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I have a complicated non-linear first order homogeneous differential equation for coherent states $\psi(t)$. Via perturbation theory I obtained a linear non-homogeneous first order recursive differential equation \begin{align} (1)\qquad\frac{d}{dt}\psi^n&=-i\omega\psi^n+f^{n-1}\\ \end{align} where $n$ denotes the order of perturbation and the functions $g^{n-1}$ and $f^{n-1}$ consist of the previous solutions up to $n-1$-th order and some other functions $\phi^i(t)$ from another differential equation I solved, so they are known. I solved these via matrix calculus for $n=0,1$ and obtained the following solutions $$ (1.1)\qquad\psi^0=A^0e^{-i\omega t}\qquad \psi^1=A^1(e^{i\omega t}-e^{-3i\omega t})\, . $$ In case of $n=0$, one can directly read out the dispersion $\omega$ because the amplitude is time-independent. In case of $n=1$, it is not possible to read out the dispersion directly since it is a combination of $\omega$ and $3\omega$.

My goal is to write the first order solution such that one can read out the dispersion directly. To this end, I separated the real and imaginary parts and wrote the polar form: $$ \psi^1=2A^1sin(2\omega t)e^{icot(\omega t)}\, . $$ Now, the time dependence entered into the amplitude, which is qualitatively different from the $n=0$ case, such that the exponent is not the dispersion anymore.

To find an expression similar to the zeroth order case, I reformulated the differential equation while knowing the solutions $\psi^1$ which I then again solved and hoped for a "nice" result. Because it is $A^1=A^1(A^0)$, one cane write $\psi^0$ in terms of $\psi^1$ and consequently it is $f^1=p(t)\psi^1$. And plug this into the differential equation gives: $$ (2)\qquad\frac{d}{dt}\psi^1=-i\omega\psi^1+f=(-i\omega+p(t))\, \psi^1. $$
Here, I solved the equation and obtained: $$ (2.1)\qquad\psi^1=A'^{1}sin(2\omega t)e^{-i\omega t}\, . $$ Unfortunately, the amplitude is still time-dependent. So I again reformulated the differential equation and obtain several other forms of differential equations hoping to encounter a desired solution. One of the other differential equations had the following form ($y\equiv\psi\, ,x\equiv t$): $$ x^2 y^{\prime\prime}(x)+2xy^{\prime}(x)-2y+axy(x)+bx^2y(x)=0\, $$ where $y: \mathbb{R}\rightarrow\mathbb{C}$ and $a,b\in\mathbb{C}$. After several failed attempts, I found out about the Laplace transform. So I transformed it to: $$ (s^2+b)Y''(s)+(2s-a)Y'(s)-2Y(s)=0\, $$ where $Y(s)=\int_0^{\infty}e^{-sx}y(x)dx$.

The solutions to this equation as mentioned by @paulplusx is even worse.

Now this brings me to the following two questions:

1.) How can I get the desired structure, i.e., $\psi^1\sim e^{-i\omega' t}$? (maybe this is not even possible, after all if $z=r(t)^{i\theta(t)}$, then why should I always be able to find a $\theta'$ such that $z=const.\, e^{i\theta'(t)}$, $r$ and $\theta$ are independent.) If not possible, how can I obtain the dispersion of $\psi^1$?

2.) Is the way, I was proceeding even consistent, I have strong doubts? Meaning: If I have a recursive differential equation $(1)$ that I solved at $n$-th order. Then take the solution at $n-1$-th order and rearrange it by writing it in terms of the $n$-th order solution. Then obtain the differential equation $(2)$ and then solve it. It seems as if the two differential equations, do not have the same set of solutions because the solutions in $(2.1)$ does not solve $(1)$, even though $(2)$ was derived from $(1)$.

I hope this is not too lengthy,....

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    $\begingroup$ Where did you get this question? I am afraid you might not like the answer. $\endgroup$ – paulplusx Sep 21 '18 at 16:12
  • $\begingroup$ Well, I had two coupled non-linear non homogeneous first order differential equations, and after several steps, decoupled them into the above differential equation...... $\endgroup$ – mr. curious Sep 21 '18 at 16:52
  • $\begingroup$ why don't you add those as well to the question as an added context. $\endgroup$ – paulplusx Sep 21 '18 at 17:43
  • $\begingroup$ Ok, I'll edit the question and provide a full context.. $\endgroup$ – mr. curious Sep 24 '18 at 12:08
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    $\begingroup$ Question is now edited @paulplusx $\endgroup$ – mr. curious Sep 24 '18 at 13:28

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