Hi everybody,
how can I solve this equation?
I found problems because of the derivative function squared. I need to collect the term derivative not squared...
Thanks
Bye
Solved! Go to Solution.
You can turn the non-linear y'(t)^2 + 4y'(t) = 5ty(t) into two simultaneous linear ode's:
Alan
Hmm squared.... Seems more like to the 3rd power.
You have (y'(t))^2 and mutiply that with 4 y'(t), that gives 4*(y'(t))^3...
Do you need a symbolic solution, or can you settle for a numeric approximation.
Either way: what are the initial values? [ y(0), y'(0)]
Luc
Found it.
Try y(t)=+/-(1/8)*sqrt(10)*t^2.
The third solution is y(t)=0, but I guess you weren't looking for that.
Success!
Luc
Here's a possible workaround for a numerical solution, both Prime and real Mathcad:
I'm so sorry because I wrote "(y'(t)^2) * 4y'(t)" instead of "(y'(t)^2) + 4y'(t)" that it changes a lot...
From Mathcad's documentation of "odesolve":
The ODE must be linear in its highest derivative term,
Does y(t) describe a physical phenomenon as a function of time? Or is this a made-up DE?
Luc
You can turn the non-linear y'(t)^2 + 4y'(t) = 5ty(t) into two simultaneous linear ode's:
Alan
How to find y(0)=1?
z(0)=0, How about z(0)=-4?
Tokoro.
y(0)=1 was an arbitrary choice on my part, as no initial conditions were specified in the original question.
z(0)=-4 gives rise to a rapidly decreasing y with time (and with y(0)=1). If this is what is expected from the model then fine! Personally, I think it’s probably safer to stick with z(0)=0.
Alan