Question

Show that y = ae^2x + be^−x is a solution of the differential equation $\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}-2y=0$.

Answer 1

We have,
$y=a{e}^{2x}+b{e}^{-x}$ ...(1)
Differentiating both sides of equation (1) with respect to $x$, we get
$\frac{dy}{dx}=2a{e}^{2x}-b{e}^{-x}$ ...(2)
Differentiating both sides of equation (2) with respect to $x$, we get
$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=4a{e}^{2x}+b{e}^{-x} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=2a{e}^{2x}-b{e}^{-x}+2a{e}^{2x}+2b{e}^{-x} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\left(2a{e}^{2x}-b{e}^{-x}\right)+2\left(a{e}^{2x}+b{e}^{-x}\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{dy}{dx}+2y\left[\text{Using equations}\left(1\right)\text{and}\left(2\right)\right] \end{gathered}$$
$\Rightarrow$$\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}-2y=0$
Hence, the given function is the solution to the given differential equation.