how to find f"(0) where f=y(1),f'=y(2),f"=y(3)

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MINATI
MINATI am 16 Apr. 2018
Kommentiert: MINATI am 16 Apr. 2018
dydx=[y(2); y(3); 2*a1*y(2)*(y(2)+y(5))-a1*y(3)*(y(1)+y(4));
y(5); y(6); 2*a1*y(5)*(y(2)+y(5))-a1*y(6)*(y(1)+y(4));
y(8); A*Pr*a2*y(7)*(y(2)+y(5))-Pr*a2*y(8)*(y(1)+y(4))];
res=[ya(1); ya(2)-1; ya(4); ya(5)-p; ya(7)-1; yb(2); yb(5); yb(7)];
  2 Kommentare
Walter Roberson
Walter Roberson am 16 Apr. 2018
Is res your initial conditions?
MINATI
MINATI am 16 Apr. 2018
yes

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Akzeptierte Antwort

Torsten
Torsten am 16 Apr. 2018
f0 = deval(sol,0);
f0(3)

Best wishes

Torsten.

Weitere Antworten (1)

Walter Roberson
Walter Roberson am 16 Apr. 2018
%construct initial values any way suitable for you
a1 = rand();
a2 = rand();
A = rand();
Pr = rand();
ya = rand(1,7);
yb = rand(1,7);
p = rand();
%now find the values
dxdy = @(t,y) [y(2); y(3); 2*a1*y(2)*(y(2)+y(5))-a1*y(3)*(y(1)+y(4));
y(5); y(6); 2*a1*y(5)*(y(2)+y(5))-a1*y(6)*(y(1)+y(4));
y(8); A*Pr*a2*y(7)*(y(2)+y(5))-Pr*a2*y(8)*(y(1)+y(4))];
res=[ya(1); ya(2)-1; ya(4); ya(5)-p; ya(7)-1; yb(2); yb(5); yb(7)];
tspan = [0 5];
[t, y] = ode45(dxdy, tspan, res);
f = y(:,1);
f_prime = y(:,2);
f_primeprime = y(:,3);

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