5. Solve the equation $x^{2}-x y-6 y^{2}+2 x+19 y=18$ in integers.

Rewrite the equation as

$$
x^{2}-x(y-2)-6 y^{2}+19 y+k=18+k
$$

and we will choose the number $k$ so that the discriminant of the quadratic trinomial on the left side of the equation becomes a perfect square. We have

$$
D=25 y^{2}-80 y+4-4 k
$$

To obtain a perfect square, we need $k=-15$. Now we have the opportunity to factor the left side of our equation

$$
x_{1,2}=\frac{y-2 \pm(5 y-8)}{2} \Rightarrow(x-3 y+5)(x+2 y-3)=3
$$

The factors on the left side are integers, so we get four systems

$$
\left\{\begin{array}{l}
x-3 y+5=1 \\
x+2 y-3=3
\end{array},\left\{\begin{array}{l}
x-3 y+5=-1 \\
x+2 y-3=-3
\end{array},\left\{\begin{array}{l}
x-3 y+5=3 \\
x+2 y-3=1
\end{array},\left\{\begin{array}{l}
x-3 y+5=-3 \\
x+2 y-3=-1
\end{array}\right.\right.\right.\right.
$$

Solve these systems and we get the answer.
Answer: $(2 ; 2),(-2 ; 2)$.