91影视

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

Given that,

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

Calculate the integrating factor of $渭\left(x\right)$.

Where $\mathrm{P}\left(\mathrm{x}\right)=\frac{\sin 2\mathrm{x}}{2\left(1+{\sin }^2\mathrm{x}\right)}$.

Then,

$\begin{array}{rcl}\mathrm{渭}\left(\mathrm{x}\right)& =& {\mathrm{e}}^{{鈭�}_{}^{}\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{{鈭�}_{}^{}\frac{\sin 2\mathrm{x}}{2\left(1+{\sin }^2\mathrm{x}\right)}\mathrm{dx}}\end{array}$

Find the value of role="math" localid="1664187557755" ${}^{{鈭�}_{}^{}\frac{\sin 2\mathrm{x}}{2\left(1+{\sin }^2\mathrm{x}\right)}dx}$separately.

Then,

${鈭�}_{}^{}\frac{\sin 2\mathrm{xdx}}{1+{\sin }^2\mathrm{x}}={鈭�}_{}^{}\frac{2\mathrm{cosxsinxdx}}{2\left(1+{\sin }^2\mathrm{x}\right)}={鈭�}_{}^{}\mathrm{cosx}\frac{\mathrm{sinxdx}}{\left(1+{\sin }^2\mathrm{x}\right)}$

Let us take $u=\sin x鈫�du=\cos xdx$.

${鈭�}_{}^{}\mathrm{cosx}\frac{\mathrm{sinxdx}}{\left(1+{\sin }^2\mathrm{x}\right)}={鈭�}_{}^{}\mathrm{cosx}\frac{\mathrm{u}}{1+{\mathrm{u}}^2}\frac{1}{\mathrm{cosx}}\mathrm{du}={鈭�}_{}^{}\frac{\mathrm{u}}{1+{\mathrm{u}}^2}\mathrm{du}$

Let us take $\mathrm{v}=1+{\mathrm{u}}^2鈫�\mathrm{dv}=2\mathrm{u}\mathrm{du}$.

$\begin{array}{rcl}{鈭�}_{}^{}\frac{\mathrm{u}}{1+{\mathrm{u}}^2}\mathrm{du}& =& {鈭�}_{}^{}\frac{\mathrm{v}}{\mathrm{u}}\frac{1}{2\mathrm{u}}\mathrm{du}\\ & =& \frac{1}{2}{鈭�}_{}^{}\frac{1}{\mathrm{v}}\mathrm{dv}\\ & =& \frac{1}{2}\ln \left|\mathrm{v}\right|\end{array}$

Substitute the value of v and u.

$\frac{1}{2}\ln \left|\mathrm{v}\right|=\frac{1}{2}\ln \left|1+{\mathrm{u}}^2\right|=\frac{1}{2}\ln \left|1+{\sin }^2\mathrm{x}\right|$

Substitute the value in $渭\left(x\right)$.

$\begin{array}{rcl}渭\left(x\right)& =& e^{\frac{1}{2}\left|1+{\sin }^{2}x\right|}\\ & =& \sqrt{1+{\sin }^{2}x}\end{array}$

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

role="math" localid="1664186609875" $$\begin{gathered} \sqrt{1+{\sin }^{2}x}\frac{\mathrm{dy}}{\mathrm{dx}}+\sqrt{1+{\sin }^{2}x}\frac{{\sin }^2\mathrm{x}}{2(1+{sin}^2\mathrm{x})}\mathrm{y}=\sqrt{1+{\sin }^{2}x} \\ \sqrt{1+{\sin }^{2}x}\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{{\sin }^2\mathrm{x}}{2\sqrt{(1+{\sin }^2\mathrm{x})}}\mathrm{y}=\sqrt{1+{\sin }^{2}x} \\ \frac{d}{dx}\left[y\sqrt{1+{\sin }^{2}x}\right]=\sqrt{1+{\sin }^2\mathrm{t}} \end{gathered}$$

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.

$\begin{array}{rcl}{鈭�}_{}^{}\frac{d}{dx}\left[y\sqrt{1+{\sin }^{2}x}\right]dx& =& {鈭�}_{}\sqrt{1+{\sin }^2\mathrm{t}}dt\\ y\sqrt{1+{\sin }^{2}x}& =& {鈭�}_0^x\sqrt{1+{\sin }^2\mathrm{t}}\mathrm{dt}\\ \mathrm{y}& =& {\left(1+{\sin }^2\mathrm{x}\right)}^{-\frac{1}{2}}{鈭�}_0^{\mathrm{x}}\sqrt{1+{\sin }^2\mathrm{t}}\mathrm{dt}\end{array}$

Hence the solution of the given equation isrole="math" localid="1664186866233" $\begin{array}{rcl}\mathrm{y}& =& {\left(1+{\sin }^2\mathrm{x}\right)}^{-\frac{1}{2}}{鈭�}_0^{\mathrm{x}}\sqrt{1+{\sin }^2\mathrm{t}}\mathrm{dt}........\left(2\right)\end{array}$

To approximate the solution at x = 1.

Since the Simpson鈥檚 rule is shown below.

${鈭�}_a^b\mathrm{f}\left(\mathrm{x}\right)\mathrm{x}\text{'}鈮�\frac{鈭�\mathrm{x}}{3}(\mathrm{f}\left({\mathrm{x}}_0\right)+4\mathrm{f}\left({\mathrm{x}}_1\right)+2\mathrm{f}\left({\mathrm{x}}_2\right)+.......4\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}-1}\right)+\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}}\right)).......\left(3\right)$

Then, using 4 intervals,

Substitute the value in equation (3)

$\begin{array}{rcl}{鈭�}_0^1{\sin }^{2}tdt& =& \frac{1}{12}\left(1+2\underset{n=1}{\overset{2-1}{鈭�}}\sqrt{1+{\sin }^2\left(\frac{n}{2}\right)}+4\underset{n=1}{\overset{2}{鈭�}}\sqrt{1+{\sin }^2\left(\frac{1}{4}\left(-1+2n\right)\right)}+\sqrt{1+{\sin }^2\left(1\right)}\right)\\ & 鈮�& 1.12387\end{array}$Substitute the value in equation (2).

$\begin{array}{rcl}\mathrm{y}& =& {\left(1+{\sin }^2\mathrm{x}\right)}^{-\frac{1}{2}}\left(1.12387\right)\\ & 鈮�& 0.860\end{array}$

Using 6 intervals,

Substitute the value in equation (3)

$\begin{array}{rcl}{鈭�}_0^1{\sin }^{2}tdt& =& \frac{1}{36}\left(1+2\underset{n=1}{\overset{6-1}{鈭�}}\sqrt{1+{\sin }^2\left(\frac{n}{6}\right)}+4\underset{n=1}{\overset{6}{鈭�}}\sqrt{1+{\sin }^2\left(\frac{1}{12}\left(-1+2n\right)\right)}+\sqrt{1+{\sin }^2\left(1\right)}\right)\\ & 鈮�& 1.12389\end{array}$Substitute the value in equation (2).

$\begin{array}{rcl}\mathrm{y}& =& {\left(1+{\sin }^2\mathrm{x}\right)}^{-\frac{1}{2}}\left(1.12389\right)\\ & 鈮�& 0.860\end{array}$

Similarly, we can find the value for other intervals as pretty much get the same approximate value.

Since the approximate value of $\mathbf{y}\left(1\right)$ is 0.860.