• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q 3.7-9E

Expert-verified Found in: Page 139 ### 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 fourth-order Runge–Kutta subroutine with h = 0.25 to approximate the solution to the initial value problem ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{x}}{\mathbf{+}}{\mathbf{1}}{\mathbf{-}}{\mathbf{y}}{\mathbf{,}}{\mathbf{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{1}}$, at x = 1. Compare this approximation with the one obtained in Problem 5 using the Taylor method of order 4.

$\mathbit{\varphi }\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3679}$

See the step by step solution

## Step 1: Find the values of ki, i = 1, 2, 3, 4

Since $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}+1-\mathrm{y}$ and x = 0, y = 1, and h = 0.25

role="math" localid="1664324055813" ${\mathrm{k}}_{1}=\mathrm{h}\text{f}\left(\mathrm{x},\mathrm{y}\right)=0.25\left(0+1-1\right)=0\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{2}=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_{1}}{2}\right)=0.03125\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{3}=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_{2}}{2}\right)=0.0273438\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{4}=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_{3}\right)=0.0556641$

## Step 2: Find the values of x and y

$\mathrm{x}=0+0.25=0.25\phantom{\rule{0ex}{0ex}}\mathrm{y}=1+\frac{1}{6}\left({\mathrm{k}}_{1}+2{\mathrm{k}}_{2}+2{\mathrm{k}}_{3}+{\mathrm{k}}_{4}\right)\phantom{\rule{0ex}{0ex}}=1+\frac{1}{6}\left(0-2\left(0.031125\right)-2\left(0.0273438\right)-0.0556641\right)\phantom{\rule{0ex}{0ex}}=1.02881$

## Step 3: Use values of x and y for finding values of ki, i = 1, 2, 3, 4

${\mathrm{k}}_{1}=\mathrm{h}\text{f}\left(\mathrm{x},\mathrm{y}\right)=0.25\left(0.25+1-1.02881\right)=0.0552975\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{2}=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_{1}}{2}\right)=0.0796353\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{3}=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_{2}}{2}\right)=0.0765931\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{4}=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_{3}\right)=0.0986392$

Now

$\mathrm{x}=0.25+0.25=0.5\phantom{\rule{0ex}{0ex}}\mathrm{y}=1.02881+\frac{1}{6}\left(0.0552975-2\left(0.0796353\right)-2\left(0.0765931\right)-0.0986492\right)\phantom{\rule{0ex}{0ex}}=1.10654\phantom{\rule{0ex}{0ex}}$

## Step 4: Repeat the procedure for two times

${\mathrm{k}}_{1}=\mathrm{h}\text{f}\left(\mathrm{x},\mathrm{y}\right)=0.25\left(0.5+1-1.10654\right)=0.098365\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{2}=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_{1}}{2}\right)=0.117319\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{3}=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_{2}}{2}\right)=0.11495\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{4}=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_{3}\right)=0.132127\phantom{\rule{0ex}{0ex}}\mathrm{x}=0.5+0.25=0.75\phantom{\rule{0ex}{0ex}}\mathrm{y}=1.10654+\frac{1}{6}\left({\mathrm{k}}_{1}+2{\mathrm{k}}_{2}+2{\mathrm{k}}_{3}+{\mathrm{k}}_{4}\right)\phantom{\rule{0ex}{0ex}}=1.10654+\frac{1}{6}\left(0.098365-2\left(0.117319\right)-2\left(0.11495\right)-0.132127\right)\phantom{\rule{0ex}{0ex}}=1.22238\phantom{\rule{0ex}{0ex}}$

And

${\mathrm{k}}_{1}=\mathrm{h}\text{f}\left(\mathrm{x},\mathrm{y}\right)=0.25\left(0.75+1-1.22238\right)=0.131905\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{2}=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_{1}}{2}\right)=0.146667\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{3}=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_{2}}{2}\right)=0.144822\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{4}=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_{3}\right)=0.1582\phantom{\rule{0ex}{0ex}}\mathrm{x}=0.75+0.25=1\phantom{\rule{0ex}{0ex}}\mathrm{y}=1.22238+\frac{1}{6}\left({\mathrm{k}}_{1}+2{\mathrm{k}}_{2}+2{\mathrm{k}}_{3}+{\mathrm{k}}_{4}\right)\phantom{\rule{0ex}{0ex}}=0.52761+\frac{1}{6}\left(0.131905-2\left(0.146667\right)-2\left(0.144822\right)-0.1582\right)\phantom{\rule{0ex}{0ex}}=1.36789\phantom{\rule{0ex}{0ex}}$

Therefore $\varphi \mathbf{\left(}1\mathbf{\right)}\mathbf{=}1.3679$

Hence the solution is $\mathbit{\varphi }\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3679}$ ### Want to see more solutions like these? 