Given different equation is 

(D² + 4)y = 4 sec² 2x

It,s auxiliarly equation is

m² + 4 = 0

= m = \(\pm\) 2i

\(\therefore\) C . F = c₁ cos 2x + c₂ sin 2x

Let y₁ = cos 2x and y₂ = sin 2x

R = 4 sec² 2x

W(y₁,y₂) = \(\begin{vmatrix} y_1 & y_2 \\[0.3em] y_1^1 & y_2^1 \\[0.3em] \end{vmatrix}\)

= \(\begin{vmatrix} cos2x & sin2x \\[0.3em] -2sin2x & 2cos2x \\[0.3em] \end{vmatrix}\)

= 2cos² 2x + 2sin² 2x

= 2(sin² 2x + cos² 2x)

= 2

\(\therefore\) u₁ = \(\int \frac{\begin{vmatrix} 0& sin2x \\[0.3em] 4sec^2 2x & 2cos2x \\[0.3em] \end{vmatrix}}{w(y_1,y_2)}dx\) 

= \(\int \frac{\begin{vmatrix} -4\frac{sin2x}{cos2x}.sec2x \\[0.3em] \end{vmatrix}}{2}dx\)

= -2 \(\int\) sec 2x . tan 2x dx 

= - 2 \(\cfrac{sec\,2x}2\) = - sec 2x

u₂ = \(\int \frac{\begin{vmatrix} cos2x&0 \\[0.3em] -2sin2x & 4sec^22x \\[0.3em] \end{vmatrix}}{w(y_1,y_2)}dx\)

= \(\int\cfrac{4cos2x.sec^2\,2x}2 dx\)

= 2 \(\int\) sec 2x dx

= 2 \(\cfrac{log|sec2x+tan2x|}2\)

= log |sec 2x x tan 2x |

P . I = u₁y₁ + u₂y₂

= - sec2x . cos 2x + log |sec 2x + tan 2x| sin 2x

= - 1+ log |sec2x + tan 2x| sin2x

\(\therefore\) complete solution of given differential equation is 

y = C . F . + P . I .

= c₁ cos 2x + c₂ sin 2x - 1 + log |sec 2x + tan 2x| sin 2x