Cy =0
Loser66
  • Loser66
where y = ax^2 +bx +c is the general solution of it
ganeshie8
  • 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
  • Empty
y'''=0 is correct then, I don't really see the issue here.
ganeshie8
  • ganeshie8
me neither
Loser66
  • 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
  • Empty
a,b, and c are all arbitrary real numbers.
ganeshie8
  • ganeshie8
we're done with finding the ode, it is y''' = 0
Loser66
  • Loser66
@Empty the reason we cannot use y"' =0 is
Loser66
  • 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
  • ganeshie8
3 arbitrary constants require the ode to be of 3rd order, yes ?
Loser66
  • Loser66
Continue, please.
dan815
  • dan815
|dw:1436379643346:dw|
Loser66
  • Loser66
My thought: if y"'=0, I can say y =0 is a solution of it, why y = ax^2 +bx+c?
Empty
  • Empty
that is a particular solution with a=0, b=0, and c=0. No prblem!
ganeshie8
  • ganeshie8
y = 0 is a solution when a=b=c=0 you get to choose the initial conditions
dan815
  • dan815
none of spotted that mistake!
dan815
  • dan815
you
Loser66
  • Loser66
Hey, barbecue, use English, pleae
ganeshie8
  • ganeshie8
y''' = 0 models the path of a projectile under constant acceleration do not belittle it just because it looks innocent ;p
Empty
  • Empty
I was going to say that example haha
Loser66
  • 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
  • Loser66
sorry y"(0) = 2A....
ganeshie8
  • 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
  • Loser66
I got a = A, b = B, c = C
ganeshie8
  • ganeshie8
I see why you were having second thoughts..
Loser66
  • Loser66
but what is the differential equation. hihihihi...
Empty
  • Empty
y'''=0
Empty
  • Empty
;)
Loser66
  • Loser66
hahahah... lol, it seems all of us are stubborn.
ganeshie8
  • 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
  • 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
  • dan815
it feels weird to say y'''=0 becasuse the same solution can also be worked down to y'''' =0 and y'''''''''''''''''=0
dan815
  • dan815
it doesnt doesnt give enough information
dan815
  • dan815
y'''=0 can also arise from y''=0 0 is just weird
ganeshie8
  • ganeshie8
3 arbitrary constants cannot make a general solution to a 4th order ode
dan815
  • dan815
im not saying its that order ODE just saiyng y'''=0 lacks info imo
Empty
  • Empty
y''''=0 has the general solution of \(y=ax^3+bx^2+cx+d\)
ganeshie8
  • ganeshie8
what info is it lacking
Loser66
  • Loser66
I think it is y" +xy' -2y = 2a-bx-c
Empty
  • 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
  • dan815
well on 2nd thought
Loser66
  • Loser66
I derive from y" = 2a y' = 2ax +b y = ax^2 +bx +c
dan815
  • 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
  • dan815
y=ax^2+bx+c y'=2ax+b y''=2a y'''=0 but from here on y''''=0 y''''''''''''''''''=0
dan815
  • 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
  • Empty
Yes but the y'''''''''''''''''''''''''''''''''''''''''''''''''''''=0 has a different general solution which is not the same as what the question is asking.
dan815
  • dan815
so i feel like y''''=0 just lacks info
ganeshie8
  • ganeshie8
Ahh right, y'''' = 0 tells us that the jerk is constant, but tells nothing about the acceleration
ganeshie8
  • ganeshie8
so we do lose info when we differentiate more than necessary
Empty
  • 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
  • dan815
lol we are just trolling lets face it
dan815
  • dan815
ive been trolling since 2014
ganeshie8
  • ganeshie8
been wondering what is it on ur profile pic
dan815
  • dan815
ribs lol
dan815
  • dan815
its a secret experiment to see if this pic will increase the average weight of OS over time
ganeshie8
  • ganeshie8
lol just so it increases ur appetite is it
ganeshie8
  • ganeshie8
i mean indirectly ur weight
dan815
  • dan815
working in the advertising industry must be fun, they get to do experiments on people all the time

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