# Second order homogenous differential equation with variable coefficients

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,....

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