Question

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

Answer 1

We have
\[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0 . . . . . \left( 1 \right)\]
\[Now, \]
\[y = e^{- 3x} \]
\[ \Rightarrow \frac{dy}{dx} = - 3 e^{- 3x} \]
\[\Rightarrow\frac{d^2y}{dx^2}=9e^{-3x}\]

\[\text{Putting the values of }\frac{d^2 y}{d x^2}, \frac{dy}{dx}\text{ and y in (1), we get}\]

\[LHS = 9 e^{- 3x} - 3 e^{- 3x} - 6 e^{- 3x} \]
\[ = 0\]
\[ = RHS\]

Thus, y = e^−3x is the solution of the given differential equation.