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Answers without the blur. Sign up and see all textbooks for free! Q5.3-21E

Expert-verified Found in: Page 260 ### 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 # Fluid Ejection. In the design of a sewage treatment plant, the following equation arises: $\mathbf{60}\mathbf{-}\mathbf{H}\mathbf{=}\mathbf{\left(}\mathbf{77}\mathbf{.}\mathbf{7}\mathbf{\right)}\mathbf{H}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{\left(}\mathbf{19}\mathbf{.}\mathbf{42}\mathbf{\right)}\mathbf{\left(}\mathbf{H}\mathbf{\text{'}}{\mathbf{\right)}}^{\mathbf{2}}\mathbf{;}\mathbf{H}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{H}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ where H is the level of the fluid in an ejection chamber, and t is the time in seconds. Use the vectorized Runge–Kutta algorithm with h = 0.5 to approximate ${\mathbf{H}}\left(t\right)$ over the interval [0, 5].

The result can get by the Runge-Kutta method.

See the step by step solution

## Transform the equation

Here the equation is $60-\mathrm{H}=\left(77.7\right)\mathrm{H}\text{'}\text{'}+\left(19.42\right)\left(\mathrm{H}\text{'}{\right)}^{2};\mathrm{H}\left(0\right)=\mathrm{H}\text{'}\left(0\right)=0.$

The system can be written as:

${\mathrm{x}}_{1}=\mathrm{H}\left(\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{2}=\mathrm{H}\text{'}=\mathrm{x}{\text{'}}_{1}\phantom{\rule{0ex}{0ex}}\mathrm{H}\text{'}\text{'}=\mathrm{x}{\text{'}}_{2}$

The transform equation:

$\mathrm{x}{\text{'}}_{1}={\mathrm{x}}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{x}{\text{'}}_{2}=\frac{\left[60-{\mathrm{x}}_{1}-\left(19.42\right){{\mathrm{x}}^{2}}_{2}\right]}{77.7}$

The initial conditions are:

${\mathrm{x}}_{1}\left(0\right)=\mathrm{H}\left(0\right)=0\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{2}\left(0\right)=\mathrm{H}\text{'}\left(0\right)=0$

## Apply the Runge-Kutta method.

Here h=0.5, N=10 steps, ${\mathrm{x}}_{1,0}=0,{\mathrm{x}}_{2,0}=0$ then;

${x}_{1,0}=0,{x}_{2,0}=0\phantom{\rule{0ex}{0ex}}{k}_{1,1}=h{x}_{2,0}=0.5\left(0\right)=0\phantom{\rule{0ex}{0ex}}{k}_{2,1}=\frac{h\left[60-{x}_{1,0}-\left(19.42\right){{x}^{2}}_{2,0}\right]}{77.7}=0.38610\phantom{\rule{0ex}{0ex}}{k}_{1,2}=0.09653\phantom{\rule{0ex}{0ex}}{k}_{2,2}=0.38144\phantom{\rule{0ex}{0ex}}{k}_{1,3}=0.09536\phantom{\rule{0ex}{0ex}}{k}_{2,3}=0.38124\phantom{\rule{0ex}{0ex}}{k}_{1,4}=0.19062\phantom{\rule{0ex}{0ex}}{k}_{2,4}=0.36732$

## Apply the recursive formula

${\mathrm{t}}_{1}={\mathrm{t}}_{0}+\mathrm{h}\phantom{\rule{0ex}{0ex}}0+0.5=0.5\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{1}\left(0.5\right)={\mathrm{x}}_{1,1}=0.09573\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{2}\left(0.5\right)={\mathrm{x}}_{2,1}=0.37980$

Apply the same procedure for n=1, 2,…., 9.

 N ${t}_{n}$ ${\mathrm{x}}_{1,\mathrm{n}}=\mathrm{H}\left({\mathrm{t}}_{\mathrm{n}}\right)$ 0 0 0 1 0.5 0.09573 2 1.0 0.37389 3 1.5 0.81045 4 2.0 1.37361 5 2.5 2.03111 6 3.0 2.7549 7 3.5 3.52322 8 4.0 4.319 9 4.5 5.13307 10 5.0 5.95554

This is the required result. ### Want to see more solutions like these? 