Find $f'(x)$ given that $2f(x) + 3f(-x) = x^2 -x + 1 \space \space\forall x \in \mathbb{R}$

Question

Find $f'(x)$ given that $2f(x) + 3f(-x) = x^2 -x + 1 \space \space\forall x \in \mathbb{R}$

My solution

Substitute $x \rightarrow -x$ to get $2f(-x) + 3f(x) = x^2 +x + 1$. We now have 2 equations in 2 unknowns: $f(x)$ and $f(-x)$.

Solve for $f(x)$ to get $f(x) = \frac{1}{5}(x^2 + 5x + 1) \implies f'(x) = \frac{1}{5}(2x + 5)$

Please advice

Is there a more insightful solution? Somehow, my solution seems a bit mechanical.

Answer 1

I think this maybe simple a little:

$$-2f'(-x)+3f'(x)=2x+1\tag{1}$$ and since $$2f(x)+3f(-x)=x^2-x+1$$ so $$2f'(x)-3f'(-x)=2x-1\tag{2}$$ $(2)\times 2-(1)\times 3$ so $$f'(x)=\dfrac{1}{5}(2x+5)$$