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Q5.3-17E

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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

# In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†Using the vectorized Runge–Kutta algorithm, approximate the solution to the initial value problem$\frac{\mathbf{du}}{\mathbf{dx}}\mathbf{=}\mathbf{3}\mathbf{u}\mathbf{-}\mathbf{4}\mathbf{v}\mathbf{;}\mathbf{u}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{\text{'}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}\mathbf{2}\mathbf{u}\mathbf{-}\mathbf{3}\mathbf{v}\mathbf{;}\mathbf{v}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$at x = 1. Starting with h=1, continue halving the step size until two successive approximations of u(1)and v(1) differ by at most 0.001.

The solution is $u\left(1\right)=0.36789$ and $v\left(1\right)=0.36789$.

See the step by step solution

## Transform the equation

Write the equation as $u\text{'}=3u-4v$ and $v\text{'}=2u-3v$.

The transformation of the equation is:

$\mathrm{u}\text{'}\left(\mathrm{t}\right)=\mathrm{x}{\text{'}}_{2}\left(\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}\mathrm{v}\text{'}\left(\mathrm{t}\right)={\mathrm{x}}_{3}\left(\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{2}\left(\mathrm{t}\right)=\mathrm{v}\left(\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{1}=\mathrm{u}\left(\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}\mathrm{x}{\text{'}}_{2}\left(\mathrm{t}\right)=3{\mathrm{x}}_{1}-4{\mathrm{x}}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{x}{\text{'}}_{3}\left(\mathrm{t}\right)=2{\mathrm{x}}_{1}-3{\mathrm{x}}_{2}$

The initial conditions are

$u\left(0\right)=1\phantom{\rule{0ex}{0ex}}v\left(0\right)=1$

## Apply Runge –Kutta method

For the solution apply the Runge-Kutta method in MATLAB for h=0.25, and the solution is u(1)=0.36789 and v(1)=0.36789.

## Find that u1 and v(1) differ by at most 0.001.

Subtracting the values of $u\left(1\right)$ and $v\left(1\right)$ then

$u\left(1\right)-v\left(1\right)=0.36789-0.36678\phantom{\rule{0ex}{0ex}}=0.001101$

Therefore, $u\left(1\right)$ and $v\left(1\right)$ differ by at most 0.001.

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