In order to find the engenvalues of the laplacian, this is what I did:
$$\nabla u = -\lambda u, \ (x^2 + y^2 <1)\\u = 0, (x^2 + y^2 =1)$$
In order to solve this problem, I worked with the polar coordinate change of variables:
$$u(r,\theta) = R(r)\Theta(\theta)$$
then the problem becomes $$u_{rr}+\frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} = -\lambda(R(r)\Theta(\theta))$$
which becomes
$$\frac{R''(r)}{R(r)}+\frac{1}{r}\frac{R'(r)}{R(r)}+\frac{1}{r^2}\frac{\Theta''(\theta)}{\Theta(\theta)} = -\lambda$$
Now do
$$\frac{\Theta''}{\Theta} = -\gamma\implies \Theta'' +\gamma\Theta = 0$$ $$R''+\frac{1}{r}R'+(\lambda-\frac{\gamma}{r^2})R=0$$
The characteristic equation for $\Theta''+ \gamma\Theta = 0$ is $p^2 + \gamma=0$ or $p=\pm\sqrt{-\gamma}$. We have a feasible solution only when $\gamma>0$ (WHY?) thus $$\Theta(\theta) = A\cos\sqrt{\gamma}\theta + B\sin\sqrt{\gamma}\theta$$
which implies that $\sqrt{\gamma} = n\in \mathbb{N}$ due to the $2\pi$-periodicity Finally we arrive at $$\Theta(\theta) = \begin{cases}\frac{1}{2}A_0,& n=0\\A_n\cos n\theta + B_n\sin n\theta,& n\in \mathbb{N}\end{cases}$$ for appropriate constants $A_0, A_n, B_n$
Next we solve $\frac{R''(r)}{R(r)}+\frac{1}{r}\frac{R'(r)}{R(r)}+\frac{1}{r^2}\frac{\Theta''(\theta)}{\Theta(\theta)} = -\lambda$ for $0\le r < 1$. We impose that at the origin, $R(0)$ is finite. Also, the Dirichlet Boundary conditions require $R(1)=0$. We know that the Dirichlet-Laplacian eigenvalues are positive, so $\lambda >0$
Now let us use the change of variable: $\rho = \sqrt{\lambda} r$ which results in $R_r = R_{\rho}\frac{d\rho}{dr} = \sqrt{\lambda}R_{\rho}, R_{rr} = \lambda R_{\rho\rho}$
Now the equation can be rewritten as $$R_{\rho\rho}+\frac{1}{\rho}R_{\rho} + (1-\frac{n^2}{\rho^2})R=0$$
which is the Bessel Equation which has solution $R(\rho) = J_n(\rho)$ where
$$J_n(\rho) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!(n+k)!}\left(\frac{\rho}{2}\right)^{n+2k}$$
So for $n$ in general $$u(r,\theta) = R(\frac{\rho}{\sqrt{\lambda}})\Theta(\theta) = J_n(\frac{\rho}{\sqrt{\lambda}})(A_n\cos n\theta + B_n\sin n\theta)$$
and for the case $n=0$:
$$u(r,\theta) = R(\frac{\rho}{\sqrt{\lambda}})\Theta(\theta) = J_0(\frac{\rho}{\sqrt{\lambda}})(\frac{1}{2}A_0)$$
Is this solution right? It seems kinda different from this solution in the page 5 where it shows the eigenvalues and eigenvectors
Your solution is correct up to the line where you introduce the Bessel functions. There you should have written $$ R(\fracρ{\sqrtλ})=J_n(ρ)\iff R(r)=J_n(\sqrtλr). $$ For greater clarity start earlier, when substituting $r$ with $ρ$ do not reuse the same function name for two different functions, write $R(r)=\tilde R(ρ)$ where $ρ=\sqrtλr$, $R_r=\sqrtλ\tilde R_ρ$ etc., so that the Bessel equation is in $\tilde R$ and its derivatives. Then its solution is $\tilde R(ρ)=J_n(ρ).$ With that you get a clean back substitution $$R(r)=\tilde R(ρ)=\tilde R(\sqrtλr)=J_n(\sqrtλr).$$
And as $R(1)=0$ we need that $\sqrtλ$ is one of the roots of $J_n$.