Question

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

$$\begin{gathered} \frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e} \\ \mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{\pounds }\mathbf{\times }\mathbf{\pounds }\mathbf{3} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\pounds }\mathbf{>}\mathbf{3}\mathbf{\}} \end{gathered}$$

Short answer

So, the required solution is $$\begin{gathered} y\left(x\right)=\left\{\frac{1}{2}-\frac{1}{2}{e}^{-2x}0⩽x⩽3\right\} \\ \left(\frac{1}{2}{e}^6-\frac{1}{2}\right)e^{-2x}x>3 \end{gathered}$$. 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,

$\frac{dy}{dx}+2y=f\left(x\right),y\left(0\right)=0$

Here, the function f(x) is defined as below.

$f\left(x\right)=\{\underset{0,x>0}{1,0\le x\le 3}$

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

Evaluation

The differential equation $\frac{dy}{dx}+2y=f\left(x\right)$is in the standard form of a linear equation.

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

Determine the integrating factor.

$$\begin{gathered} \mu \left(x\right)=e^{\int p\left(x\right)dx} \\ =e^{\int \left(2\right)dx} \\ =e^{2x} \end{gathered}$$

Multiply the given differential equation with the integrating factor

$$\begin{gathered} e^{2x}\left(\frac{dy}{dx}+2y\right)=e^{2x}·f\left(x\right) \\ {e}^{2x}·\left(\frac{dy}{dx}\right)+\left(2e^{2x}\right)·y=e^{2x}·f\left(x\right) \\ \frac{d}{dx}\left(e^{2x}y\right)=e^{2x}·f\left(x\right) \end{gathered}$$

The above differential equation can be represented as follows:

$$\begin{gathered} \frac{d}{dx}\left(e^{2x}y\right)=\{e^{2x},0⩽x⩽3 \\ 0,x>3 \end{gathered}$$

Integrate both sides

Integrate on both sides of the above equation.

$$\begin{gathered} e^{2x}y=\{\frac{1}{2}{e}^{2x}+c_1,0⩽x⩽3 \\ {c}_2,x>3 \\ y=\{\frac{1}{2}+c_1{e}^{-2x},0⩽x⩽3, \\ {c}_2{e}^{-2x},x>3 \end{gathered}$$

Apply initial condition to y(0) =0 solution $y=\frac{1}{2}+c_1{e}^{-2x}$.

$$\begin{gathered} 0=\frac{1}{2}+c_1\left(e^0\right) \\ {c}_1=-\frac{1}{2} \end{gathered}$$

Find the general solution

The function, y(x) is continuous at x=3 ; which gives $\underset{x\to 3^{+}}{lim}y\left(x\right)=y\left(3\right)$.

It implies that

$$\begin{gathered} c_2{e}^{-6}=\frac{1}{2}-\frac{1}{2}{e}^{-6} \\ {c}_2=\frac{1}{2}{e}^6-\frac{1}{2} \end{gathered}$$

Required sketch of continuous function y(x) is shown as below: