91影视

Since $\mathrm{P}\left(\mathrm{t}\right)=\sqrt{1+{sin}^2\mathrm{t}}$ is a continuous real valued function on an open interval which contains points 0 we can use fundamental theorem of calculus to obtain

$\frac{\mathrm{d}}{\mathrm{dx}}\left[{鈭�}_0^{\mathrm{x}}\mathrm{P}\left(\mathrm{t}\right)\mathrm{dt}\right]=\mathrm{P}\left(\mathrm{x}\right)={鈭�}_0^{\mathrm{x}}\mathrm{P}\left(\mathrm{t}\right)\mathrm{dt}=鈭�\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}$

We can write integrating factor as $\mathrm{渭}\left(\mathrm{x}\right)=\exp \left({鈭�}_0^{\mathrm{x}}\sqrt{1+{\sin }^2\mathrm{t}}\mathrm{dt})\right)$

Multiplying by $渭\left(x\right)$

$\frac{\mathrm{d}}{\mathrm{dx}}\overline{)\left(\mathrm{渭}\left(\mathrm{x}\right)\mathrm{y}\left(\mathrm{x}\right)\right)}=\mathrm{虫渭}\left(\mathrm{x}\right)=\mathrm{诲渭}\left(\mathrm{x}\right)\mathrm{y}\left(\mathrm{x}\right)=\mathrm{虫渭}\left(\mathrm{x}\right)\mathrm{y}\left(\mathrm{x}\right)$

Now,

$$\begin{gathered} 渭\left(x\right){\overline{)y\left(x\right)}}_0^s={鈭�}_0^s\left(x渭\left(x\right)\right)=渭\left(s\right)y\left(s\right)-渭\left(0\right)y\left(0\right)={鈭�}_0^{s}x渭\left(x\right) \\ 渭\left(s\right)y\left(s\right)={鈭�}_0^{s}x渭\left(x\right)dx+2 \\ y\left(s\right)=\frac{1}{渭\left(s\right)}{鈭�}_0^{s}x渭\left(x\right)dx+\frac{2}{渭\left(s\right)} \end{gathered}$$

Here $y\left(s\right)=\frac{1}{渭\left(s\right)}{鈭�}_0^{s}x渭\left(x\right)dx+\frac{2}{渭\left(s\right)}$

Put x=1 then $y\left(1\right)=\frac{1}{渭\left(1\right)}{鈭�}_0^{s}x渭\left(s\right)dx+\frac{2}{渭\left(1\right)}$

We find the value of${鈭�}_0^1渭\left(s\right)sdx$.

Using Simpson鈥檚 rule

$$\begin{gathered} {鈭�}_a^{b}f\left(x\right)dx=\frac{鈭�x}{3}\underset{k=1}{\overset{n}{鈭�}}\left[f\left(x_{2k-2}\right)+4f\left(x_{2k-2}\right)+f\left(x_{2k}\right)\right] \\ 鈭�x=\frac{b-a}{2n},x_k=a+k鈭�x,k=0.1,0.2,...,1 \end{gathered}$$

So, $\mathrm{渭}\left(\mathrm{x}\right)=\exp \left({鈭�}_0^{\mathrm{x}}\sqrt{1+{\sin }^2\mathrm{t}}\mathrm{dt})\right)$at x=0.1,0.2, 鈥︹��1

The values are

x = 0, =1

x = 0.1, =1.105354

x = 0.2, =1.223010

x = 0 .3, =1.355761

x = 0 .4, =1.506975

x = 0.5, =1.680635

x = 0 .6, =1.881401

x = 0 .7, =2.114679

x = 0 .8, =2.386713

x = 0 .9, =2.70670

x = 1, =3.076723

Now, using the previous conclusions

$鈭�x=\frac{1}{2n}=0.1=n=5$

So,

role="math" localid="1663932388162" $\begin{array}{rcl}{鈭�}_0^{1}s渭\left(s\right)ds& =& \frac{0.1}{3}[0路渭\left(0\right)+4\left(0.1\right)路渭\left(0.1\right)+2\left(0.2\right)路渭\left(0.2\right)+4\left(0.3\right)路渭\left(0.3\right)+2\left(0.4\right)路渭\left(0.4\right)+4\left(0.5\right)路渭\left(0.5\right)+2\left(0.6\right)路渭\left(0.6\right)\\ & & +4\left(0.7\right)路渭\left(0.7\right)+2\left(0.8\right)路渭\left(0.8\right)+4\left(0.9\right)路渭\left(0.9\right)+1路渭\left(1\right)]\\ & =& 1.064539\end{array}$role="math" localid="1663932537871" $\begin{array}{rcl}y\left(1\right)& =& \frac{1}{渭\left(1\right)}{鈭�}_0^{s}x渭\left(s\right)dx+\frac{2}{渭\left(1\right)}\\ & =& 0.9960\end{array}$

Therefore, $\mathit{y}\left(1\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9960}$

The differential equation is $y\text{'}=x-\sqrt{1+{\sin }^{2}xy}$

Use the recursive formula

localid="1663933001609" $\begin{array}{rcl}{x}_{n+1}& =& x_n+h\\ y_{n+1}& =& y_n+hf\left(x_n{y}_n\right)\\ & =& y_n+hf\left(x_n-\sqrt{1+{\sin }^2{x}_n{y}_n}\right)\end{array}$

Where,

$x_0=0$and $y_0=0$, h=0.1, N=10 steps at x=1

The values are

$$\begin{gathered} x_1=0.1,y_1=1.8 \\ {x}_2=0.2,y_2=1.6921 \\ {x}_3=0.3,y_3=1.4830 \\ {x}_4=0.4,y_4=1.3584 \\ {x}_5=0.5,y_5=1.2526 \\ {x}_6=0.6,y_6=1.1637 \\ {x}_7=0.7,y_7=1.09 \\ {x}_8=0.8,y_8=1.0304 \\ {x}_9=0.9,y_9=0.9836 \\ {x}_{10}=1,y_{10}=0.9586 \end{gathered}$$

Therefore,$\mathit{y}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9586}$

$x_0=0$and$y_0=0$, h=0.1, N=20 steps at x=1

$$\begin{gathered} x_1=0.05,y_1=1.9 \\ {x}_2=0.1,y_2=1.8074 \\ {x}_3=0.15,y_3=1.4830 \\ {x}_4=0.2,y_4=1.642 \\ {x}_5=0.25,y_5=1.5683 \\ {x}_6=0.3,y_6=1.5 \\ {x}_7=0.35,y_7=1.4368 \\ {x}_8=0.4,y_8=1.3784 \\ {x}_9=0.45,y_9=1.3244 \\ {x}_{10}=0.5,y_{10}=1.2747 \\ {x}_{11}=0.55,y_{11}=1.229 \\ {x}_{12}=0.6,y_{12}=1.187 \\ {x}_{13}=0.65,y_{13}=1.149 \\ {x}_{14}=0.7,y_{14}=1.1144 \\ {x}_{15}=0.75,y_{15}=1.0831 \\ {x}_{16}=0.8,y_{16}=1.0551 \\ {x}_{17}=0.85,y_{17}=1.0301 \\ {x}_{18}=0.9,y_{18}=1.0082 \\ {x}_{19}=0.95,y_{19}=0.9892 \\ {x}_{20}=1,y_{20}=0.9729 \end{gathered}$$

Therefore, $\mathit{y}\left(1\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9729}$$\mathit{y}\left(1\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9729}$