Answer
$$y = 11\sqrt {{x^2} - 16} + C$$
Work Step by Step
$$\eqalign{
& \sqrt {{x^2} - 16} y' = 11x \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \sqrt {{x^2} - 16} \frac{{dy}}{{dx}} = 11x \cr
& {\text{Separate the variables}} \cr
& dy = \frac{{11x}}{{\sqrt {{x^2} - 16} }}dx \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {\frac{{11x}}{{\sqrt {{x^2} - 16} }}} dx \cr
& y = \int {\frac{{11x}}{{\sqrt {{x^2} - 16} }}} dx \cr
& {\text{Let }}u = {x^2} - 16,{\text{ }}du = 2xdx,{\text{ }}dx = \frac{1}{{2x}}du \cr
& {\text{Substituting}} \cr
& y = 11\int {\frac{x}{{\sqrt u }}\left( {\frac{1}{{2x}}} \right)} du \cr
& y = \frac{{11}}{2}\int {\frac{1}{{\sqrt u }}} du \cr
& y = \frac{{11}}{2}\int {{u^{ - 1/2}}} du \cr
& y = \frac{{11}}{2}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& y = 11\sqrt u + C \cr
& y = 11\sqrt {{x^2} - 16} + C \cr} $$
