BvU said:
is satisfied for ##y(x) = 0##
With nonzero initial conditions it looks like a candidate for numerics to me:
View attachment 220733
[edit] ah,
@fresh_42 did the same but faster.
I'd go for a runge kutta and see what it brings, but I suppose ODE45 is equivalent
fresh_42 said:
I find the plots in the first and last row give a good answer:
http://www.wolframalpha.com/input/?i=y'''+++(1-x^2)y=0
You only have to draw that ##y=1## line at ##y=0## for the first row plot. It doesn't appear to be solvable analytically. It is already a complicated formula without the ##1-x^2## factor. Maybe you can get an expression with
Lambert's W-function, but note, that this isn't a function.
Thanks for answers.
I have tried ODE45:
fun = @(x,y)[y(2);y(3);y(1)*(2+x^2)];
y0 = [0 0 0];
xspan = [0 5];
[X,Y] = ode45(fun,xspan,y0);
plot(X,Y(:,1),X,Y(:,2),X,Y(:,3));
It is a rather boring plot, and it doesn't appear as what I was looking for.
About the Runge-Kutta, can you send me a link to an example ?
@Fresh42, I got that result earlier indeed, however, I got an even nicer results (but only once!):
Z = (1./3).*(exp(x) + 2*exp(-x/2)*cos(w))
where w = (sqrt(3)*x)./2Note the peculiar properties of this function, it is the same as it third derivative, its sixth derivative and its ninth derivative. It cycles back every third derivation. Unusual for such a long form, I expect.
I am however NOT able to reproduce that solution in Wolfram alpha, as it appeared once in the result as a solution, in addition to the plots. Ttrying to test it in the ODE, with IC y(0)=0, y'(0)=0, y''(0)=0 does not work. So I wonder where that Z function came from..
Thanks