Solving a 2nd order ODE for specific parameters

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Job on 4 Jun 2011
Answered: John D'Errico on 7 Feb 2023
So i've been trying for over an hour now to solve this equation in matlab -> m(d^2x/dt^2) = −kx − β(dx/dt), and I'm new to the software so I am unable to do it, I tried it with ODE45 with no success. The goal is to solve this (find value of B) for when λ^2 − ω^2 > 0, with 2λ = β/m, and ω^2 = k/m, any help would be greatly appreciated thank you!!

Answers (1)

John D'Errico
John D'Errico on 7 Feb 2023
Did you try it? What did you try? For example, I started with what you said. YOU CANNOT use ODE45 to solve a problem where the parametters are not given numerically. However a symbolic solution almost falls out on the floor. It just drops out directly. So my guess is you thought that a numerical solver could solve such a problem. It cannot.
syms x(t)
syms B k lambda omega m
ODE = diff(x,2) == -k/m*x - B/m*diff(x)
ODE(t) = 
Next, replace k/m with omega^2.
ODE = subs(ODE,k/m,omega^2)
ODE(t) = 
And B/m is now just 2*lambda.
ODE = subs(ODE,B/m,2*lambda)
ODE(t) = 
You don't tell us anything about initial values. So a numeerical solver would not apply. But dsolve should have no serious problem.
ans = 
If I look at the solution as found, we see that if (lambda^2 - omega^2) is greater than zero, then look carefully in the radicals. Do you recognize that
lambda^2 - omega^2 = (lambda+omega)*(lambda-omega) > 0
This tells us the solutions to your problem are ALWAYS real valued as long as that inequality applies, because the value inside the radical is always positive.
Anyway, the solution seems straightforward. If you had initial values, etc., or values for those parameters, you could now get the complete solution, as just a sum of the two exponential components.

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