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Answers without the blur. Sign up and see all textbooks for free! Q 3.6-12E

Expert-verified Found in: Page 130 ### Fundamentals Of Differential Equations And Boundary Value Problems

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 ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{-}}{\mathbf{sin}}{ }{\mathbf{y}}{\mathbf{,}}{ }{ }{\mathbf{y}}{ }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, 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 $\varphi \left(\pi \right)=1.09580$

See the step by step solution

## Step 1: Important formula.

The required Euler’s formula,

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

## Step 2: Find the equation of approximation value

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

For value of $\epsilon =0.01$ , x = 0, y0 = 0, $c=\pi$ , M = 10, h = 3.141593 then

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

## Step 3: solve for x and y

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 $\varphi \left(\pi \right)=y\left(1,3.141593\right)=3.14159$

## Step 4: Evaluate the value of  x and y

Now, for the values of $x=0, y=0, h=1.570796$

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

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

## Step 5: Determine the value of x and t for the conditions

Now, for the values of F and G

$x=1.570796, y=0.785398, h=1.570796$

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

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

## Step 6: Determine the all-other values.

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 $\varphi \left(\pi \right)=1.09580$ ### Want to see more solutions like these? 