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- #1

$\begin{align}

&{{u}_{t}}=K{{u}_{xx}},\text{ }0<x<L,\text{ }t>0, \\

&{{u}_{x}}(0,t)={{g}_{0}}(t),\text{ }{{u}_{x}}(L,t)={{g}_{1}}(t),\text{ for }t>0, \\

&u(x,0)=f(x),\text{ for }0<x<L.

\end{align}

$

2) Same as above:

$\begin{align}

& {{u}_{t}}=K{{u}_{xx}},\text{ }0<x<L,\text{ }t>0, \\

& u(0,t)={{h}_{0}}(t)\text{ for }t>0, \\

& {{u}_{x}}(L,t)+\alpha u(L,t)={{h}_{1}}(t),\text{ for }t>0, \\

& u(x,0)=f(x),\text{ for }0<x<L.

\end{align}

$

Well first one is pretty much alike to other one I posted, but it now contains first derivative in the boundary conditions so I'd like to know how to proceed. Second one looks harder, I don't see how to start.