using ode 45 to estimate the derivative

Question: Use ode45 to estimate y'(3), where y is the solution to the initial value problem y" + (1/t)*y = 0 ; y(0) = 0, y'(0) = 2. Note I'm asking for an estimate of the derivative, not the function itself.

Attempted code:

ode = @(t, y) [y(2) ; (1/t)*y(1)];
[t, y] = ode45(ode, [-1, 3], [0, 2]);
y(end, 1);

I do not think this actual estimating the derivative y'(3)?

[Torsten]

1. y''=-1/t*y, thus

ode = @(t, y) [y(2) ; -(1/t)*y(1)];

2. Why do you start integration at t=-1 if your initial conditions are given at t=0 ?

3. Your y(1) is the solution of the ODE y''+1/t*y=0, your y(2) is its derivative ... So to estimate the derivative at t=3, you will have to evaluate y(2) there.

Best wishes

Torsten.

Answer 1

[Azzi Abdelmalek]

ode=@(t, x) [x(2) ; -(1/t)*x(1)];
[t, x] = ode45(ode, [-1, 3], [0, 2]);
y=x(:,1)
dy=x(:,2)
out=dy(3)