Solve nonlinear 2nd order ODE numerically

1 visualizzazione (ultimi 30 giorni)
I need to solve the following nonlinear 2nd order ODE, that is, find such that
1-x=-\frac{y''(x)}{(1+(y'(x))^2)^{{3/2}}
I tried using
>> syms y(x)
>> ode = -diff(y,x,2)/(1+(diff(y,x))^2)^(3/2) == 1-x;
>> ySol(x) = dsolve(ode)
but it doesn't work since apparently there is no anaylitical solution (if I rearrange the terms it does find a system of complex solutions, but I think the it is not right).
Isn't there a command to solve ODEs numerically? I am expeting something like the family of plots from here https://www.wolframalpha.com/input?i=f%27%27%28t%29%2F%28%281%2B%28f%27%28t%29%29%5E2%29%5E%283%2F2%29%29+%3D+-%281-0.25t%29
Many thanks oin advance!
  2 Commenti
Torsten
Torsten il 28 Lug 2022
What are your initial/boundary conditions for y ?
Lucas
Lucas il 29 Lug 2022
My idea was to screen these conditions to find one that satisfies my problem.

Accedi per commentare.

Risposta accettata

Sam Chak
Sam Chak il 28 Lug 2022
You can follow the example here
and try something like this:
tspan = [0 1.15];
y0 = [1 0]; % initial condition
[t,y] = ode45(@(t, y) odefcn(t, y), tspan, y0);
plot(t, y(:,1)), grid on, xlabel('t')
function dydt = odefcn(t, y)
dydt = zeros(2,1);
c = 0.25;
dydt(1) = y(2);
dydt(2) = - (1 - c*t)*(1 + y(2)^2)^(3/2);
end

Più risposte (2)

James Tursa
James Tursa il 28 Lug 2022

MOSLI KARIM
MOSLI KARIM il 12 Ago 2022
function pvb_pr13
tspan=[0 1.5];
y0=[1 0];
[x,y]=ode45(@fct,tspan,y0);
figure(1)
hold on
plot(x,y(:,1),'r-')
grid on
function yp=fct(x,y)
c=0.25;
yp=[y(2);-(1-c*x)*((1+(y(2))^2)^(3/2))];
end
end

Categorie

Scopri di più su Symbolic Math Toolbox in Help Center e File Exchange

Prodotti


Release

R2018a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by