Question

Proceed as in Example 6 to solve the given initial-value problem. Use a graphing utility to graph the continuous function y(x).

$\begin{array}{l}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{}\\ \mathbf{where}\mathbf{}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{}\mathbf{\ge }\mathbf{x}\mathbf{}\mathbf{\ge }\mathbf{2}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{5}\mathbf{,}\mathbf{}\mathbf{x}\mathbf{>}\mathbf{2}\end{array}$

Short answer

So, the required solution of the given initial value problem is$$\begin{gathered} y\left(x\right)=\{4e^{-x},0\le x\le 2 \\ 4e^{-5x+8},x>2 \end{gathered}$$

Step-by-step solution

Definition of Continuous function

Continuous functions are functions that have no restrictions throughout their domain or a given interval. Their graphs won't contain any asymptotes or signs of discontinuities as well.

Evaluation 

The interval is considered for which p(x) = 1 .

So, the given differential equation, can be written as

$\begin{array}{l}\frac{\mathrm{d}y}{\mathrm{d}x}+y=0\\ \frac{\mathrm{d}y}{\mathrm{d}x}=-y\end{array}$

Separate the variables x and y in the differential equation

$\frac{\mathrm{d}y}{y}=-\mathrm{d}x$

Integrating the equation

Take integration on both sides

$\begin{array}{l}\int \frac{\mathrm{d}y}{y}=-\int \mathrm{d}x\\ y=e^{-x+c}\left(e^c-c_1\right)\\ y=e^{-x}{c}_1\end{array}$

Use the initial condition y(0) =4, to get c₁

$\begin{array}{l}y\left(0\right)=e^0{c}_1\left(e^0-1\right)\\ 4=c_1\end{array}$

Substitute the value c₁=4 in the equation (1).

$y\left(x\right)=4e^{-x}$

Integrate the function

Now is considering $x>2$, for which $P\left(x\right)=5$So, the given differential equation, can be written as

$\begin{array}{l}\frac{\mathrm{d}y}{\mathrm{d}x}+5y=0\\ \frac{\mathrm{d}y}{\mathrm{d}x}=-5y\end{array}$

Separate the variables and in the differential equation

$\frac{\mathrm{d}y}{y}=-5\mathrm{d}x$

Take integration on both sides

$\begin{array}{l}\int \frac{\mathrm{d}y}{y}=-5\int \mathrm{d}x\\ y=e^{-5x+c}\left(e^c-c_1\right)\\ y=e^{-5x}{c}_1\end{array}$

Substitution

Substitute the value x =2 in (2) and (3).

$\begin{array}{l}y\left(2\right)=4e^{-2}\\ y\left(2\right)=e^{-10}{c}_1\end{array}$

Then,

$\begin{array}{l}4e^{-2}=e^{-10}{c}_1\\ c_1=4e^{-2}{e}^{10}{c}_1\\ =4e^8\end{array}$

Substitute the value $c_1=4e^8$ in the equation (3).

$y\left(x\right)=4e^{-5x+8}$

Finally, we have

$$\begin{gathered} y\left(x\right)=\{4e^{-x},0\le x\le 2 \\ 4e^{-5x+8},x>2 \end{gathered}$$.