Explanation

(a)
Since,
$y = e^{2x}(a\cos x + b\sin x)$ …(i)
$\Rightarrow$ $y_{1} = e^{2x}\left( - {asin}x + b\cos x \right) + \left( a\cos x + b\sin x \right)2e^{2x}$
$\Rightarrow$ $y_{1} = e^{2x}\left( - a\sin x + b\cos x \right) + 2y$ …(ii)
$\Rightarrow$ $y_{2} = e^{2x}\left( - a\cos x - b\sin x \right) + \left( - a\sin x + b\cos x \right)e^{2x}2 + 2y_{1}$
= $- y + 2e^{2x}\left( - a\sin x + b\cos x \right) + 2y_{1}$ (using eq.(ii))
$\Rightarrow$ $y_{2} = - y + 2\left( y_{1} - 2y \right) + 2y_{1}$ (using eq.(ii))
$\Rightarrow$ $y_{2} = - y + 4y_{1} - 4y$
$\Rightarrow$ $y_{2} - 4y_{1} + 5y = 0$