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DivGradCurl

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[tex]y^{\prime} = \frac{3t^2}{3y^2 - 4} \mbox{,} \qquad y(1) = 0\mbox{.}[/tex]

(a) Use Euler's method with [tex]h=0.1[/tex] to obtain approximate values of the solution at [tex]t=1.2\mbox{, }1.4\mbox{, }1.6\mbox{, and } 1.8[/tex].

(b) Repeat part (a) with [tex]h=0.05[/tex].

(c) Compare the results of parts (a) and (b). Note that they are reasonably close for [tex]t=1.2\mbox{, }1.4\mbox{, and }1.6[/tex], but are quite different for [tex]t=1.8[/tex]. Also note (from the differential equation) that the line tangent to the solution is parallel to the y-axis when [tex]y=\pm 2/\sqrt{3}\approx \pm 1.155[/tex]. Explain how this might cause such a difference in the calculated values.

**My work: (PARTS A & B ARE OK)**(a)

Approximate values of the solution, which were found using the Euler method, follow below:

[tex]\begin{equation*}\begin{array}{|c|r|} \hline \multicolumn{1}{|c|}{t} & \multicolumn{1}{c|}{h=0.1} \\ \hline 1.2 & -0.166134 \\ 1.4 & -0.410872 \\ 1.6 & -0.804660 \\ 1.8 & 4.15867 \\ \hline \end{array} \end{equation*}[/tex]

(b)

Approximate values of the solution, which were found using the Euler method, follow below:

[tex]\begin{equation*} \begin{array}{|c|r|r|} \hline \multicolumn{1}{|c|}{t} & \multicolumn{1}{c|}{h=0.1} & \multicolumn{1}{c|}{h=0.05} \\ \hline 1.2 & -0.166134 & -0.174652 \\ 1.4 & -0.410872 & -0.434238 \\ 1.6 & -0.804660 & -0.889140 \\ 1.8 & 4.15867 & -3.09810 \\ \hline \end{array} \end{equation*}[/tex]

(c)

I've plotted the direction field with the solution. You can find it at: http://myplot.cjb.net

The graph shows that the line tangent to the solution is parallel to the y-axis when [tex]y=\pm 2/\sqrt{3}\approx \pm 1.155[/tex]. That's when the denominator of the fraction in the given differential equation equals zero.

My approximate solution (found with the aid of mathematica's NDSolve) is valid for [tex]t\leq 1.5978094538975247[/tex]. So, it seems to me that the values found should not be even taken into consideration from that point on. However, if I were to consider values beyond [tex]t = 1.5978094538975247[/tex], I'd say that the significant difference in step size may lead each particular approximation to tangent lines with completely different slopes, which ultimately gives those values. I'm not sure.

Any help is highly appreciated.