Suggested languages for you:

Americas

Europe

Q 3.6-8E

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 subroutine with step size h = 0.2 to approximate the solution to the initial value problem ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{x}}{\mathbf{\left(}}{{\mathbf{y}}}^{{\mathbf{2}}}\mathbf{+}\mathbf{y}\mathbf{\right)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}$ at the points x = 1.2, 1.4, 1.6, and 1.8. (Thus, input N = 4.) Compare these approximations with those obtained using Euler’s method (see Exercises 1.4, Problem 6, page 28).

 xn yn 1.2 1.48 1.4 2.24788 1.6 3.6518 1.8 6.88733
See the step by step solution

## Step 1: Find the equation of approximation value

Here, $\mathrm{y}\text{'}=\frac{1}{\mathrm{x}}\left({\mathrm{y}}^{2}+\mathrm{y}\right),\mathrm{y}\left(1\right)=0$ for $1⩽x⩽1.8$

For h=0.2, x=1, y=1, N=4

$\mathrm{F}=\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\frac{1}{\mathrm{x}}\left({\mathrm{y}}^{2}+\mathrm{y}\right)\phantom{\rule{0ex}{0ex}}\mathrm{G}=\mathrm{f}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+\mathrm{hF}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{x}+0.2}\left({\left(\mathrm{y}+\frac{0.2}{\mathrm{x}}\left({\mathrm{y}}^{2}+\mathrm{y}\right)\right)}^{2}+\mathrm{y}+\frac{0.2}{\mathrm{x}}\left({\mathrm{y}}^{2}+\mathrm{y}\right)\right)\phantom{\rule{0ex}{0ex}}$

## Step 2: Solve for x1 and y1

Apply initial points ${\mathrm{x}}_{\mathrm{o}}=1,{\mathrm{y}}_{\mathrm{o}}=1,\mathrm{h}=0.2$

$\mathrm{F}\left(1,1\right)=2\phantom{\rule{0ex}{0ex}}\mathrm{G}\left(1,1\right)=2.8\phantom{\rule{0ex}{0ex}}$

${\mathrm{x}}_{1}=1+0.2\phantom{\rule{0ex}{0ex}}=1.2\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{1}=1+\frac{0.2}{2}\left(2+2.8\right)\phantom{\rule{0ex}{0ex}}=1.48\phantom{\rule{0ex}{0ex}}$

## Step 3: Evaluate the value of  x2 and y2

$\mathrm{F}\left(1.2,1.48\right)=3.05867\phantom{\rule{0ex}{0ex}}\mathrm{G}\left(1.2,1.48\right)=4.61934\phantom{\rule{0ex}{0ex}}$

${\mathrm{x}}_{2}=1.2+0.2\phantom{\rule{0ex}{0ex}}=1.4\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{2}=1.48+0.1\left(3.05867+4.61934\right)\phantom{\rule{0ex}{0ex}}=2.24788\phantom{\rule{0ex}{0ex}}$

## Step 4: Determine the value of  x3 and y3

$\mathrm{F}\left(1.4,2.24788\right)=5.21489\phantom{\rule{0ex}{0ex}}\mathrm{G}\left(1.4,2.2488\right)=8.82538\phantom{\rule{0ex}{0ex}}$

${\mathrm{x}}_{3}=1.4+0.2\phantom{\rule{0ex}{0ex}}=1.6\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{3}=2.24788+0.1\left(5.21489+8.82538\right)\phantom{\rule{0ex}{0ex}}=3.6518\phantom{\rule{0ex}{0ex}}$

## Step 5: Determine the value of  x4 and y4

$\mathrm{F}\left(1.6,3.6518\right)=10.6172\phantom{\rule{0ex}{0ex}}\mathrm{G}\left(1.6,3.6518\right)=21.7381\phantom{\rule{0ex}{0ex}}$

${\mathrm{x}}_{4}=1.6+0.2\phantom{\rule{0ex}{0ex}}=1.8\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{4}=3.6518+0.1\left(10.6172+21.7381\right)\phantom{\rule{0ex}{0ex}}=6.88733\phantom{\rule{0ex}{0ex}}$

Hence the solution is

 xn yn 1.2 1.48 1.4 2.24788 1.6 3.6518 1.8 6.88733