Question

Determine the exact intervalof definition by analytical methods. Use a graphing utility to plot the graph of the solution.

${\mathit{e}}^{\mathbf{x}}\mathit{d}\mathit{x}\mathbf{-}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}$

Short answer

The explicit solution & exact interval are$y=1-\sqrt{x^3+2x^2+2x+4}andl=\left(2,\infty \right)$ respectively.

Step-by-step solution

Definition

A first-order differential equation of the form$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{g}\left(x\right)\mathit{h}\left(y\right)$is said to be separable or to have separable variables.

Separate differential equation

Consider the differential equation$e^{x}dx-e^{-x}dy=0$

Separate the variables by first adding theterm to the right hand side and then diving by e^y and e^-x.

$$\begin{gathered} e^{x}dx-e^{-x}dy=0 \\ {e}^{x}dx=e^{-x}dy \\ \frac{dx}{e^{-x}}=\frac{dy}{e^y}{e}^x \\ dx=e^{-x}dy \end{gathered}$$

Integrating

Now we integrate the left hand side in terms of x and the right hand side in terms of y.

$\begin{array}{rcl}{e}^x+c& =& -e^{-y}\\ -e^x-c& =& e^{-y}\\ \ln \left(-e^x-c\right)& =& \ln \left(e^{-y}\right)\\ \ln \left(-e^x-c\right)& =& -y\\ -\ln \left(-e^x-c\right)& =& y\end{array}$

Apply initial condition

Substitute the initial condition y(0)=0 in order to solve for c.

$\begin{array}{rcl}-\ln \left(-e^0-c\right)& =& 0\\ \ln (-1-c)& =& 0\end{array}$

Recall that In 1 = 0 so,

$\begin{array}{rcl}-1-c& =& 1\\ c& =& -2\end{array}$

The solution becomes$-\ln \left(-e^x+2\right)=y$

Graph

The graph is as given below:

Find interval

In order to find the interval of definition we need so solve where$\ln (-e^x+2)$is defined. Since is only defined for positive values,

$\begin{array}{rcl}-e^x+2& >& 0\\ 2& >& e^x\\ \ln 2& >& \ln e^x\\ \ln 2& >& x\end{array}$

The interval of definition is$(-\infty ,\ln 2)$.