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### 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 with step size h = 0.1 to approximate the solution to the initial value problem ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{x}}{\mathbf{-}}{{\mathbf{y}}}^{{\mathbf{2}}}$, y (1) = 0 at the points ${\mathbf{x}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{1}}{\mathbf{,}}{}{\mathbf{1}}{\mathbf{.}}{\mathbf{2}}{\mathbf{,}}{}{\mathbf{1}}{\mathbf{.}}{\mathbf{3}}{\mathbf{,}}{}{\mathbf{1}}{\mathbf{.}}{\mathbf{4}}{}{\text{and}}{}{\mathbf{1}}{\mathbf{.}}{\mathbf{5}}$ .

 ${x}_{n}$ 1.1 1.2 1.3 1.4 1.5 ${y}_{n}$ 0.1 0.209 0.325 0.444 0.564
See the step by step solution

## Write the recursive formula

For the solution use the Euler’s formula ${y}_{n+1}={y}_{n}+h.f\left({x}_{n},{y}_{n}\right)$

## Apply recursive formula

We have, $f\left(x,y\right)=x-{y}^{2}, {x}_{0}=1, {y}_{0}=0, h=0.1$

Then, ${y}_{n+1}={y}_{n}+h.f\left({x}_{n},{y}_{n}\right)\phantom{\rule{0ex}{0ex}}={y}_{n}+\left(0.1\right)\left(x-{y}^{2}\right)$

## Put  n=0 to find  y1

Now, find the value of ${y}_{1}$

${y}_{1}={y}_{0}+\left(0.1\right)\left({x}_{0}-{y}_{0}^{2}\right)\phantom{\rule{0ex}{0ex}}=0+\left(0.1\right)\left(1\right)\phantom{\rule{0ex}{0ex}}=0.1$

Hence, the value of ${{\mathbf{y}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{1}}$ when ${{\mathbf{x}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{1}}$

## Put  n=1 to find  y2

The value of ${y}_{2}$ is

${y}_{2}={y}_{1}+\left(0.1\right)\left({x}_{1}-{y}_{1}^{2}\right)\phantom{\rule{0ex}{0ex}}=0.1+\left(0.1\right)\left(1.1-0.01\right)\phantom{\rule{0ex}{0ex}}=0.1+0.109\phantom{\rule{0ex}{0ex}}=0.209$

Consequently, the value is ${{\mathbf{y}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{209}}$ when ${{\mathbf{x}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{2}}$

## Put  n=2 to find  y3

Now the value of ${y}_{3}$

${y}_{3}={y}_{2}+\left(0.1\right)\left({x}_{2}-{y}_{2}^{2}\right)\phantom{\rule{0ex}{0ex}}=0.209+\left(0.1\right)\left(1.2-0.043\right)\phantom{\rule{0ex}{0ex}}=0.209+0.116\phantom{\rule{0ex}{0ex}}=0.325$

So, the value of ${{\mathbf{y}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{325}}$ when ${{\mathbf{x}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{3}}$

## Put  n=3 to find  y4

The value of ${y}_{4}$ is

${y}_{4}={y}_{3}+\left(0.1\right)\left({x}_{3}-{y}_{3}^{2}\right)\phantom{\rule{0ex}{0ex}}=0.325+\left(0.1\right)\left(1.3-0.106\right)\phantom{\rule{0ex}{0ex}}=0.325+0.119\phantom{\rule{0ex}{0ex}}=0.444$

Thus, the value is ${{\mathbf{y}}}_{{\mathbf{4}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{444}}$ when ${{\mathbf{x}}}_{{\mathbf{4}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{4}}$

## Put  n=4 to find  y5

The value of ${y}_{5}$ is

${y}_{5}={y}_{4}+\left(0.1\right)\left({x}_{4}-{y}_{4}^{2}\right)\phantom{\rule{0ex}{0ex}}=0.444+\left(0.1\right)\left(1.4-0.197\right)\phantom{\rule{0ex}{0ex}}=0.444+0.12\phantom{\rule{0ex}{0ex}}=0.564$

Therefore, the value is ${{\mathbf{y}}}_{{\mathbf{5}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{564}}$ when ${{\mathbf{x}}}_{{\mathbf{5}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{5}}$

Therefore the solution is

 role="math" localid="1663939345323" ${{\mathbit{x}}}_{{\mathbf{n}}}$ 1.1 1.2 1.3 1.4 1.5 ${{\mathbit{y}}}_{{\mathbf{n}}}$ 0.1 0.209 0.325 0.444 0.564

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