Question

If y = ae^2x + be^−x, show that, $\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}-2y=0$.

Answer 1

Here,
$$\begin{gathered} y=a{e}^{2x}+b{e}^{-x} \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=2a{e}^{2x}-b{e}^{-x} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^2}=4a{e}^{2x}+b{e}^{-x} \\ \Rightarrow \frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^2}=2a{e}^{2x}-b{e}^{-x}+2\left(a{e}^{2x}+b{e}^{-x}\right) \\ \Rightarrow \frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^2}=\frac{\mathit{d}y}{\mathit{d}x}+2y \\ \Rightarrow \frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^2}-\frac{\mathit{d}y}{\mathit{d}x}-2y=0 \end{gathered}$$

Hence proved.