Question

Why is \[y=e^{-3x}\] a solution of \[\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}-6y=0\]?

Options

• Because the relation contains two arbitrary constants

• Because \[9e^{-3x}-3e^{-3x}-6e^{-3x}=9e^{-3x}-9e^{-3x}=0\]

• Because \[9e^{-3x}-3e^{-3x}-6e^{-3x}=6e^{-3x}\]

• Because \[\frac{dy}{dx}=9e^{-3x}\]

Answer 1

Substitution makes the left-hand side equal to zero: \[9e^{-3x}-3e^{-3x}-6e^{-3x}=0\]. Hence the function and its derivatives satisfy the differential equation.