ordinary differential equation
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i have an equation d(theta)/dt=((P*L^2)/(E*I))*sin(theta*D/L)
L=55.863; D=15.484; E=200*10^9; I=138; P=440*10^6
The initial theta at 1198 years is 0.010degrees.
i have to find theta from the year 1198 to 1990
can someone please help me.its urgent
回答（1 个）
Grzegorz Knor
2011-9-29
Use ode45:
L=55.863; D=15.484; E=200*10^9; I=138; P=440*10^6;
ode45(@(t,theta)((P*L^2)/(E*I))*sin(theta*D/L),[1198 1990],0.01)
7 个评论
12ar.af
2011-9-29
Grzegorz Knor
2011-9-29
Just add two output arguments, and colon in tspan:
[T THETA] = ode45(@(t,theta)((P*L^2)/(E*I))*sin(theta*D/L),[1198:1990],0.01)
Grzegorz Knor
2011-9-29
I don't recomend you inline function, but solution is as follows:
fun = inline('((440*10^6*55.863^2)/(200*10^9*138))*sin(theta*15.484/55.863)','t','theta');
ode45(fun,[1198 1990],0.01)
12ar.af
2011-9-29
Grzegorz Knor
2011-9-29
Solve your equation from 1198 to 1990+200:
[T Y] = ode45(@(t,theta)((P*L^2)/(E*I))*sin(theta*D/L),[1198 1990+200],0.01);
disp(sprintf('year: %i, theta: %.3d',T(end),Y(end)))
12ar.af
2011-9-29
Grzegorz Knor
2011-9-29
Yes, read one of my comment above.
[T THETA] = ode45(@(t,theta)((P*L^2)/(E*I))*sin(theta*D/L),[1198:1990],0.01);
disp([T THETA])
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