Solving one dimensional first order equation which is periodic

Question

Does below equation have an answer? $\theta$ changes from $0$ to $2 \pi$ and $a$ and $b$ constants:

$$(\sin \theta + a \sin 2\theta )F(\theta) - \partial_{\theta} F(\theta)-b=0$$

How can I solve it for $F(\theta)$?

Answer 1

Hint: This is a first order linear ODE

$$\dfrac{dF}{d\theta}=(\sin \theta +a\sin 2\theta)F-b.$$

There is a closed form solution for it.

Use separation of variables for the homogeneous solution

$$\dfrac{dF_h}{d\theta}=(\sin \theta +a\sin 2\theta)F_h \implies \dfrac{dF_h}{F_h}=(\sin \theta +a\sin 2\theta)d\theta$$

$$\implies \ln F_h=\int(\sin \theta +a\sin 2\theta)d\theta+\ln c$$ $$\implies F_h=c\exp\left(\int(\sin \theta +a\sin 2\theta)d\theta\right)$$ $$\implies F_h=c\exp\left(-\cos(\theta)-\frac a2\cos 2\theta\right).$$ and add the particular solution to it.

In order to determine the particular solution $F_p$ you assume $F_p=k(\theta)F_h$ as ansatz.

Note that the ansatz $F_p=k(\theta)F_h$ for the ODE will give you:

$$\dfrac{dk}{d\theta}F_h+c\dfrac{dF_h}{d\theta}=(\sin \theta + a\sin 2\theta)cF_h-b$$

$$\implies \dfrac{dk}{d\theta}F_h+c(\sin \theta + a\sin 2\theta)F_h=(\sin \theta + a\sin 2\theta)cF_h-b$$

$$\implies \dfrac{dk}{d\theta}F_h=-b \implies k(\theta)=-\int\frac{b}{F_h}d\theta$$

$$\implies k(\theta)=-\int\frac{b}{c\exp\left(-\cos(\theta)-\frac a2\cos 2\theta\right)}d\theta=-\frac bc\int\exp\left(\cos(\theta)+\frac a2\cos 2\theta\right)d\theta.$$

I think the last integral is not solvable by a closed form solution of elementary functions.