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

^{$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}-6y=0.....\left(1\right) \\ \mathrm{Now}, \\ y=e^{-3x} \\ \Rightarrow \frac{dy}{dx}=-3e^{-3x} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=9e^{-3x} \end{gathered}$$}
$\mathrm{Putting}\mathrm{the}\mathrm{values}\mathrm{of}\frac{d^{2}y}{d{x}^2},\frac{dy}{dx}\mathrm{and}y\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}$

$$\begin{gathered} \mathrm{LHS}=9e^{-3x}-3e^{-3x}-6e^{-3x} \\ =0 \\ =\mathrm{RHS} \end{gathered}$$

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