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Q5.3-27E

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Found in: Page 261

### 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

# Generalized Blasius Equation. H. Blasius, in his study of the laminar flow of a fluid, encountered an equation of the form $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{yy}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{\left(}\mathbf{y}\mathbf{\text{'}}{\mathbf{\right)}}^{\mathbf{2}}{\mathbf{-}}{\mathbf{1}}$. Use the Runge–Kutta algorithm for systems with h = 0.1 to approximate the solution that satisfies the initial conditions $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}{\mathbf{.}}{\mathbf{32824}}$. Sketch this solution on the interval [0, 2].

The result can get by the Runge-Kutta method, and the result is y(2)=1.6001.

See the step by step solution

## Transform the equation

Here the equation is $\mathrm{y}\text{'}\text{'}\text{'}+\mathrm{yy}\text{'}\text{'}=\left(\mathrm{y}\text{'}{\right)}^{2}-1$.

The system can be written as:

${\mathrm{x}}_{1}=\mathrm{y}\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{2}=\mathrm{y}\text{'}=\mathrm{x}{\text{'}}_{1}\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{3}=\mathrm{y}\text{'}\text{'}=\mathrm{x}{\text{'}}_{2}$

The transform equation is:

$\mathrm{x}{\text{'}}_{1}={\mathrm{x}}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{x}{\text{'}}_{2}={\mathrm{x}}_{3}\phantom{\rule{0ex}{0ex}}\mathrm{x}{\text{'}}_{3}=-{\mathrm{x}}_{1}{\mathrm{x}}_{3}+{{\mathrm{x}}^{2}}_{2}-1$

The initial conditions are:

${\mathrm{x}}_{1}\left(0\right)=\mathrm{y}\left(0\right)=0\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{2}\left(0\right)=\mathrm{y}\text{'}\left(0\right)=0\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{3}\left(0\right)=\mathrm{y}\text{'}\text{'}\left(0\right)=1.32824$

## Apply the Runge-Kutta method

For h=0.1

 t Y T Y 0 0 1.1 0.599 0.1 0.00647 1.2 0.69515 0.2 0.0252 1.3 0.79515 0.3 0.0553 1.4 0.89926 0.4 0.0957 1.5 1.0072 0.5 0.1456 1.6 1.1189 0.6 0.20407 1.7 1.234 0.7 0.27032 1.8 1.3526 0.8 0.34363 1.9 1.4747 0.9 0.4233 2 1.6001 1 0.50882

## Graph

Therefore, the value of y(2)=1.6001.

Thus, this is the required result.