So, when we have this problem we can see easily that general solution

$$

X= A\cosh (cx)+ B\cos (cx) +C\sinh (cx) +D\sin (cx)

$$

satisfies conditions at $0$ as $A=B=0$ as

$$

c^{-2}X''= A\cosh (cx)- B\cos (cx) +C\sinh (cx) -D\sinh (cx)

$$

and therefore

$$

X=C\sinh (cx) +D\sinh (cx)

$$

satisfies conditions at $0$ as $A=B=0$ as

$$

c^{-2}X''= C\sinh (cx) -D\sin (cx).

$$

To satisfy conditions on the right end we need

\begin{align*}

& C\sinh (cl) +D\sin (cl) =0,\\

& C\sinh (cl) -D\sin (cl) =0

\end{align*}

which is possible for $(X,D)\ne 0$ iff $C=0$, $\sin (cl)=0$ i.e. we get $\lambda_n= n^2 \pi^2/l^2$, $X_n=\sin (n\pi x/l)$.