Question

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

where$$\begin{gathered} \frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{} \\ \mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{\le }\mathit{x}\mathbf{\le }\mathbf{2} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\frac{\mathbf{2}}{\mathbf{x}}\mathbf{,}\mathbf{}\mathbf{x}\mathbf{>}\mathbf{1} \end{gathered}$$

Short answer

So, the required solution of the given initial value problem is

$$\begin{gathered} y\left(x\right)=\{4e^{-2x}+2x-1,0\le x\le 1 \\ {x}^2\left(4e^{-2}+1\right)+4x^2\ln |x|,x>1 \end{gathered}$$

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.

Given Data

The interval [01] is considered for which p(x) = 2.

So, the given differential equation, can be written as

$\frac{\mathrm{d}y}{\mathrm{d}x}+2y=4x$

Linear differential equation of the first order

$y^{\text{'}}+P\left(x\right)y=f\left(x\right)$

Based on (1) and (2), we obtain

$P\left(x\right)=2\mathrm{and}f\left(x\right)=4x$

Evaluation

Now, substitute the value and in the formula

$y\left(x\right)=e^{-\int P\left(x\right)\mathrm{d}x}\left[c+\int f\left(x\right)e^{\int P\left(x\right)\mathrm{d}x}\mathrm{d}x\right]$

to get

$y\left(x\right)=e^{-\int 2\mathrm{d}x}\left[c+\int 4x{e}^{\int 2\mathrm{d}x}\mathrm{d}x\right]$

$\begin{array}{l}y\left(x\right)=e^{-2x}\left[c+4\int x{e}^{2x}\mathrm{d}x\right]\\ y\left(x\right)=e^{-2x}\left[c+4·\frac{e^{2x}}{4}(2x-1)\right]\\ y\left(x\right)=e^{-2x}c+2x-1\end{array}$

Find the value of C

Use the initial condition $y\left(0\right)=3$, to get c.

role="math" localid="1664254111783" $$\begin{gathered} \begin{array}{l}y\left(0\right)=e^{0}c-1\left(e^0-1\right)\\ 3=c-1\\ c=4\end{array} \\ Substitutethevaluec=4intheequation\left(3\right). \\ y\left(x\right)=4e^{-2x}+2x-1 \end{gathered}$$

Integrate the function

Now is considering x>1, for which $P\left(x\right)=-\frac{2}{x}$.

So, the given differential equation, can be written as

$\frac{\mathrm{d}y}{\mathrm{d}x}-\frac{2}{x}y=4x$

Linear differential equation of the first order

$y^{\text{'}}+P\left(x\right)y=f\left(x\right)$

Based on (5) and (6), we obtain

$P\left(x\right)=-\frac{2}{x}\mathrm{and}f\left(x\right)=4x$

Substituting values in the formula

Now, substitute the value and in the formula

$y\left(x\right)=e^{-\int P\left(x\right)\mathrm{d}x}\left[c+\int f\left(x\right)e^{\int P\left(x\right)\mathrm{d}x}\mathrm{d}x\right]$

to get

$\begin{array}{l}y\left(x\right)=e^{\int \frac{2}{x}\mathrm{d}x}\left[c+\int 4x{e}^{-\int \frac{2}{x}\mathrm{d}x}\mathrm{d}x\right]\\ y\left(x\right)=e^{2\ln \left|x\right|}\left[c+4\int x{e}^{-2\ln \left|x\right|}\mathrm{d}x\right]\\ y\left(x\right)=e^{\ln |x{|}^2}\left[c+4\int x{e}^{\ln |x{|}^{-2}}\mathrm{d}x\right]\\ y\left(x\right)=x^2\left(c+4\int x·x^{-2}\mathrm{d}x\right)\\ y\left(x\right)=x^2(c+4\ln |x\left|\right)\\ y\left(x\right)=x^{2}c+4x^2\ln |x\end{array}$

Substitution

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

$y\left(1\right)=4e^{-2}+1$

And substitute the value x= 1 in the equation ( 7 ).

$\begin{array}{l}y\left(1\right)=1^2·c+4·1^2·\ln \left|1\right|(\ln 1-0)\\ y\left(1\right)=c\end{array}$

Then, based on (8) and (9), we get

$c=4e^{-2}+1$

Substitute the value in the equation (7).

$y\left(x\right)=x^2\left(4e^{-2}+1\right)+4x^2\ln \left|x\right|$

Finally, we have $$\begin{gathered} y\left(x\right)=\{4e^{-2x}+2x-1,4e^{-2x}+2x-1, \\ {x}^2\left(4e^{-2}+1\right)+4x^2\ln |x|,x>1 \end{gathered}$$