I solved this by hand to confirm Maple solution.
Solve $u_{t}=u_{xx}$ with initial conditions $u\left( x,0\right)
=\sin\left( x\right) $ and B.C. $u_{x}\left( 0,t\right) =1,u_{x}\left(
\pi,t\right) =-1$. For $t>0,0<x<\pi$.
Since the boundary conditions are not homogeneous, we can't use separation of
variables. Let the solution be
$$
u=v\left( x,t\right) +r\left( x\right)
$$
Where $v\left( x,t\right) $ is the solution to $v_{t}=v_{xx}$ and homogeneous
B.C. $v_{x}\left( 0,t\right) =0,v_{x}\left( \pi,t\right) =0$ and $r\left(
x\right) $ is any reference solution which only needs to satisfy the
non-homogeneous boundary conditions: $r^{\prime}\left( 0\right) =1,r^{\prime
}\left( \pi\right) =-1$. By guessing, let $r\left( x\right) =Ax+Bx^{2}$.
Let see if this satisfies the boundary conditions. $r^{\prime}=A+2Bx$. At
$x=0$ this implies $1=A$. Hence $r=x+Bx^{2}$. Now $r^{\prime}=1+2Bx$. At
$x=\pi$ this gives $-1=1+2B\pi$ or $B=-\frac{1}{\pi}$. Therefore
$$
r\left( x\right) =x-\frac{1}{\pi}x^{2}
$$
Substituting $u=v\left( x,t\right) +r\left( x\right) $ into the
PDE $u_{t}=u_{xx}$ and noting that $r^{\prime\prime}\left( x\right)
=-\frac{2}{\pi}$ gives
\begin{equation}
v_{t}=v_{xx}-\frac{2}{\pi}\tag{1}
\end{equation}
PDE (1) is now solved using eigenfunction expansion. We need to find
eigenfunctions and eigenvalues of $v_{t}=v_{xx}$ with $v_{x}\left(
0,t\right) =0,v_{x}\left( \pi,t\right) =0$. This is known PDE and have
eigenfunctions and eigenvalues as follows (ref 1). For zero eigenvalue, the
eigenfunction is an arbitrary constant. Say $\beta$. let $\beta=1$ since scale
is not important.
$$
\Phi_{0}\left( x\right) =1
$$
And for $n=1,2,3,\cdots$
\begin{align*}
\Phi_{n}\left( x\right) & =\cos\left( \sqrt{\lambda_{n}}x\right) \\
& =\cos\left( nx\right)
\end{align*}
with eigenvalues $\lambda_{n}=n^{2}$ for $n=1,2,3,\cdots$. Now we can use
eigenfunction expansion and assume the solution to (1) is
\begin{equation}
v\left( x,t\right) =\sum_{n=0}^{\infty}A_{n}\left( t\right) \Phi
_{n}\left( x\right) \tag{2}
\end{equation}
Plugging this into the PDE (1) gives
$$
\sum_{n=0}^{\infty}A_{n}^{\prime}\left( t\right) \Phi_{n}\left( x\right)
=\sum_{n=0}^{\infty}A_{n}\left( t\right) \Phi_{n}^{\prime\prime}\left(
x\right) -\frac{2}{\pi}
$$
But $\Phi_{n}^{\prime\prime}\left( x\right) =-\lambda_{n}\Phi_{n}\left(
x\right) $ and the above simplifies to
$$
\sum_{n=0}^{\infty}A_{n}^{\prime}\left( t\right) \Phi_{n}\left( x\right)
=-\sum_{n=0}^{\infty}A_{n}\left( t\right) \lambda_{n}\Phi_{n}\left(
x\right) -\frac{2}{\pi}
$$
Since eigenfunctions are complete, we can expand $\frac{2}{\pi}$ using them
and the above becomes
\begin{align}
\sum_{n=0}^{\infty}A_{n}^{\prime}\left( t\right) \Phi_{n}\left( x\right)
& =-\sum_{n=0}^{\infty}A_{n}\left( t\right) \lambda_{n}\Phi_{n}\left(
x\right) -\sum_{n=0}^{\infty}C_{n}\Phi_{n}\left( x\right) \nonumber\\
A_{n}^{\prime}\left( t\right) \Phi_{n}\left( x\right) +A_{n}\left(
t\right) \lambda_{n}\Phi_{n}\left( x\right) & =-C_{n}\Phi_{n}\left(
x\right) \nonumber\\
A_{n}^{\prime}\left( t\right) +A_{n}\left( t\right) \lambda_{n} &
=-C_{n}\tag{3}
\end{align}
To find $C_{n}$
$$
\sum_{n=0}^{\infty}C_{n}\Phi_{n}\left( x\right) =\frac{2}{\pi}
$$
For $n=0$
$$
C_{0}\Phi_{0}\left( x\right) =\frac{2}{\pi}
$$
But $\Phi_{0}\left( x\right) =1$, hence
$$
C_{0}=\frac{2}{\pi}
$$
All other $C_{m}$ for $m>0$ are zero.
