let the following problem $$ \begin{cases} \dfrac{\partial u}{\partial t}= \dfrac{\partial^2 u}{\partial x^2}, 0<x<1, t>0\\ u(0,t)=0, u(1,t)+ \dfrac{\partial u}{\partial x}(1,t)=0\\ u(x,0)=f(x) \end{cases} $$
I want to calculate the solution with separation of variables method. This is what I try: we put $$u(x,t)= X(x) T(t) \neq 0$$ then we have $$\dfrac{T'(t)}{T(t)}= \dfrac{X''(x)}{X(x)}$$ So we obtain two equations $$ T'(t) + \lambda T(t)= 0 $$ and $$ X''(x)+ \lambda X(x)=0, X(0)=0, X(1)+ X'(1)=0 $$ this second equation admits eigenvalues $\lambda=\alpha^2$ where $\alpha$ is solution of equation $\mathrm{tg\ }x = - \dfrac{1}{x}$, and the eigenfunctions associated are $y_{\alpha}(x)= C \sin(\alpha x)$. The solutions of the first equations are $$T_{\alpha}(t)= e^{-\alpha^2 t}$$ Then we have $$u(x,t)= \sum_{\alpha} C_{\alpha} e^{- \alpha t} \sin(\alpha x)$$ My question is: is my solution correct? And what about $\alpha?$ It must be in $\Bbb Z$. Or not? Thanks for the help.
Apart from a few typos, your derivation looks correct. There should be $e^{-\alpha^2 t}$ in the final equation and I get $x + \tan(x) = 0$ (which is $\cot(x) = -\frac{1}{x}$), as the defining equation for the $\alpha_n$'s. This follows from $$X(x) = \sin(\alpha x) \implies 0 = X(1) + X'(1) = \sin(\alpha) + \alpha \cos(\alpha) \implies \tan(\alpha) + \alpha = 0$$ I would double check this to be sure.
As for allowed values of $\alpha$: The function $\tan(x) + x$ is increasing on $I_n = (-\frac{\pi}{2} + \pi n, \frac{\pi}{2} + \frac{\pi}{2}n)$ for $n\in\mathbb{Z}$ and since $\lim_{x\to (-\frac{\pi}{2} + \pi n)^+}f(x) = -\infty < 0$ and $\lim_{x\to (\frac{\pi}{2} + \pi n)^-}f(x) = \infty > 0$ there is a single zero in the interval $I_n$. Thus there is a countable number of solutions to $\tan(x) + x = 1$ which we label as $\alpha_n$ for $n=1,2,3,\ldots$. The exact value of these numbers is not that important for an analytical treatment of the problem. There is no simple formula for these numbers, but if you want you can easily compute them one by one numerically using your favoritte root-finding method.
We can write the final solution as $$u(x,t) = \sum_{n=1}^\infty C_n e^{-\alpha_n^2 t}\sin(\alpha_n x)$$
The only thing missing now is to express the $C_n$ in terms of the initial condition $f(x) = u(x,0)$. To do so we can first show that the set of functions $\{\sin(\alpha_n x)\}_{n=1}$ is orthogonal on $[0,1]$, i.e. $$\int_0^1\sin(\alpha_n x)\sin(\alpha_m x){\rm d}x = \left\{\matrix{A_n & n= m\\0 & n \not= m}\right.$$ To prove this and derive an expression for $A_n$ simply perform the integral and use $\tan(\alpha_n) + \alpha_n = 1$ to simplify.
Since $u(x,0) = f(x)$ we have $f(x) = \sum_{n=1}^\infty C_n \sin(a_nx)$ and by integrating this over $[0,1]$ we get
$$\int_0^1f(x)\sin(a_m x){\rm d}x = \sum_{n=1}^\infty C_n\int_0^1 \sin(a_nx)\sin(a_mx){\rm d}x$$
Now just use the orthogonality condition above to simplify and you will get an expression for $C_n$ in terms of $\alpha_n$ completing the solution.