You have to solve the non-linear time variant differential equation:
(cos(y)+2)*y''+atan(y)*sin(0.01*t)^2+y*sin(0.5*t)=0
with initial condition as input of the function
[y(0);y'(0)]=[a,b]
and return the time 'te' when 'y' crosses zero from the negatives values to the positive values.
te=zero_crossing(a,b)
tip: use the 'Events' option in odeset to detect the crossing, stop integration and get the time 'te'. See how to use it here : http://www.mathworks.com/help/matlab/math/ode-event-location.html?searchHighlight=ballode
Additional info:
- in the test case, the solver ode45 is used, with options 'RelTol'=1e-8 and 'AbsTol'=1e-10.
- because you don't know in advance the final time of the crossing, you can solve again further in time if no crossing is detected, starting from the point of the final time of previous integration.
Solution Stats
Problem Comments
Solution Comments
Show commentsProblem Recent Solvers24
Suggested Problems
-
Replace NaNs with the number that appears to its left in the row.
3069 Solvers
-
Project Euler: Problem 9, Pythagorean numbers
1400 Solvers
-
Back to basics 12 - Input Arguments
626 Solvers
-
Make an awesome ramp for a tiny motorcycle stuntman
768 Solvers
-
Solving Quadratic Equations (Version 1)
506 Solvers
Problem Tags
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
It looks like [y;y']=[a,b] would mean y(0)=a and y'(0)=b, but a first order differential equation should have one initial condition y(t0)=y0.
Thanks Tim, yes, I did a mistake in the equation, its a 2nd order.