Question

Suppose that ${\mathit{y}}_{\mathbf{1}}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathbf{}\mathbf{and}\mathbf{}{\mathit{y}}_{\mathbf{2}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$are two solutions of a homogeneous linear differential equation. Explain why${\mathit{y}}_{\mathbf{3}}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathbf{}\mathit{x}\mathbf{}\mathbf{and}\mathbf{}{\mathit{y}}_{\mathbf{4}}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{}\mathit{x}$ are also solutions of the equation.

Short answer

Proved.

Step-by-step solution

Derive the homogeneous differential equation

We have the two solutions of a homogeneous differential equation as$y_1=e^{x}and{y}_2=e^{-x}$ , and we have to show that $y_3=\cosh xand{y}_4=\sinh x$are also solutions for this differential equation as the following:

Since we have$y_1=e^{x}and{y}_2=e^{-x}$, then we can have the solution of the given D.E as

$y=c_1{e}^x+c_2{e}^{-x}······\left(1\right)$

Where $c_{1}and{c}_2$ are arbitrary constants

Also, since we have $y=c_3\cosh x+c_4\sinh x$ then we can have

$y=c_3\cosh x+c_4\sinh x$

After that, we can simplify this function as

$\begin{array}{rcl}y& =& c_3\left(\frac{e^x+e^{-x}}{2}\right)+c_4\left(\frac{e^x-e^{-x}}{2}\right)\\ & =& \frac{c_3}{2}{e}^x+\frac{c_3}{2}{e}^{-x}+\frac{c_4}{2}{e}^x-\frac{c_4}{2}{e}^{-x}\\ & =& \left(\frac{c_3+c_4}{2}\right)e^x+\left(\frac{c_3-c_4}{2}\right)e^{-x}\\ & =& c_1{e}^x+c_2{e}^{-x}······\left(2\right)\end{array}$

Where$c_1=\frac{c_3+c_4}{2}and{c}_2=\frac{c_3-c_4}{2}$

Then from equations (1) and (2), we find that $y_3=\cosh xand{y}_4=\sinh x$are a two solutions of the given differential equation.