Find all continuous functions $f:[0, \infty) \to (0, \infty)$ such that $(\forall x > 0)$ $ 2x \int_0^x f(t) dt = f(x) $ Find all continuous functions $f:[0, \infty) \to (0, \infty)$ such that $(\forall x > 0)$ $$ 2x \int_0^x f(t) dt = f(x) $$
My work: First, for $x \not = 0$ given equality can be written down as
$$\int_0^x f(t) dt = \frac{f(x)}{2x}$$
When we differentiate that we get
$$f(x) = \frac{1}{2} \frac{xf'(x) - f(x)}{x^2}$$
which is
$$2x^2 f(x) = f'(x) \cdot x - f(x)$$
My idea was to try to find function whose derivative is $2x^2 f(x) - f'(x) \cdot x + f(x)$, but closest I have got was this function:
$$e^{x^2} \cdot x \cdot f(x)$$
but it is not exactly what I am looking for.
Second, what I have tried, was the following:
Let $$ F(x) = \int_0^x f(t) dt$$
Task condition then becomes
$$2x F(x) = F'(x)$$
I tried to integrate $F(x)$ using partial integration I haven't got anything useful.
Can someone please give me some hint or any other help? Thanks!

You have
$$
2x^2 f(x) = f'(x) \cdot x - f(x)
$$
and therefore
$$
f'(x) = \frac{2x^2+1}{x} f(x)
$$
and you can change variables to $u = \ln f(x)$ and $u' = f'(x)/f(x)$ with a much easier result. Can you finish it?
UPDATE
For your question, here is the check on Wolfram Alpha, which seems to agree with you result.
