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Okay So ill start by writing the question out, as maybe I'm interpreting it wrong:

One end A of an insulated metal bar AB of length 2m is kept at 0c while the other end B is maintained at 50c until a steady sate temperature along the bar is reached, At t=0 the end B is suddenly reduced to 0c and kept at that temperature. Using the heat equation $\frac{1}{c^2}u_t=u_{xx}$ determine an expression for the temperature at any point in the bar distance x from A at any time t.

$u(0,t)=0$ , $u(2,t) = 50$, $u(2,0) = 0$ , $u(x,0) = 0 (Not,sure)$

So because it isn't homogeneous boundary conditions:

$u(x,t) = v(x) + w(x,t)$

$v(x) = Ax + B ,\space \space v(0)=B=0,\space\space v(2)=2A=50,\space A=25$

$v(x)=25x$

$u(x,t) - v(x) = w(x,t)$

$u(0,t) - v(0) = 0 \space \space, \space u(2,t) - v(2) = 0$

As the boundary conditions are now homogeneous i can proceed with separation of variables which produces the following:

$w(x,t)= \sum_{r=1}^{\infty}Q_r\sin(\frac{r\pi x}{2})\mathrm{e}^{-\lambda^2t} , \space\space\space where \space \lambda=\frac{r\pi c}{2}$

I'm not proficient enough in LaTeX commands to show all the steps but if the above is wrong i guess i need to !

This is kind of where i have a problem, im not sure which boundary conditions i should apply, so if i applied $u(x,0) = 0$ then $v(x)=25x$ therefore:

$-25x = w(x,0)= \sum_{r=1}^{\infty}Q_r\sin(\frac{r\pi x}{2})$

$Q_r = \frac{2}{2} \int_{0}^{2}-25xsin(\frac{r\pi x}{2})dx$

$Q_r = \left[\frac{50x}{r\pi}\cos(\frac{r\pi x}{2})\right]_0^{2} $

$Q_r = -\frac{100}{r\pi}\space when\space r = 1,3,5$ or $Q_r = \frac{100}{r\pi}\space when\space r = 2,4,6$

So my solution to u(x,t) would be

$u(x,t) = 25x + \frac{100}{\pi}\sum_{r=1}^{\infty}\frac{1}{r}(-1)^r\sin(\frac{r\pi x}{2})\mathrm{e}^{-\lambda^2t}$

But the book has $u(x,t) = \frac{100}{\pi}\sum_{r=1}^{\infty}\frac{1}{r}(-1)^{r+1} \sin(\frac{r\pi x}{2})\mathrm{e}^{-\lambda^2t} $

So it doesn't even take into account the steady state part of the solution, im confused, maybe Ive interpreted boundary conditions wrong?

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The steady state solution for the case where one end is being held at Temp=$0$ and the other at Temp=$50$ would be a linear temperature distribution ranging from $0$ to $50$. That definitely would be a solution where $u_{xx}=0$, and $u_{t}=Ku_{xx}=0$. So that's steady state. That means the next part starts with $u(x,0)=25x$, where $x$ is in meters. Assuming the end at $B$ is suddenly set to $0$, then, setting $t=0$ at that instant, you'll have the new problem $$ u_{t}(x,t)=u_{xx}(x,t),\;\;\; 0 \le x \le 2,\;\; t \ge 0,\\ u(0,t)=u(2,t)=0,\;\;\; t \ge 0,\\ u(x,0)=25x,\;\;\; 0 \le x < 2. $$ The idea is that you instantly set $u(2,0)=0$, which still leaves $u(x,0)=25x$ for $0 \le x < a$. Of course that's out of equilibrium, which is why you must solve a new heat equation. I think that's a reasonable intepretation based on the description. You'll want a solution of the form $$ u(x,t) =\sum_{n=1}^{\infty}C_{n}\sin(n\pi x/2)e^{-n^{2}\pi^{2}c^{2} t/4}. $$ The $n\pi x/2$ comes from making sure that $\sin(n\pi x/2)=0$ at $x=2$ for all $n\ge 1$. Differentiating twice in $x$ and multiplying by $c^{2}$ must give the same thing as differentiation once in $t$. So the exponent of $e$ must be $-c^{2}(n\pi/2)^{2}t$. (Check my constants; I'm always getting things like that wrong.) The only missing piece is to make sure that $u(x,0)=25x$ by expanding $25x$ in a Fourier sine series. So you need to find the constants $C_{n}$ such that $$ u(x,0)=\sum_{n=1}^{\infty}C_{n}\sin(n\pi x/2). $$ Using orthogonality of the $\sin$ functions (which you do have,) $$ \int_{0}^{2}(25x)\sin(n\pi x/2)\,dx = C_{n}\int_{0}^{2}\sin^{2}(n\pi x/2)\,dx. $$ You're wanting to add the extra $25x$ in, but that will just lead to the previous steady state solution because all of the coefficients of the separated solutions must be 0 in order to match $u(x,0)=25x$. And that solution won't satisfy $u(2,t)=0$ for $t > 0$.

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