Answer: Start with  $y=\dfrac{e^x}{2} 2cos2xcosx=\dfrac{e^x}{2}[cos3x+cosx]$  [Since $2cosAcosB=cos(A+B)+cos(A-B)$ ]

                               $y= \dfrac{1}{2} e^x cos 3x+\dfrac{1}{2} e^x cos x$  

Let $u=\dfrac{1}{2} e^x cos 3x, v=\dfrac{1}{2} e^x cos x,$then $y_n=u_n+v_n$.........................(1)

We know, if  $y=e^{ax}cosbx,$ then $y_n=(a^2+b^2)^{n/2} e^{ax}cos(bx+n tan^{-1}(b/a))$ (From successive differentiation formula).Thus we calculate $u_n$ and $v_n$ using this concept.

Comparing $u$ with $y=e^{ax}cosbx$,we can see that in $u$, $a=1, b=3$, then $u_n=\dfrac{1}{2}(1^2+3^2)^{n/2} e^xcos(3x+ntan^{-1}3)=\dfrac{1}{2}(10)^{n/2}e^xcos(3x+ntan^{-1}3)$ ........(2) And similarly comparing $v$with $y=e^{ax}cosbx$ we get, $a=1, b=1,$then $v_n=\dfrac{1}{2}(1^2+1^2)^{n/2}e^xcos(x+ntan^{-1}1)=\dfrac{1}{2}(2)^{n/2}e^xcos(x+n{\pi/4}).$......................(3) Now substituting the values of $u_n$ and $v_n$ from (2) and (3) in (1), we get  $y_n={1\over 2}(10)^{n/2} e^{x}cos(3x+n tan^{-1}3)+{1\over 2}(2)^{n/2} e^{x}cos(x+n \pi/4).$           $=\dfrac{e^x}{2}[(10)^{n/2} cos(3x+n tan^{-1}3)+(2)^{n/2} cos(x+n \pi/4)].$             Answer.