Differential equation $\int_0^1f(tx)dt=nf(x),x>0$. 
$$\int_0^1f(tx)dt=nf(x),x>0$$


I tried the following:


*

*I know this is a bad move:
$$\int_0^1f(t(x/t))dt=nf(x/t)\implies f(x)=nf(x/t)$$

*I know this too is a bad move: $$nf(x)=\int_0^1f(tx)dt=\int_0^1\int_0^1f(t^2x)dtdt=\ldots=\lim_{j\to\infty}\underbrace{\int_0^1\int_0^1\cdots\int_0^1}_{j\;times}f(t^jx)(dt)^j$$

*$$nf'(x)=\int_0^1tf'(tx)dt$$ which is also getting me nowhere.

*$$\lim_{j\to\infty}\sum_{k=0}^{j}\frac1jf(xk/j)=nf(x)\implies \lim_{j\to\infty}f(x)+f(2x)+\ldots+f(jx)=njf(jx)$$ which probably too is useless.


Additional Question:
Explaination on why the first two are bad/wrong/a fallacy.
 A: Hint Note that $$x\int_0^1f(tx)dt=\int_0^1 f(tx)d(tx)=\int_0^x f(t)dt=xnf(x)$$
A: Following @Pedro Tamaroff's excellent hint, you have $\int_0 ^x f(t) \, dt = x (n f(x) )$. Hence,
$$
f(x) = nf(x) + xn f'(x) 
$$
and so $f(x) = A x^r$. That is, introducing the integrating factor
$$
\mu(x) = e^{ \frac{n-1}{n} \int \frac{1}{x} \, dx} = e^{ \frac{n-1}{n} \ln x} = x^{ \frac{n-1}{n} }, 
$$
we have 
$$
0 = \frac{(n-1)}{n} x^{ \frac{n-1}{n} - 1 } f(x) + x^{ \frac{n-1}{n} } f'(x) = \frac{d}{dx} ( x^{\frac{n-1}{n}} f(x) ).
$$
Thus, 
$$
f(x) = A x^{ \frac{1-n}{n} }.
$$
We see by inspection that the function satisfies the original condition: 
$$
n \cdot A x^{ \frac{1-n}{n} }
=
\int_0 ^1 A (xt)^{ \frac{1-n}{n} } \, d t 
=
A  x^{ \frac{1-n}{n} } \cdot \int_0 ^1 t^{ \frac{1-n}{n} } \, d t 
=
A  x^{ \frac{1-n}{n} } \cdot n .
$$
A: Following Karl's Comment I did this:
$$nf'(x)=\int_0^1tf'(tx)dt=f(x)/x-\int_0^1f(tx)/x\;.dt\implies xf'(x)=\frac{1-n}nf(x)$$
Or:
$$f'(x)/f(x)=\frac{1-n}{nx}\implies (f(x))=c.x^{\frac{1-n}n}$$
