Verify that the function $y=e^{-3x}$ is a solution of the differential equation $\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}-6y=0$.

(N/A) Given function is $y=e^{-3x}$.
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = -3e^{-3x}$ ... $(1)$
Now,differentiating $(1)$ with respect to $x$ again,we get:
$\frac{d^{2}y}{dx^{2}} = 9e^{-3x}$
Substituting the values of $\frac{d^{2}y}{dx^{2}}$,$\frac{dy}{dx}$,and $y$ into the left-hand side $(L.H.S.)$ of the given differential equation:
$L.H.S. = \frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} - 6y$
$L.H.S. = 9e^{-3x} + (-3e^{-3x}) - 6(e^{-3x})$
$L.H.S. = 9e^{-3x} - 3e^{-3x} - 6e^{-3x}$
$L.H.S. = 9e^{-3x} - 9e^{-3x} = 0$
Since $L.H.S. = R.H.S. = 0$,the function $y=e^{-3x}$ is a solution of the given differential equation.