Finding $f$ from $f$'s derivative

Question

I have a function $f:\mathbb{R}\rightarrow\mathbb{R}$ which can derives with $f(0)=0$ and: $$2xf(x)=(x^2+1)f(x)-(x^2+1)f'(x)+1$$ I want to find the $f$, so I am thinking of using the theorem that: $$F'(x)=G'(x)\iff F(x)=G(x)+c$$ But I don't know to apply this exactly in my case. Any ideas?

Answer 1

\begin{align} 2xf(x)&=(x^2+1)f(x)-(x^2+1)f'(x)+1 \\ 2xf(x)+(x^2+1)f'(x)&=(x^2+1)f(x)+1 \\ \end{align} Note that \begin{align} \int 2xf(x)+(x^2+1)f'(x)dx=(x^2+1)f(x) \end{align} by product rule.

Answer 2

On rearrangement, you will get a linear differential equation. Writing $f(x)$ as $y$.

$$(x^2+1)y'-(x-1)^2y=1$$

Using integration factor $\exp\left(\int \tfrac{-(x-1)^2}{x^2+1}dx\right) = e^{-x}(x^2+1)$, the solution is given as:

$$y(e^{-x})(x^2+1)=\int e^{-x} dx=-e^{-x}+c$$