Solve a non-linear ODE of third order

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SUMMARY

The discussion centers on solving the third-order non-linear ordinary differential equation (ODE) given by y''' + (1-x^2)y=0 with initial conditions y(0)=0, y'(0)=0, and y''(0)=0. Users report that analytical solutions are elusive, with suggestions to utilize numerical methods such as MATLAB's ODE45 and Runge-Kutta techniques. One participant mentions a potential solution involving the Lambert W-function, but acknowledges its complexity. The consensus is that numerical approaches are more viable for this equation due to its intricate nature.

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  • Understanding of third-order ordinary differential equations
  • Familiarity with MATLAB ODE45 function
  • Knowledge of numerical methods, specifically Runge-Kutta
  • Basic concepts of power series and their convergence
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  • Research the implementation of Runge-Kutta methods for solving ODEs
  • Explore the Lambert W-function and its applications in differential equations
  • Study the properties of power series and their convergence in the context of ODEs
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Mathematicians, engineers, and students working with differential equations, particularly those interested in numerical solutions and the complexities of non-linear ODEs.

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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