Hence (3) becomes, for
$n=0$ (since $\lambda_{0}=0$)
\begin{align*}
A_{0}^{\prime}\left( t\right) & =- \frac{2}{\pi}\\
A_{0}\left( t\right) & =-\frac{2}{\pi}t+B_{0}
\end{align*}
Where $B_{0}$ is integration constant. For $n>0$ (3) becomes
$$
A_{n}^{\prime}\left( t\right) +A_{n}\left( t\right) n^{2}=0
$$
This has the solution
$$
A_{n}\left( t\right) =B_{n}e^{-n^{2}t}
$$
Where $B_{n}$ is constant of integration. Hence from (2)
\begin{align*}
v\left( x,t\right) & =\sum_{n=0}^{\infty}A_{n}\left( t\right) \Phi
_{n}\left( x\right) \\
& =A_{0}\left( t\right) +\sum_{n=1}^{\infty}A_{n}\left( t\right) \Phi
_{n}\left( x\right) \\
& =-\frac{2}{\pi}t+B_{0}+\sum_{n=1}^{\infty}B_{n}e^{-n^{2}t}\cos\left(
nx\right)
\end{align*}
Since $u=v\left( x,t\right) +r\left( x\right) $ then the solution becomes
\begin{equation}
u\left( x,t\right) =\left( x-\frac{1}{\pi}x^{2}\right) -\frac{2}{\pi
}t+B_{0}+\sum_{n=1}^{\infty}B_{n}e^{-n^{2}t}\cos\left( nx\right) \tag{4}
\end{equation}
At $t=0$
\begin{equation}
\sin\left( x\right) =\left( x-\frac{1}{\pi}x^{2}\right) +B_{0}+\sum
_{n=1}^{\infty}B_{n}\cos\left( nx\right) \tag{5}
\end{equation}
case $n=0$
$$
\int_{0}^{\pi}\sin\left( x\right) \cos\left( \sqrt{\lambda_{0}}x\right)
dx=\int_{0}^{\pi}\left( x-\frac{1}{\pi}x^{2}\right) \cos\left(
\sqrt{\lambda_{0}}x\right) dx+\int_{0}^{\pi}B_{0}\cos\left( \sqrt
{\lambda_{0}}x\right) dx
$$
But $\lambda_{0}=0$ hence
\begin{align*}
0 & =\int_{0}^{\pi}\left( x-\frac{1}{\pi}x^{2}\right) dx+\int_{0}^{\pi
}B_{0}dx\\
0 & =\frac{\pi^{2}}{6}+B_{0}\pi\\
B_{0} & =\frac{-\pi}{6}
\end{align*}
For $n>0$, Multiplying both sides of (5) by $\cos\left( mx\right) $ and integrating
$$
\int_{0}^{\pi}\sin\left( x\right) \cos\left( mx\right) dx=\int_{0}^{\pi
}\left( x-\frac{1}{\pi}x^{2}\right) \cos\left( mx\right) dx+\sum
_{n=1}^{\infty}B_{n}\int_{0}^{\pi}\cos\left( nx\right) \cos\left(
mx\right) dx
$$
For $m=1$
\begin{align*}
0 & =0+B_{1}\frac{\pi}{2}\\
B_{1} & =\frac{2}{\pi}
\end{align*}
For $m>1$
\begin{align*}
-\frac{1+\left( -1\right) ^{m}}{m^{2}\left( -1+m^{2}\right) } &
=\frac{\pi}{2}B_{n}\\
B_{n} & =\frac{-2}{\pi}\left( \frac{1}{m^{2}}\frac{\left( -1\right)
^{m}+1}{m^{2}-1}\right)
\end{align*}
Hence solution (4) becomes
\begin{align*}
u\left( x,t\right) & =\left( x-\frac{1}{\pi}x^{2}\right) -\frac{2}{\pi
}t-\frac{\pi}{6}+\frac{2}{\pi}+\sum_{n=1}^{\infty}B_{n}e^{-n^{2}t}\cos\left(
nx\right) \\
& =\left( x-\frac{1}{\pi}x^{2}\right) -\frac{2}{\pi
}t-\frac{\pi}{6}+\frac{2}{\pi}+\sum_{n=2}^{\infty}\frac{-2}{\pi}\left(
\frac{1}{m^{2}}\frac{\left( -1\right) ^{m}+1}{m^{2}-1}\right) e^{-n^{2}
t}\cos\left( nx\right)
\end{align*}
Maple gives
heqn := diff(u(x, t), t) = diff(u(x, t), x$2):
ic := u(x, 0) = sin(x):
bc := eval(diff(u(x,t),x),x=0)=1, eval( diff(u(x,t),x),x=Pi)=-1:
sol := pdsolve({heqn, ic, bc}, u(x, t))
[![enter image description here][1]][1]
Here is verification
mysol=(x-1/Pi x^2)-2/Pi t-Pi/6+2/Pi-
2/Pi Inactivate[Sum[((-1)^n+1)/(n^2(n^2-1)) Cos[n x]Exp[-n^2 t],{n,2,Infinity}]];
D[mysol/.Infinity->20//Activate,x]/.t->0/.x->0
(*1*)
D[mysol/.Infinity->20//Activate,x]/.t->0/.x->Pi
(*-1*)
Plot[{ic,Sin[x]},{x,0,Pi},AxesOrigin->{0,0}]
[1]: https://i.sstatic.net/cN3nQ.png
[2]: https://www.12000.org/my_notes/boundary_values/report.htm#x1-70001.5
ref (1): [boundary_values][2]