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
2 Comments
Solution Comments
Show comments
Loading...
Problem Recent Solvers24
Suggested Problems
-
The Goldbach Conjecture, Part 2
2413 Solvers
-
524 Solvers
-
Find Index of maximum Value and maximum Value of a vector
167 Solvers
-
636 Solvers
-
10246 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.