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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

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# In problems 1-4 Use Euler’s method to approximate the solution to the given initial value problem at the points ${\mathbf{x}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{1}}{\mathbf{,}}{\mathbf{0}}{\mathbf{.}}{\mathbf{2}}{\mathbf{,}}{\mathbf{0}}{\mathbf{.}}{\mathbf{3}}{\mathbf{,}}{\mathbf{0}}{\mathbf{.}}{\mathbf{4}}$ , and ${\mathbf{0}}{\mathbf{.}}{\mathbf{5}}$ , using steps of size ${\mathbf{0}}{\mathbf{.}}{\mathbf{1}}\left(\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\right)$ .$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}{\mathbf{=}}\frac{\mathbf{x}}{\mathbf{y}}{\mathbf{ }}{\mathbf{ }}{\mathbf{,}}{\mathbf{ }}{\mathbf{ }}{\mathbit{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{-}}{\mathbf{1}}$

 ${x}_{n}$ 0.1 0.2 0.3 0.4 0.5 ${y}_{n}$ -1 -1.01 -1.029 -1.085 -1.096
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

One has ,$f\left(x,y\right)=\frac{-x}{y}, {x}_{0}=0, {y}_{0}=-1, h=0.1$

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

## Put  n=0  to find  y1

Now, find the value of ${y}_{1}$ when n = 0, then

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

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

## Put  n=1 to find  y2

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

${y}_{2}={y}_{1}+\left(0.1\right)\left(\frac{{x}_{1}}{{y}_{1}}\right)\phantom{\rule{0ex}{0ex}}=-1+\left(0.1\right)\left(\frac{0.1}{-1}\right)\phantom{\rule{0ex}{0ex}}=-1.01$

Thus, the value of ${{\mathbit{y}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{1}}{\mathbf{.}}{\mathbf{01}}$ when ${{\mathbit{x}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{2}}$

## Put   n=2 to find  y3

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

${y}_{3}={y}_{2}+\left(0.1\right)\left(\frac{{x}_{2}}{{y}_{2}}\right)\phantom{\rule{0ex}{0ex}}=-1.01+\left(0.1\right)\left(\frac{0.2}{-1.01}\right)\phantom{\rule{0ex}{0ex}}=-1.01+\left(-0.019\right)\phantom{\rule{0ex}{0ex}}=-1.029$

So, the value is ${{\mathbit{y}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{1}}{\mathbf{.}}{\mathbf{029}}$ when ${{\mathbit{x}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{3}}$

## Put  n=3 to find  y4

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

${y}_{4}={y}_{3}+\left(0.1\right)\left(\frac{{x}_{3}}{{y}_{3}}\right)\phantom{\rule{0ex}{0ex}}=-1.029+\left(0.1\right)\left(\frac{0.3}{-1.029}\right)\phantom{\rule{0ex}{0ex}}=-1.029+\left(-0.029\right)\phantom{\rule{0ex}{0ex}}=-1.058$

Consequently, the value is ${{\mathbit{y}}}_{{\mathbf{4}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{1}}{\mathbf{.}}{\mathbf{058}}$ when ${{\mathbit{x}}}_{{\mathbf{4}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{4}}$

## Put n=4  to find  y5

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

${y}_{5}={y}_{4}+\left(0.1\right)\left(\frac{{x}_{4}}{{y}_{4}}\right)\phantom{\rule{0ex}{0ex}}=-1.058+\left(0.1\right)\left(\frac{0.4}{-1.058}\right)\phantom{\rule{0ex}{0ex}}=-1.058+\left(-0.0378\right)\phantom{\rule{0ex}{0ex}}=-1.096$

Therefore, the value is ${{\mathbit{y}}}_{{\mathbf{5}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{1}}{\mathbf{.}}{\mathbf{096}}$ when ${{\mathbf{x}}}_{{\mathbf{5}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{5}}$

Therefore the solution is

 ${{\mathbit{x}}}_{{\mathbf{n}}}$ 0.1 0.2 0.3 0.4 0.5 ${{\mathbit{y}}}_{{\mathbf{n}}}$ -1 -1.01 -1.029 -1.058 -1.096

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