For the heat equation
$$u_t = 4 u_{xx}$$ $$u(x,0) = 1$$ $$u_x(0,t)=u_x(1,t)$$ $$0 \leq x \leq 1$$
we know that the general solution for Neumann boundary conditions is
$$u(x,t) = \frac{A_0}{2} + \sum_{n=1}^\infty A_n e^{-(n \pi)^2 4t} \cos n \pi x$$
but substituting the initial condition we get
$$u(x,0) = 1 = \frac{A_0}{2} + \sum_{n=1}^\infty A_n \cos n \pi x$$
But the Fourier cosine series of $1$ is just $1$, so $A_0 = 2$ and $A_i = 0$ for $i > 0$
Then $$u(x,t) = 1$$
The fact that the solution is just $1$ makes me think that I did something wrong