y" + y = x + cos x

It's auxiliary equation is

m² + 1 = 0

m = \(\pm i\)

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

Let y₁ = cos x & y₂ = sin x are two solution of given differential equation

R = x + cos x

W(y₁, y₂) = \(\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}\) = \(\begin{vmatrix}cosx&sin x\\-sin x&cosx\end{vmatrix}\)

 = cos²x + sin²x = 1

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

\(=\int\frac{\begin{vmatrix}0&sin x\\x+cos x&cos x\end{vmatrix}}{1}dx\) 

\(= \int(x sin x + sin x cos x)dx\)

\(=-\int xsin xdx-\frac12\int sin 2xdx\) 

\(=-[x\int sin x-\int(\frac{dx}{dx}\int sin xdx)dx)+\frac{cos 2x}4\) 

 = x cos x + \(\int-cos x dx+\frac{cos 2x}4\)

 = x cos x - sin x + \(\frac{cos 2x}4\)

u₂ = \(\int\frac{\begin{vmatrix}cos x&0\\-sin x&x+cos x\end{vmatrix}dx}{1}\)

\(=\int(xcosx+cos^2x)dx\) 

= \(\int\) x cos x dx + \(\int\)cos²x dx

 = x sin x - \(\int\) sin x dx + \(\int\) \(\frac{1+cos 2x}2dx\)

 = x sin x + cos x + x/2 + \(\frac{sin 2x}4\) 

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

y = u₁y₁ + u₂y₂ + C. F.

 = C₁ cos x + C₂ sin x + (x cos x - sin x + \(\frac{cos 2x}4\)) cos x +

 (x sin x + cos x + x/2 + \(\frac{sin 2x}4\)) sin x

 = C₁ cos x + C₂ sin x + x cos²x - sin x cos x + \(\frac{cos xcos 2x}4\) + x sin²x + sin x cos x + \(\frac{xsin x}2+\frac{sinx sin2x}4\)

 = C₁ cos x + C₂ sin x + x + \(\frac{xsin x}2+\frac14(sin x sin 2x + cosx cos2x)\)