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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.