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Answers without the blur. Sign up and see all textbooks for free! Q 3.6-17E

Expert-verified Found in: Page 131 ### 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 Euler’s method (4) with h = 0.1 to approximate the solution to the initial value problem ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{20}}{\mathbf{y}}{\mathbf{,}}{\mathbf{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{1}}$, on the interval ${\mathbf{0}}{\mathbf{⩽}}{\mathbf{x}}{\mathbf{⩽}}{\mathbf{1}}$ (that is, at x = 0, 0.1, . . . , 1.0).Compare your answers with the actual solution y = e-20x. What went wrong? Next, try the step size h = 0.025 and also h = 0.2. What conclusions can you draw concerning the choice of step size?

 xn yn(h = 0.2) yn(h = 0.1) yn(h = 0.025) 0.1 -1 0.06250 0.2 -3 1 0.00391 0.3 -1 0.00024 0.4 9 1 0.00002 0.5 -1 0.00000 0.6 -27 1 0.00000 0.7 -1 0.00000 0.8 81 1 0.00000 0.9 -1 0.00000 1.0 -243 1 0.00000

The result is for h=0.1 the values are 1 and -1. The values for h=0.2 the results are increasing and decreasing with different signs. For h=0.025 is the best approximation result.

See the step by step solution

## Step 1: Find the equation of approximation value

Here f(x,y)=-20y then

${\mathrm{y}}_{\mathrm{n}}={\mathrm{y}}_{\mathrm{n}-1}+\mathrm{h}\left(-20{\mathrm{y}}_{\mathrm{n}-1}\right)\phantom{\rule{0ex}{0ex}}=\left(1-20\mathrm{h}\right){\mathrm{y}}_{\mathrm{n}-1}\phantom{\rule{0ex}{0ex}}.\phantom{\rule{0ex}{0ex}}.\phantom{\rule{0ex}{0ex}}.\phantom{\rule{0ex}{0ex}}=\left(1-20\mathrm{h}{\right)}^{\mathrm{n}}{\mathrm{y}}_{\mathrm{o}}\phantom{\rule{0ex}{0ex}}\left({\mathrm{y}}_{0}=1\right)\left(1-20\mathrm{h}{\right)}^{\mathrm{n}}\phantom{\rule{0ex}{0ex}}$

## Step 2: Solve for x and y

Apply initial points ${\mathrm{x}}_{\mathrm{n}}={\mathrm{x}}_{\mathrm{o}}+\mathrm{nh},{\mathrm{x}}_{\mathrm{o}}=0$ for

$\mathrm{h}=0.1\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{\mathrm{n}}=0.1\mathrm{n},{\mathrm{y}}_{\mathrm{n}}=\left(-1{\right)}^{\mathrm{n}},\mathrm{n}=1,2,.....10\phantom{\rule{0ex}{0ex}}\mathrm{h}=0.025\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{\mathrm{n}}=0.025\mathrm{n},\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{\mathrm{n}}=\left(0.5{\right)}^{\mathrm{n}},\phantom{\rule{0ex}{0ex}}\mathrm{n}=1,2,...40\phantom{\rule{0ex}{0ex}}\mathrm{h}=0.2\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{\mathrm{n}}=0.2\mathrm{n},{\mathrm{y}}_{\mathrm{n}}=\left(-3{\right)}^{\mathrm{n}},\mathrm{n}=1,2,...5\phantom{\rule{0ex}{0ex}}$

Where $\mathrm{n}=1,2,3,....\frac{1}{\mathrm{h}}$

## Step 3: Determine the all other values for h=0.1 and substituting the value of n

$\left({\mathrm{x}}_{1}=0.1,{\mathrm{y}}_{1}=-1\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{2}=0.2,{\mathrm{y}}_{2}=1\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{3}=0.3,{\mathrm{y}}_{3}=-1\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{4}=0.4,{\mathrm{y}}_{4}=1\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{5}=0.5,{\mathrm{y}}_{5}=-1\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{6}=0.6,{\mathrm{y}}_{6}=1\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{7}=0.7,{\mathrm{y}}_{7}=-1\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{8}=0.8,{\mathrm{y}}_{8}=1\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{9}=0.9,{\mathrm{y}}_{9}=-1\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{10}=1,{\mathrm{y}}_{10}=1\right)\phantom{\rule{0ex}{0ex}}$

## Step 4: Evaluate the all other values for h=0.025 and substituting the value of n

$\left({\mathrm{x}}_{1}=0.1,{\mathrm{y}}_{1}=0.0625\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{2}=0.2,{\mathrm{y}}_{2}=0.003906\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{3}=0.3,{\mathrm{y}}_{3}=0.000244\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{4}=0.4,{\mathrm{y}}_{4}=0.000015\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{5}=0.5,{\mathrm{y}}_{5}=0.000001\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{6}=0.6,{\mathrm{y}}_{6}=0\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{7}=0.7,{\mathrm{y}}_{7}=0\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{8}=0.8,{\mathrm{y}}_{8}=0\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{9}=0.9,{\mathrm{y}}_{9}=0\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{10}=1,{\mathrm{y}}_{10}=0\right)\phantom{\rule{0ex}{0ex}}$

## Step 5: Find the all-other values for h=0.2 and substituting the value of n

$\left({\mathrm{x}}_{2}=0.2,{\mathrm{y}}_{2}=-3\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{4}=0.4,{\mathrm{y}}_{4}=9\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{6}=0.6,{\mathrm{y}}_{6}=-27\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{8}=0.8,{\mathrm{y}}_{8}=81\right)\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}_{10}=1,{\mathrm{y}}_{10}=-243\right)\phantom{\rule{0ex}{0ex}}$

Hence the solution is

 xn yn(h = 0.2) yn(h = 0.1) yn(h = 0.025) 0.1 -1 0.06250 0.2 -3 1 0.00391 0.3 -1 0.00024 0.4 9 1 0.00002 0.5 -1 0.00000 0.6 -27 1 0.00000 0.7 -1 0.00000 0.8 81 1 0.00000 0.9 -1 0.00000 1.0 -243 1 0.00000

The result is for h=0.1 the values are 1 and -1. The values for h=0.2 the results are increasing and decreasing with different signs. For h=0.025 is the best approximation result. ### Want to see more solutions like these? 