Cartesiantheater
Feb 12th, 2009, 12:02 PM
Engineer people (this means you, Fire Ant) get in here! I know you know Series! There hasn't been a single (successful) engineer in the history of modern engineering that doesn't know series well! So, here goes...
I am certain I utterly failed an exam over differential equations. I got screwed on the series solution part. Now, I can do easy ones, like the one I will show in a second, but the hard ones, with unknown functions of variable coefficients? LOST.
Okay, here is what I CAN do.
Find a solution to
y' = 2xy
Using series. Piece o' cake. Series isn't needed, but I can do it with series.
Assume a solution of the form:
y = a0 + a1x + a2x^2 + a3x^3 + ... = Σ an x^n
differentiating term by term gives:
y' = 0 + a1 + 2a2x + 3a3x^3 + ... Σ nan x^(n-1)
Now plugging each into the original DE:
Σ nan x^(n-1) = 2x Σ an x^n
a1 + 2a2x + 3a3x^3 + ... =
2x*a0 + 2a1x^2 + 2a2x^3 + 2a3x^4 + ...
a1 = 0, because there is no constant term by itself in y
lining up powers of x and solving gives:
a2 = a0
a3 = (2/3)a2 = (2/3)a0
a4 = (1/2)a0
a5 = (4/15)a1 = 0
and so on until you get cΣx^2n/n!
I know this is right because using normal methods I get:
dy/dx = 2xy
dy/y = 2xdx
ln y = x^2 + ln c
y = ce^x^2 ==> y = c( 1 + x^2 + x^4 /2! + ...) = cΣx^2n/n!
But that is easy! (relatively speaking... sigh)
Here is where I need MONSTROUS help: when you have unknown variable coefficients, for example.
What would I do here?
y'' + 2x^(-1) y' + 2 q(x) y = 1
such that
y(1) = 1 and y'(1) = -1
Notice the q(x) term. Wtf do I do with it? Anyone who knows this stuff well, please lend a brotha a hand.
Of course, with the scarcity of our math people ever showing up, I doubt this will even be read, lol. But here's hoping someone sees it before next week (by then it will be... too late!)
Thanks!
I am certain I utterly failed an exam over differential equations. I got screwed on the series solution part. Now, I can do easy ones, like the one I will show in a second, but the hard ones, with unknown functions of variable coefficients? LOST.
Okay, here is what I CAN do.
Find a solution to
y' = 2xy
Using series. Piece o' cake. Series isn't needed, but I can do it with series.
Assume a solution of the form:
y = a0 + a1x + a2x^2 + a3x^3 + ... = Σ an x^n
differentiating term by term gives:
y' = 0 + a1 + 2a2x + 3a3x^3 + ... Σ nan x^(n-1)
Now plugging each into the original DE:
Σ nan x^(n-1) = 2x Σ an x^n
a1 + 2a2x + 3a3x^3 + ... =
2x*a0 + 2a1x^2 + 2a2x^3 + 2a3x^4 + ...
a1 = 0, because there is no constant term by itself in y
lining up powers of x and solving gives:
a2 = a0
a3 = (2/3)a2 = (2/3)a0
a4 = (1/2)a0
a5 = (4/15)a1 = 0
and so on until you get cΣx^2n/n!
I know this is right because using normal methods I get:
dy/dx = 2xy
dy/y = 2xdx
ln y = x^2 + ln c
y = ce^x^2 ==> y = c( 1 + x^2 + x^4 /2! + ...) = cΣx^2n/n!
But that is easy! (relatively speaking... sigh)
Here is where I need MONSTROUS help: when you have unknown variable coefficients, for example.
What would I do here?
y'' + 2x^(-1) y' + 2 q(x) y = 1
such that
y(1) = 1 and y'(1) = -1
Notice the q(x) term. Wtf do I do with it? Anyone who knows this stuff well, please lend a brotha a hand.
Of course, with the scarcity of our math people ever showing up, I doubt this will even be read, lol. But here's hoping someone sees it before next week (by then it will be... too late!)
Thanks!