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 already have $2x F(x) = F'(x)$ which is preferred over using $f'$ because you have only given that $f$ is continuous, but we don't know whether where $f'$.
To solve the equation for $F$ is possible because $F'/F$ is just the so-called logarithmic derivative of $F$: Just integrate on both sides over an inteval where $F(x)\neq0$:
$$2x=\frac{F'(x)}{F(x)}$$
thus
$$\int_a^b \!2x\, dx = \int_a^b\frac{F'(x)}{F(x)} dx $$
Then substitute $y=F(x)$ and $dy = F'(x) dx$:



$$b^2-a^2 = \int_{F(a)}^{F(b)} \frac{dy}{y} = \ln|y|\Bigg|_{y=F(a)}^{y=F(b)} = \ln|F(b)| - \ln|F(a)| $$
or
$$\ln|F(x)| = x^2 + C$$
$$|F(x)| = \exp(x^2+C) = K\cdot\exp (x^2)$$
and finally absorbing the $\pm$ from the absolute value into the integration constant $K$.
$$\begin{align}
F(x) &= K\cdot\exp (x^2) \\
F'(x) &= f(x) = 2Kx\cdot\exp (x^2)
\end{align}$$
