The PDE is the following: $$ω_ξ + \frac{1}{ξ}ω=f(ξ) \quad :(1)$$ where $ω=ω(ξ,n)$ and $ω,f,g \in \mathbb{C}^2$
My solution is: $$(1) \Rightarrow ξ \frac{dω}{dξ} + \frac{dξ}{dξ}ω=ξ f(ξ) \Rightarrow \frac{d}{dξ}(ξ \cdot ω)=ξ f(ξ) \Rightarrow \int\frac{d}{dξ}(ξ \cdot ω)dξ=\int ξf(ξ)dξ +g(n)$$
So we have: $$ξ \cdot ω=\int ξf(ξ)dξ +g(n) \Rightarrow ω=\frac{1}{ξ}\int ξf(ξ)dξ +\frac{1}{ξ}g(n)$$ $$\Rightarrow ω=\frac{1}{ξ}φ(ξ)+\frac{1}{ξ}g(n)$$
Is my solution correct ? Can I say that? $$φ(ξ)=\int ξf(ξ)dξ$$
