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Q 3.6-12E

Expert-verifiedFound in: Page 130

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**Use the improved Euler’s method with tolerance to approximate the solution to** ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{-}}{\mathbf{sin}}{\u200a}{\mathbf{y}}{\mathbf{,}}{\u200a}{\u200a}{\mathbf{y}}{\u200a}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$**, at ** ${\mathbf{x}}{\mathbf{=}}{\mathbf{\pi}}$**. For a tolerance of ** ${\mathbf{\epsilon}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{01}}$**, use a stopping procedure based on the absolute error.**

The required result is $\varphi \left(\pi \right)=1.09580$

The required Euler’s formula,

** ** $x=\left(x+h\right)\phantom{\rule{0ex}{0ex}}y=x+\frac{h}{2}\left(F+G\right)\phantom{\rule{0ex}{0ex}}$

Here given $y\text{'}=1-\mathrm{sin}y,\u200a\u200ay\left(0\right)=0$ ,

For value of $\epsilon =0.01$ , x = 0, y_{0} = 0, $c=\pi $ , M = 10, h = 3.141593 then

$F=f\left(x,y\right)\phantom{\rule{0ex}{0ex}}=1-\mathrm{sin}\left(y\right)\phantom{\rule{0ex}{0ex}}G=f\left(x+h,\u200a\u200ay+hF\right)\phantom{\rule{0ex}{0ex}}=1-\mathrm{sin}\left(y+hF\right)\phantom{\rule{0ex}{0ex}}=1-\mathrm{sin}\left(y+h\left(1-\mathrm{sin}\u200ay\right)\right)\phantom{\rule{0ex}{0ex}}$

Apply initial points

$F\left(0,0\right)=1\phantom{\rule{0ex}{0ex}}G\left(0,0\right)=1\phantom{\rule{0ex}{0ex}}$

$x=\left(x+h\right)\phantom{\rule{0ex}{0ex}}y=x+\frac{h}{2}\left(F+G\right)\phantom{\rule{0ex}{0ex}}x=3.141593\phantom{\rule{0ex}{0ex}}y=3.14159\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence, the value is $\varphi \left(\pi \right)=y\left(1,3.141593\right)=3.14159$

Now, for the values of $x=0,\u200a\u200ay=0,\u200a\u200ah=1.570796$

$F\left(0,0\right)=1\phantom{\rule{0ex}{0ex}}G\left(0,0\right)=5.34017\times {10}^{-14}\phantom{\rule{0ex}{0ex}}$

$dx=0+1.570796\phantom{\rule{0ex}{0ex}}=1.570796\phantom{\rule{0ex}{0ex}}y=0+\frac{1.570796}{2}\left(1+5.34017\times {10}^{-14}\right)\phantom{\rule{0ex}{0ex}}=0.785398\phantom{\rule{0ex}{0ex}}\varphi \left(1.570796\right)=0.785398\phantom{\rule{0ex}{0ex}}$

Now, for the values of F and G

$x=1.570796,\u200a\u200ay=0.785398,\u200a\u200ah=1.570796$

$F\left(1.570796,0.785398\right)=0.292893\phantom{\rule{0ex}{0ex}}G\left(1.570796,0.785398\right)=0.0524523\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}x=\left(1.570796+0.570796\right)\phantom{\rule{0ex}{0ex}}=3.14159\phantom{\rule{0ex}{0ex}}y=1.05663\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

The value of $\varphi \left(\pi \right)=y\left(1,0.570796\right)=1.05663$

Apply the same procedure for all other values and the values are

$\varphi \left(\pi \right)=y\left(1,0.785398\right)\phantom{\rule{0ex}{0ex}}=1.07575\phantom{\rule{0ex}{0ex}}\varphi \left(\pi \right)=y\left(1,0.392699\right)\phantom{\rule{0ex}{0ex}}=1.09229\phantom{\rule{0ex}{0ex}}\varphi \left(\pi \right)=y\left(1,0.196350\right)\phantom{\rule{0ex}{0ex}}=1.09580\phantom{\rule{0ex}{0ex}}$

Since the value is

role="math" localid="1664308176524" $\left|y\left(1,0.196350\right)-y\left(1,0.392699\right)\right|\phantom{\rule{0ex}{0ex}}=\left|1.09580-1.09229\right|\phantom{\rule{0ex}{0ex}}=0.00351<0.01\phantom{\rule{0ex}{0ex}}\varphi \left(\pi \right)=1.09580$

Therefore, the result is $\varphi \left(\pi \right)=1.09580$

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