Question

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

$$\begin{gathered} \left(\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\right)\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \mathbf{where} \\ \mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\mathit{x}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{⩽}\mathbf{x}\mathbf{<}\mathbf{1} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathit{x}\mathbf{,}\mathbf{x}\mathbf{⩾}\mathbf{1} \end{gathered}$$

Short answer

So, the required solution is $$\begin{gathered} y\left(x\right)=\{\frac{x^2}{2\left(1+x^2\right)}0⩽x<1 \\ -\frac{x^2}{2\left(1+x^2\right)}+\frac{1}{1+x^2}x⩾1 \end{gathered}$$. 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 initial value problem,

$\left(1+x^2\right)\frac{dy}{dx}+\left(2x\right)y=f\left(x\right),y\left(0\right)=0$

Here,

$$\begin{gathered} f\left(x\right)=\{00⩽x<1 \\ -xx⩾1 \end{gathered}$$

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

Solve the initial value problem

Rewrite the differential equation as follows:

$\left(1+x^2\right)\frac{dy}{dx}+\left(2x\right)y=f\left(x\right)$

Divide on the both sides with respect to $\left(1+x^2\right)$.

$\frac{dy}{dx}+\frac{\left(2x\right)}{\left(1+x^2\right)}y=\frac{f\left(x\right)}{\left(1+x^2\right)}\dots \dots \left(1\right)$

The standard form of the linear differential equitation is $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$.

Comparison

Compare the differential equation (1) with $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$.

Find the integrating factor I.F as follows:

$$\begin{gathered} I·F=e^{\int pxdx} \\ =e^{\int \frac{2x}{1+x^2}dx} \\ =e^{\ln |1+x^2\mid } \\ =1+x^2 \end{gathered}$$

Finding Integrated factor

Multiplied the differential equation (1), with the integrating factor 1+x² .

$$\begin{gathered} \left(1+x^2\right)·\left(\frac{dy}{dx}+\frac{\left(2x\right)}{\left(1+x^2\right)}y\right)=\frac{f\left(x\right)}{\left(1+x^2\right)}· \\ \left(1+x^2\right)\left(1+x^2\right)\frac{dy}{dx}+2xy=f\left(x\right) \\ \frac{d}{dx}\left(\left(1+x^2\right)y\right)=f\left(x\right) \\ \text{Since}\frac{d}{dx}\left(\left(1+x^2\right)y\right)=\left(1+x^2\right)\frac{dy}{dx}+2xy \end{gathered}$$

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

That implies,

role="math" localid="1664196815777" $$\begin{gathered} \frac{d}{dx}\left(\left(1+x^2\right)y\right)=f\left(x\right)=\{x,0⩽x<1 \\ -x,x⩾1 \end{gathered}$$

Integrating the function

Integrate on the both sides,

$$\begin{gathered} \int \frac{d}{dx}\left(\left(1+x^2\right)y\right)\{\frac{x^2}{2}+c_1,0⩽x<1 \\ -\frac{x^2}{2}+c_2,x⩾1 \\ \left(1+x^2\right)y=\frac{{\displaystyle x^2}}{{\displaystyle 2}}+c_1,0⩽x<1 \\ -\frac{{\displaystyle x^2}}{{\displaystyle 2}}+c_2,x⩾1 \end{gathered}$$

Substitute the initial condition y(0) in the above result.

$$\begin{gathered} y\left(0\right)=0\frac{0}{2}+c_1 \\ =0c_1 \\ =0 \end{gathered}$$

Substitution

Substitute C₁=0 in the equation (2)

$$\begin{gathered} y\left(x\right)=\{\frac{x^2}{2\left(1+x^2\right)}0⩽x<1 \\ -\frac{x^2}{2\left(1+x^2\right)}+\frac{c_2}{1+x^2}x⩾1 \end{gathered}$$

Finding general solution

Since, y(x) should be continuous at x=1.

$$\begin{gathered} \underset{x\to 1}{lim}y\left(x\right)=y\left(1\right)\frac{1}{2\left(1+1^2\right)} \\ =-\frac{1}{2(1+1)}+\frac{c_2}{1+1}\frac{1}{4} \\ =1 \end{gathered}$$

Thus, the general solution is

$$\begin{gathered} y\left(x\right)=\{\frac{x^2}{2\left(1+x^2\right)}0⩽x<1 \\ -\frac{x^2}{2\left(1+x^2\right)}+\frac{1}{1+x^2}x⩾1 \end{gathered}$$

Sketch the graph

Sketch the graph of the continuous function as follows: