91Ó°ÊÓ

• Write the equation in the standard form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.
• Calculate the integrating factor by $\mathbf{μ}\left(x\right)$the formula .
• Multiply the equation in standard form by $\mathbf{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and, recalling that the left-hand side is just $\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}$, obtain$$\begin{gathered} \mathit{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathit{P}\mathit{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{y}\mathbf{=}\mathit{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathit{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \end{gathered}$$

Given that,

$\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{xy}=1,\mathrm{y}\left(2\right)=1$

To prove it can be expressed as $\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{-{\mathrm{x}}^2}({\mathrm{e}}^4+{∫}_2^{\mathrm{x}}{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{dt})$.

Calculate the integrating factor of $\mathrm{μ}\left(x\right)$.

Where $\mathrm{P}(x)=2x$.

Then,

$\begin{array}{rcl}\mathrm{μ}\left(x\right)& =& e^{{∫}_{}^{}P\left(x\right)dx}\\ & =& e^{{∫}_{}^{}2xdx}\\ & =& e^{x^2}\end{array}$

Multiply $\mathrm{μ}\left(x\right)$in equation (1)

role="math" localid="1664183318615" $$\begin{gathered} e^{x^2}\frac{dy}{dx}+2e^{x^2}xy=e^{x^2} \\ \frac{d}{dx}\left[e^{x^2}y\right]=e^{x^2} \\ \frac{d}{dt}\left[e^{t^2}y\right]=e^{t^2} \end{gathered}$$

Integrating both sides. And the initial value of x is starts at 2, where y = 1. So, the limits of integration would be from 2 to x.

role="math" localid="1664183563908" $$\begin{gathered} {∫}_{}^{}\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{y}\right]\mathrm{dt}={∫}_{}^{}{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{dt} \\ {\overline{){\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{y}}}_2^{\mathrm{x}}={∫}_2^{\mathrm{x}}{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{dt} \\ {\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{y}-{\mathrm{e}}^{2^2}={∫}_2^{\mathrm{x}}{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{dt} \\ {\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{y}-{\mathrm{e}}^4={∫}_2^{\mathrm{x}}{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{dt} \\ {\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{y}={\mathrm{e}}^4+{∫}_2^{\mathrm{x}}{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{dt} \\ \mathrm{y}={\mathrm{e}}^{-{\mathrm{x}}^2}\left({\mathrm{e}}^4+{∫}_2^{\mathrm{x}}{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{dt}\right) \end{gathered}$$

Hence proved. Because the solution is expressed in the form of

$\mathrm{y}={\mathrm{e}}^{-{\mathrm{x}}^2}\left({\mathrm{e}}^4+{∫}_2^{\mathrm{x}}{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{dt}\right)......\left(2\right)$

Hence it is shown that the solution to the initial value problem $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}\mathbf{1}$can be expressed as $\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}\mathbf{(}{\mathbf{e}}^{\mathbf{4}}\mathbf{+}{\mathbf{∫}}_{\mathbf{2}}^{\mathbf{x}}{\mathbf{e}}^{{\mathbf{t}}^{\mathbf{2}}}\mathbf{dt}\mathbf{)}$.

Referring to part a: solution is $\mathrm{y}={\mathrm{e}}^{-{\mathrm{x}}^2}\left({\mathrm{e}}^4+{∫}_2^{\mathrm{x}}{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{dt}\right)$.

To approximate the solution at x = 3.

Since the Simpson’s rule is shown below.

role="math" localid="1664183811884" ${∫}_a^b\mathrm{f}\left(\mathrm{x}\right)\mathrm{x}\text{'}≈\frac{∆\mathrm{x}}{3}(\mathrm{f}\left({\mathrm{x}}_0\right)+4\mathrm{f}\left({\mathrm{x}}_1\right)+2\mathrm{f}\left({\mathrm{x}}_2\right)+.......4\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}-1}\right)+\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}}\right)).......\left(3\right)$

Then, using 4 intervals,

$\begin{array}{c}∆\mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ =\frac{3-2}{4}\\ =\frac{1}{4}\end{array}$

Substitute the value in equation (3)

$\begin{array}{rcl}{∫}_2^2{e}^{t^2}dt& ≈& \frac{1}{12}\left(e^4+2\underset{n=1}{\overset{2-1}{∑}}{e^{\frac{1}{4}\left(4+n\right)}}^2+\underset{n=1}{\overset{2}{4∑}}{e^{\frac{1}{16}\left(7+2n\right)}}^2+e^9\right)\\ & ≈& \frac{1}{12}\left(e^4+2\underset{n=1}{\overset{1}{∑}}{e^{\frac{1}{4}\left(4+n\right)}}^2+\underset{n=1}{\overset{2}{4∑}}{e^{\frac{1}{16}\left(7+2n\right)}}^2+e^9\right)\\ & ≈& 1460.35435\end{array}$

Substitute the value in equation (3).

$\begin{array}{rcl}y\left(3\right)& =& e^{-3^2}\left(e^4+1460.35435\right)\\ & ≈& 0.186960\end{array}$

Using 6 intervals,

$\begin{array}{c}∆\mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ =\frac{3-2}{6}\\ =\frac{1}{6}\end{array}$

Substitute the value in equation (3)

$\begin{array}{rcl}{∫}_2^2{e}^{t^2}dt& ≈& \frac{1}{36}\left(e^4+2\underset{n=1}{\overset{6-1}{∑}}{e^{\frac{1}{4}\left(2+\frac{n}{6}\right)}}^2+\underset{n=1}{\overset{6}{4∑}}{e^{\frac{1}{144}\left(23+2n\right)}}^2+e^9\right)\\ & ≈& \frac{1}{36}\left(e^4+2\underset{n=1}{\overset{5}{∑}}{e^{\frac{1}{4}\left(2+\frac{n}{6}\right)}}^2+\underset{n=1}{\overset{6}{4∑}}{e^{\frac{1}{144}\left(23+2n\right)}}^2+e^9\right)\\ & ≈& 1428.61685\end{array}$

Substitute the value in equation (2).

$\begin{array}{rcl}y\left(3\right)& =& e^{-3^2}\left(e^4+1428.68685\right)\\ & ≈& 0.1803043\end{array}$

Using 8 intervals,

$\begin{array}{c}∆\mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ =\frac{3-2}{8}\\ =\frac{1}{8}\end{array}$

$\begin{array}{rcl}{∫}_2^2{e}^{t^2}dt& ≈& \frac{1}{48}\left(e^4+2\underset{n=1}{\overset{8-1}{∑}}{e^{\frac{1}{4}\left(2+\frac{n}{8}\right)}}^2+\underset{n=1}{\overset{6}{4∑}}{e^{\frac{1}{256}\left(31+2n\right)}}^2+e^9\right)\\ & ≈& \frac{1}{48}\left(e^4+2\underset{n=1}{\overset{7}{∑}}{e^{\frac{1}{4}\left(2+\frac{n}{8}\right)}}^2+\underset{n=1}{\overset{6}{4∑}}{e^{\frac{1}{256}\left(31+2n\right)}}^2+e^9\right)\\ & ≈& 1428.26131\end{array}$

Substitute the value in equation (2).

$\begin{array}{rcl}y\left(3\right)& =& e^{-3^2}\left(e^4+1428.26131\right)\\ & ≈& 0.18300\end{array}$

Table of Simpson’s rule is shown below for the solution $\mathrm{y}={\mathrm{e}}^{-{\mathrm{x}}^2}\left({\mathrm{e}}^4+{∫}_2^{\mathrm{x}}{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{dt}\right)$.

Since the approximate value of $\mathit{y}\left(3\right)$1s 0.183.