Question

In Problems 3–8, determine whether the given function is a solution to the given differential equation.

$\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{2}\mathbf{x}}\mathbf{-}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}$,$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

The given function is a solution to the given differential equation.

Step-by-step solution

Differentiating the given equation w.r.t. (with respect to) x

Firstly, we will differentiate $\mathrm{y}={\mathrm{e}}^{2\mathrm{x}}-3{\mathrm{e}}^{-\mathrm{x}}$with respect to x,

$\frac{\mathrm{dy}}{\mathrm{dx}}=2{\mathrm{e}}^{2\mathrm{x}}+3{\mathrm{e}}^{-\mathrm{x}}$

Again, differentiating the given function with respect to x,

$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=4{\mathrm{e}}^{2\mathrm{x}}-3{\mathrm{e}}^{-\mathrm{x}}$

Simplification

Putting the values from step 1 in the L.H.S. (Left-hand side) of the given differential equation,

$$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-\frac{\mathrm{dy}}{\mathrm{dx}}-2\mathrm{y}=4{\mathrm{e}}^{2\mathrm{x}}-3{\mathrm{e}}^{-\mathrm{x}}-\left(2{\mathrm{e}}^{2\mathrm{x}}+3{\mathrm{e}}^{-\mathrm{x}}\right)-2\left({\mathrm{e}}^{2\mathrm{x}}-3{\mathrm{e}}^{-\mathrm{x}}\right) \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-\frac{\mathrm{dy}}{\mathrm{dx}}-2\mathrm{y}=4{\mathrm{e}}^{2\mathrm{x}}-3{\mathrm{e}}^{-\mathrm{x}}-2{\mathrm{e}}^{2\mathrm{x}}-3{\mathrm{e}}^{-\mathrm{x}}-2{\mathrm{e}}^{2\mathrm{x}}+6{\mathrm{e}}^{-\mathrm{x}} \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-\frac{\mathrm{dy}}{\mathrm{dx}}-2\mathrm{y}=0 \end{gathered}$$

which is the same as the R.HS. (Right-hand side) of the given differential equation.

Hence, $\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{2}\mathbf{x}}\mathbf{-}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}$is a solution to the differential equation$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$.