Question

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

$$\begin{gathered} \frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,} \\ \mathbf{wheref}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{⩽}\mathbf{x}\mathbf{⩽}\mathbf{1} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{x}\mathbf{>}\mathbf{1} \end{gathered}$$

Short answer

So, the required solution is$\mathrm{y}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\begin{array}{cc}1& 0\le \mathrm{x}\le 1\\ -1+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{x}},& \mathrm{x}>1\end{array}\\ \end{array}\right.$. So, the graph of a continuous function is shown below:

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.

Given Data

Consider the following initial value problem,

$\frac{dy}{dx}+y=f\left(x\right),y\left(0\right)=1$

Here, the function$f\left(x\right)$is defined as below.

$$\begin{gathered} f\left(x\right)=\{1,0⩽x⩽1 \\ -1,x>1 \end{gathered}$$

The objective is to solve the initial value problem and to draw the graph of the continuous function y(x) .

Evaluation

The differential equation $\frac{dy}{dx}+y=f\left(x\right)$ is in the standard form of a linear equation.

The standard from of a linear equation is $\frac{dy}{dx}+P\left(x\right)y=f\left(x\right)$.

This implies p(x) = 0.

Determine the integrating factor.

$$\begin{gathered} \mu \left(x\right)=e^{\int p\left(x\right)dx} \\ =e^{\int \left(1\right)dx} \\ =e^x \end{gathered}$$

Multiply the given differential equation with the integrating factor

$$\begin{gathered} e^x\left(\frac{dy}{dx}+y\right)=e^x·f\left(x\right) \\ {e}^x·\left(\frac{dy}{dx}\right)+\left(e^x\right)·y=e^x·f\left(x\right) \\ \frac{d}{dx}\left(e^{x}y\right)=e^x·f\left(x\right) \end{gathered}$$

Since the function$f\left(x\right)$is a piecewise function.

The above differential equation can be represented as follows:

$\frac{d}{dx}\left(e^{2x}y\right)=\left\{\begin{array}{l}\begin{array}{ll}{e}^x,& 0\le x\le 1,\\ -e^x,& x>1\end{array}\\ \end{array}\right.\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Since,}f\left(x\right)=\{\left\{\begin{array}{ll}\begin{array}{ll}1,& 0\le x\le 1\\ -1,& x>1\end{array}& \end{array}\right.$

Integrate both sides

Integrate on both sides of the above equation.

$$\begin{gathered} e^{x}y=\{e^x+c_1,0⩽x⩽1, \\ -e^x+c_2,x>1 \end{gathered}$$

$$\begin{gathered} y=\{1+c_1{e}^{-x},0⩽x⩽1,Di\text{vide both sides by}{e}^x \\ -1+c_2{e}^{-x},x>1 \end{gathered}$$

Apply initial condition $y\left(0\right)=1$to solution $y=1+c_1{e}^{-x}$.

$$\begin{gathered} 1=1+c_1\left(e^0\right) \\ {c}_1=0 \end{gathered}$$

So, the value of y(x) is $$\begin{gathered} e^{x}y=\{e^x+c_1,0⩽x⩽1 \\ -1+c_2{e}^{-x},x>1 \end{gathered}$$

Find continuous function

The function, $y\left(x\right)$is continuous at $\underset{x\to 1^2}{lim}y\left(x\right)=y\left(1\right)$ which gives .

It implies that

$\begin{array}{rcl}-1+c_2{e}^{-1}& =& 1\\ c_2{e}^{-1}& =& 2\\ c_2& =& 2e\text{Multiply bothsides by}e\end{array}$

Therefore, the general solution of the given initial value problem is,

$$\begin{gathered} y\left(x\right)=\left\{\begin{array}{l}10⩽x⩽1, \\ -1+2e·e^{-x},x>1\\ \end{array}\right. \\ y\left(x\right)=\left\{\begin{array}{l}10⩽x⩽1, \\ -1+2e^{1-x},x>1\\ \end{array}\right. \end{gathered}$$

So, the required sketch of the continuous function $y\left(x\right)$ is shown as below: