Graduate Solve a non-linear ODE of third order

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The discussion centers on solving the non-linear third-order ordinary differential equation (ODE) y''' + (1-x^2)y=0 with specified initial conditions. Attempts to find an analytical solution using Wolfram Alpha yielded results that could not be reproduced, and MATLAB's ODE45 provided only plots without satisfactory solutions. Participants suggest that the equation may not be solvable analytically and recommend numerical methods like Runge-Kutta for further exploration. One user mentions an intriguing function Z that appeared once in their results but could not be replicated, highlighting the complexities of the problem. Overall, the consensus leans towards numerical approaches due to the equation's challenging nature.
SemM
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Hi, I tried to solve the following in Wolfram alpha:

y''' + (1-x^2)y=0

y(0)=0
y'(0)=0
y''(0)=0

however, I got answer which cannot be reproduced (even at wolfram pages).

I have tried ODE45 in MATLAB, but it only gives a plot.

Is there any way to solve this analytically or numerically to give a solution?

Thanks
 
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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.
 
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SemM said:
y''' + (1-x^2)y=0
y(0)=0
y'(0)=0
y''(0)=0
is satisfied for ##y(x) = 0## :smile:

With nonzero initial conditions it looks like a candidate for numerics to me:

upload_2018-2-20_17-52-42.png


[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
 

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BvU said:
is satisfied for ##y(x) = 0## :smile:

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
 
The last plot on the WolframAlpha page has a big advantage: It gives the correct impression on possible solutions. As you've fixed your initial conditions around ##x=0##, the solution is pretty well determined at this point and the further away, the higher are the uncertainties. This demonstrates once more the locality of differentiation. This means in return, that only in a neighborhood of this point you can get a good convergence of the power series.
 
fresh_42 said:
The last plot on the WolframAlpha page has a big advantage: It gives the correct impression on possible solutions. As you've fixed your initial conditions around ##x=0##, the solution is pretty well determined at this point and the further away, the higher are the uncertainties. This demonstrates once more the locality of differentiation. This means in return, that only in a neighborhood of this point you can get a good convergence of the power series.
:(

Thanks Fresh!
 

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