Math time!

[That Kid with the Hair]

All budding physicists, try this out:

y''' - 2y'' - y' + 2y = 2x^2 - 6x +4

y(0) = 5
y'(0) = -5
y''(0) = 1

Too lazy to think of a physical model... maybe an electrical circuit if it helps.

[Damien (AKA AUSSIE)]

I would have said 17 = 4x +4 but I doubt thats even close,
I might be absoulotley blind as I usally am for a friday night but I dout anyone will figure that out....

[That Kid with the Hair]

Maybe I was a bit too vague...it's a third order, nonhomogeneous, ordinary differential equation. I've given three initial conditions. The solution is in the form of y = f(x).

[Red Venus Moon]

Originally Posted by That Kid with the Hair

All budding physicists, try this out:

y''' - 2y'' - y' + 2y = 2x^2 - 6x +4

y(0) = 5
y'(0) = -5
y''(0) = 1

Too lazy to think of a physical model... maybe an electrical circuit if it helps.

Jesus H. Christ, you brought physics into the forums

[Fearless™]

gimme a couple of days to figure it out cuz i'm really busy with other stuff atm. don't post the answer yet. (:

[CuteBlackBoy]

Originally Posted by Damien (AKA AUSSIE);2576***

I would have said 17 = 4x +4 but I doubt thats even close,
I might be absoulotley blind as I usally am for a friday night but I dout anyone will figure that out....

Give me a lifetime to figure it out... ONLY SUBJECT I FAILED AT SCHOOL

[Fearless™]

tell me if this is right.

y = (1/30)x^5 - (1/4)x^4 + (2/3)x^3 + 4x^2 -16x + 5

[Ğǿđfâŧħέř]

Originally Posted by Fearless™;25***80

tell me if this is right.

y = (1/30)x^5 - (1/4)x^4 + (2/3)x^3 + 4x^2 -16x + 5

That's what I came up with too.

[Fearless™]

Originally Posted by Ğǿđfâŧħέř

That's what I came up with too.

O RLY?
then you wouldn't mind posting your working out
i.e. how did you come to that answer

[That Kid with the Hair]

Well...I'd like to see how you got that answer, Fearless. I'm glad to see someone actually made it that far lol. Not bad though....

As for the work involved:

General solution of this problem would be:
yh = ae^-x + be^x + ce^2x.

Now all we need is a yp. So we try a quadratic:
yp = Kx^2 + Mx + N
y'p = 2Kx + M
y''p = 2K
y'''p = 0

Then we substitute:

-2 * 2K - (2Kx + M) + 2(Kx^2 + Mx + N) = 2x^2 - 6x + 4

If we rearrange and equate like powers...we'll get something like this:

2Kx^2 = 2x^2, (-2K +2M)x = -6x, and -4K - M + 2N = 4

So we get K = 1, M = -2, and N = 3.

And remembering that yh + yp = y, we get:

y = ae^-x + be^x + ce^2x + x^2 - 2x + 3
y' = -ae^-x + be^x + 2ce^2x + 2x - 2
y' = ae^-x + be^x + 4ce^2x + 2

Now, we have to set x=0 and use the ICs.

y(0) = a + b + c + 3 = 5
y'(0) = -a + b + 2c - 2 = -5
y''(0) = a + b + 4c + 2 = 1

I'll spare the Gaussian elimination...

a = 2, b = 1, c = -1

So finally, our answer becomes:

y = 2e^-x + e^x - e^2x + x^2 - 2x + 3

Finally...btw anyone know how to put superscripts in? "sup" tags?