$y = {e^x}(a\cos x + b\sin x)$

.: We have, $y = {e^x}(a\cos x + b\sin x)$

…(1)
Dividing both sides by ${e^x}$,

we get
${e^{ - x}}y = a\cos x + b\sin x$ …(2)

Differentiating twice w.r.t. $x$,

we get
${e^{ - x}}{y_1} + y{e^{ - x}}( - 1) = - a\sin x + b\cos x$,

${e^{ - x}}{y_2} + {y_1}{e^{ - x}}( - 1) - {e^{ - x}}{y_1} + y{e^{ - x}} = - a\cos x - b\sin x$

$\Rightarrow {e^{ - x}}{y_2} - 2{y_1}{e^{ - x}} + y{e^{ - x}} = - (y{e^{ - x}})$

(using (2))
$\Rightarrow {e^{ - x}}({y_2} - 2{y_1} + 2y) = 0$

$\Rightarrow {y_2} - 2{y_1} + 2y = 0$,

which is the required differential equation.