C_2=1+\frac{\dot{q}R^2}{4k}-ln|R|*C_1
T(r)=1+\frac{\dot{q}}{4k}(R^2-r^2)+C_1*ln|(\frac{r}{R})|
So now this mess again: T(x)=\sum_{n=0}^{\infty}b_n x^n ?
Ahhhh, I feel silly, here I was thinking it was first order.
Ok so:
\frac{dT}{dr}=-\frac{\dot{q}r}{2k}+\frac{C_1}{r}
T=-\frac{\dot{q}r^2}{4k}+ln|r|*C_1+C_2
\int \frac{-\dot{q}}{k} r dr => \frac{-\dot{q}r^2}{2k}
\int \frac{d}{dr}(r \frac{dT}{dr})dr => Tr
The dT is throwing me off, I'm not quite sure about it.
Ok now were going back to some of my more fundamental problems in this class.
\frac{1}{r}\frac{d}{dr}(r\frac{dT}{dr})=-\frac{\dot{q}}{k}
(ln|r|)rT=-\frac{\dot{q}}{k}
?
1=b_0+b_1(0)-\frac{\dot{q}0^2}{2k}
b_0=1
2=1+b_1(1)-\frac{\dot{q}}{2k}
1+\frac{\dot{q}}{2k}=b_1
T(x)=1+x+\frac{\dot{q}x}{2k}-\frac{\dot{q}x^2}{2k}
(I found an algebra mistake as I was recopying it.)
Ok: T(x)=-\frac{\dot{q}x^2}{2k}
but
bn => n(n-1)*bn*x^(n-2)
b0 => 0(0-1)*b0*x^(-2)=0
b1 => 1(1-1)*b1*x^(-1)=0
I'm assuming this is wrong, but that's my logic for it.
Not quite. In 2B_2 + 6B_3 x+ 12B_4 x^2+... where are the 2, 6, 12, etc comming from?
And why is this B_3=B_4=\ldots B_{\infty}=0 true especially when B_2 is equal to some real number?
Isn't it true that as n increases B_n decreases? Are we just assuming B_3 and on to be too small to be...
q, heat generation, is a constant. \dot{q} will change with distance, x. So no \dot{q} is not constant.
Edit: Scratch that, \dot{q} is the rate of heat transfer so it is a constant in this condition.