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Question about third-order ODE.

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Yusuke Nakamura
Yusuke Nakamura 2015 年 4 月 26 日
コメント済み: Star Strider 2015 年 4 月 26 日
Q: The exact solution to the differential equation (t^2/0.357)f'''-(1/0.3)f'+f = t^0.3+5 is given in the figure below.
Starting with an initial condition f(2) = 6.2311, f'(2) = 0.1847 and f''(2) = -0.0646, Euler's method will over predict the solution when computing f(4).
From my following codes, I found out the graph is over predict with Euler's method, but I don't know how to get draw the exactly same graph as shown above. How could I plot the same graph?
function du = dfdeta(t,f)
du = [f(2); f(3); 1.19/t^2*f(2)-0.357/t^2*f(1)+0.357*t^(-1.7)+1.785/t^2];
and
[eta,f] = eulsys(@dfdeta,[2 4],[6.2311 0.1847 -0.0646],.01);
  1 件のコメント
Star Strider
Star Strider 2015 年 4 月 26 日
The Euler Method is an approximation. As the Wikipedia article (link) discusses, you cannot expect it to match an exact solution.

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