Given differential equation is

\(\frac{d^2y}{dx^2}+4y=cos\,x\)

It's auxiliary equation is 

m² + 4 = 0

⇒ m = \(\pm2i\)

\(\therefore\) C.F. = C₁ cos2x + C₂ sin 2x

Hence, cos2x and sin 2x are solution of given differential  equation 

Let = y₁ = cos2x ⇒ y'₁ = -2 sin 2x

y₂ = sin 2x ⇒ y'2 = 2 cos 2x

W(y_1, y₂) = \(\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}\) = \(\begin{vmatrix}cos2x&sin2x\\-2sin2x&2cos2x\end{vmatrix}\)

 = 2cos²2x + 2sin²2x

 = 2(cos²2x + sin²2x)

 = 2

u₁ = \(\int\frac{\begin{vmatrix}0&y_2\\R&y_2\end{vmatrix}}{W(y_1,y_2)}dx\) 

= \(\int\frac{\begin{vmatrix}0&sin2x\\cosx&2cos2x\end{vmatrix}}{2}dx\)

 = \(\int-\frac{cosx\,sin2x}2dx\)

 = \(-\frac12\int2sinx\,cos^2xdx\)

 = -1\(\int t^2(-dt)\) (\(\because\) By taking cos x = t ⇒ -sin x dx = dt)

 = t³/3 (\(\because\) t = cos x)

 = \(\frac{cos^3x}3\) 

u₂ = \(\int\frac{\begin{vmatrix}y_1&0\\y'_1&R\end{vmatrix}}{W(y_1,y_2)}dx\) = \(\int\frac{\begin{vmatrix}cos2x&0\\-2sin2x&cosx\end{vmatrix}}{2}dx\)

= \(\frac12\int\) cosx cos2x dx

 = \(\frac12\int\) cosx(1 - 2sin²x)dx

=\(\frac12\)\([\int cosxdx-2\int sin^2x\,cosxdx]\)

 = \(\frac12\)[sin x - \(\frac23\)sin³x]

 = \(\frac{sin x}2-\frac13sin^3x\)

 \(\therefore\) P.I. = u₁y₁ + u₂y₂

 = \(\frac{cos^3x}3cos2x+(\frac{sinx}2-\frac13sin^3x)sin2x\) 

 = \(\frac{cos^3x}3(2cos62x-1)+(\frac{sinx}2-\frac13sin^3x)2sinx cosx\)

 = \(\frac23\)cos⁵x - \(\frac13\)cos³x + sin²x sinx - \(\frac23\)sin⁴x cos x

 = \(\frac23\)cos⁵x -  \(\frac13\)cos³x + (1 - cos²x)cos x - \(\frac23\)(1 - cos²x)² cosx

= \(\frac23\)cos⁵x -  \(\frac13\)cos³x + cos x - cos³x - \(\frac23\) cos x (1 - 2cos²x + cos⁴x)

 =  \(\frac23\)cos⁵x -  \(\frac43\)cos³x + cos x - cos³x - \(\frac23\) cos x + \(\frac43\)cos³x - \(\frac23\)cos⁵x

 = cos x - \(\frac23\) cos x

= \(\frac13\) cos x

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

y = C.F. + P.I.

 = C₁ cos2x + C₂ sin 2x + \(\frac13\) cos x

Alternative:(Without variation of parameter)

P.I. = \(\frac{1}{D^2+4}cosx\)

 = \(\frac{cosx}{-1^2+4}\) = \(\frac{cosx}3\)