If $y = {e^{ - x}}(A\cos x + B\sin x)$, then $y$ is a Solution f

Given that, $y = {e^{ - x}}(A\cos x + B\sin x)$

On differentiating both sides w.r.t.,

$x$ we get $\frac{{dy}}{{dx}} = - {e^{ - x}}(A\cos x + B\sin x) + {e^{ - x}}( - A\sin x + B\cos x)$

$\frac{{dy}}{{dx}} = - y + {e^{ - x}}( - A\sin x + B\cos x)$

Again, differentiating both sides w.r.t. $x$, we get

$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - dy}}{{dx}} + {e^{ - x}}( - \cos x - B\sin x) - {e^{ - x}}( - A\sin x + B\cos x)$

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{dy}}{{dx}} - y - \left[ {\frac{{dy}}{{dx}} + y} \right]$

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{dy}}{{dx}} - y - \frac{{dy}}{{dx}} - y$

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} = - 2\frac{{dy}}{{dx}} - 2y$
$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} + 2y = 0$