Verify that the function y=e^{-3x} is a solution to the

Step 1
To verify the function ${\displaystyle y=e^{-3x}}$ is a solution of the differential equations ${\displaystyle \frac{d^{2}y}{{dx}^2}+\frac{dy}{dx}-6y=0}$.
That is the function ${\displaystyle y=e^{-3x}}$ satisfies the equation ${\displaystyle \frac{d^{2}y}{{dx}^2}+\frac{dy}{dx}-6y=0}$ ...(1)
Step 2
Now, ${\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(e^{-3x}\right)=e^{-3x}(-3)=-3e^{-3x}}$
${\displaystyle \frac{d^{2}y}{{dx}^2}=\frac{d}{dx}-3\left(e^{-3x}\right)=(-3)\frac{d}{dx}\left(e^{-3x}\right)=-3\left(e^{-3x}(-3)\right)=9e^{-3x}}$
${\displaystyle -6y=-6e^{-3x}}$
Therefore, ${\displaystyle \frac{d^{2}y}{{dx}^2}+\frac{dy}{dx}-6y}$
${\displaystyle =9e^{-3x}-3e^{-3x}-6e^{-3x}}$
${\displaystyle =9e^{-3x}-9e^{-3x}}$
${\displaystyle =0}$
Hence, equation (1) satisfies.
Therefore, the given function ${\displaystyle y=e^{-3x}}$ is a solution to the given differential equation.