Linear ODEs: Resonance and Repeated Roots

[georgesadler]

Professor MathTheBeautiful, thank you for explaining and analyzing Linear ODEs with Resonance and Repeated Roots. This is an error free video/lecture on YouTube.

[ManojKumar-cjoj]

at resonance and repeated roots cond multiply our guess with t^(n+2)

[andrewszymczak]

I do have a good motivation for the guess of t^2 exp(-3t). Assuming smoothness of u, differentiate both sides

(i) u'' + 6u' + 9u = exp(-3t)
(ii) u''' + 6u'' + 9u' = -3exp(-3t)

Now take the combination 3(i) + (ii). You are left with a 3rd order homogenous equation

(iii) (u''' + 6u'' + 9u') + 3 (u'' + 6u' + 9u) = 0

If (i) has a characteristic polynomial p(λ) = (λ+3)^2, then (ii) has characteristic polynomial λ p(λ). So (iii) = 3(i)+(ii) has characteristic polynomial (λ+3) p(λ) = (λ+3)^3. Any solution u of our original ODE must also be a solution of (iii), so we find that u = (a + bt + ct^2) exp(-3t) where a, b, c can be determined by the initial conditions and by equation (i) itself. This generalizes to any RHS with terms of the form q(t) exp(at). For example, the ODE u' - 3u = (1 + bt + ct^2) exp(t) + (1 - t) exp(3t) + t^4 exp(-2t) can be converted into the 11th order homogeneous ODE with characteristic polynomial (λ-1)^3 (λ-3)^3 (λ+2)^5.

[scitwi]

I have the following ODE: (x²–9)·y' – (x³+5x²–9x–45)·y = exp(2x)
and I already found the general solution for the homogenized version: y = C·exp(x²/2 + 5x)
But now I have a trouble with matching the non-homogeneous part :q Can anyone help me?
Also I'd like to know how to solve higher-order ODEs with NON-constant coefficients, e.g. y" = x²·y

[prafulldani]

According to my understanding, for resonance, as time goes to infinity, u (displacement wrt time) should also go to infinity.
However in case of the solution of the above example : u = [1/2 t^2 e^(-3t)] + [C1 t e^(-3t)] + [C2 e^(-3t)], due to all terms being negative exponents, u will go to zero as time goes to infinity. How would that be considered as resonance ?