Determine the steady state temperature distribution for the given problems I have the following problem, where I'm kinda lost what to do:
$${y}''=-T_{0}, \quad {y}'(0)=0, \quad y(1)=0$$
how can I solve this equation that has the constant  $T_{0}$. Any kind of help will be appreciated. Thank you.
 A: Integrating twice gives
$$y'=-T_0x+A$$
$$y=-\frac{T_0}2x^2+Ax+B$$
Now use initial conditions:
$$y'(0)=0\implies A=0$$
$$y(1)=0\implies 0=-\frac{T_0}2+B\implies B=\frac{T_0}2$$
$$y=-\frac{T_0}2x^2+\frac{T_0}2=\frac{T_0}2(1-x^2)$$
A: Write $y’’(t)=-T_0$ as $ dy’(t) = -T_0dt $ and integrate over $(0,t)$ with the initial value $y’(0)=0$ to get
$$y’(t)=y’(0) -T_0\int_0^t dt=-T_0t$$
Then, integrate the resulting $dy(t)=-T_0tdt$ over $(1,t) $ with the given value $y(1)=0$ to get
$$y(t)=y(1)-T_0\int_1^t tdt =- \frac 12T_0(
t^2-1)$$
A: Well, if a solution is sought on the interval $[0, 1]$, we may integrate
$y'' = -T_0 \tag 1$
from $0$ to $x \in [0, 1]$ to obtain
$y'(x) = y'(x) - y'(0) = \displaystyle \int_0^x y''(s) \; ds$
$= -\displaystyle \int_0^x T_0 \; ds = -T_0 (x - 0) = -T_0 x, \tag 2$
where we have used the given boundary condition
$y'(0) = 0; \tag 3$
we may then intgrate (2) over this same interval one more time and find
$y(x) - y(0) = \displaystyle \int_0^x y'(s) \; ds = -T_0 \int_0^x s \; ds = -T_0 \left ( \dfrac{x^2}{2} \right ) = -\dfrac{1}{2} T_0 x^2; \tag 4$
in light of the given
$y(1) = 0 \tag 5$
we find
$-y(0) = y(1) - y(0) = -\dfrac{1}{2}T_0, \tag 6$
or
$y(0) = \dfrac{1}{2} T_0, \tag 7$
whence (4) becomes
$y(x) = \dfrac{1}{2} T_0 - \dfrac{1}{2} T_0 x^2 = \dfrac{1}{2} T_0 (1 - x^2).\tag 8$
