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Q5.3-16E

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Found in: Page 1

### 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 for systems with ${\mathbit{h}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{175}}$, approximate the solution to the initial value problem ${\mathbf{x}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{x}}{\mathbf{-}}{\mathbf{y}}{\mathbf{;}}{\mathbf{x}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}{\mathbf{,}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{x}}{\mathbf{+}}{\mathbf{6}}{\mathbf{y}}{\mathbf{;}}{\mathbf{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{-}}{\mathbf{2}}$ at ${\mathbf{t}}{\mathbf{=}}{\mathbf{1}}$. Compare this approximation to the actual solution.

The solution is $y\left(1\right)=-423.48$ and $x\left(1\right)=127.77$.

See the step by step solution

## Transform the equation

Write the equation as $x\text{'}=2x-y$ and $y\text{'}=3x+6y$

The transformation of the equation is:

$x{\text{'}}_{1}\left(t\right)={x}_{2}\left(t\right)\phantom{\rule{0ex}{0ex}}{x}_{1}\left(t\right)=y\left(t\right)\phantom{\rule{0ex}{0ex}}{x}_{2}\left(t\right)=2x-{x}_{1}\phantom{\rule{0ex}{0ex}}x{\text{'}}_{2}\left(t\right)=y\text{'}\left(\left(t\right)\phantom{\rule{0ex}{0ex}}x{\text{'}}_{2}\left(t\right)=3x+6{x}_{1}\left(t\right)$

The initial conditions are:

$\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\phantom{\rule{0ex}{0ex}}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$

## Apply Runge –Kutta method

For the solution, apply the Runge-Kutta method in MATLAB, and the solution is $\mathrm{y}\left(1\right)=-423.48$ and $\mathrm{x}\left(1\right)=127.77$.

## Compare this approximation to the actual solution x(t)=e5t-e3t,y(t)=e3t-3e5t

By putting the value of $t=1$

$x\left(1\right)={e}^{5}-{e}^{3}\phantom{\rule{0ex}{0ex}}=128.32\phantom{\rule{0ex}{0ex}}y\left(1\right)={e}^{3}-3{e}^{5}\phantom{\rule{0ex}{0ex}}=-425.15$

Therefore, the approximation solution is $x\left(1\right)=128.32$ and $y\left(1\right)=-425.15$.