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Q5.327E
ExpertverifiedGeneralized Blasius Equation. H. Blasius, in his study of the laminar flow of a fluid, encountered an equation of the form $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{yy}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{(}\mathbf{y}\mathbf{\text{'}}{\mathbf{)}}^{\mathbf{2}}{\mathbf{}}{\mathbf{1}}$. 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 RungeKutta method, and the result is y(2)=1.6001.
Here the equation is $\mathrm{y}\text{'}\text{'}\text{'}+\mathrm{yy}\text{'}\text{'}=(\mathrm{y}\text{'}{)}^{2}1$.
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:
${\mathrm{x}}_{1}\left(0\right)=\mathrm{y}\left(0\right)=0\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{2}\left(0\right)=\mathrm{y}\text{'}\left(0\right)=0\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{3}\left(0\right)=\mathrm{y}\text{'}\text{'}\left(0\right)=1.32824$
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 


Therefore, the value of y(2)=1.6001.
Thus, this is the required result.
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