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For boundary value problem \begin{align} &X''+\lambda X =0\\[3pt] &(X'-\alpha X)|_{x=0}=0,\qquad (X'+\beta X)(l)=0 \end{align} let us find positive eigenvalues $\lambda=k^2$ ($k>0$) first. We have $$ X= A\cos(k x) +B\sin (k x). $$ Then \begin{array}{rcr} -\alpha A& +& k B=0,\\ (\beta-k\tan (k l)) A& +& (k+\beta \tan(kl))B=0 \end{array} and calculating determinant and taking it equal to $0$ we get \begin{equation} \tan (kl)=\frac{(\alpha+\beta)k}{k^2-\alpha\beta}. \end{equation}
Solve it graphically:
$\alpha=\beta$, varies
$\alpha=-.5$, $\beta$ varies
Let us find negative eigenvalues $\lambda=-k^2$ ($k>0$). We have \begin{array}{rcr} -\alpha A& +& k B=0,\\ (\beta+k\tanh (k l)) A& +& (k+\beta \tanh(kl))B=0 \end{array} and calculating determinant and taking it equal to $0$ we get \begin{equation} \tanh (kl)=-\frac{(\alpha+\beta)k}{k^2+\alpha\beta}. \end{equation}
Solve it graphically:
$\alpha=\beta$, varies
$\alpha=-1$, $\beta$ varies
$\alpha=-2$, $\beta$ varies
$\alpha=-3$, $\beta$ varies