Finding a function that satisfies $xf^2(x) = \int_0^x f(t)\,dt$ I am trying to find a function $f$ for $x>0$ which satisfies the following:
$$[f(x)]^2=\frac1x \int_0^xf(t)dt$$
Multiply by $x$
$$x[f(x)]^2=\int_0^xf(t)dt$$
Take the derivative of both sides
$$2x[f(x)f'(x)]) +[f(x)]^2 =f(x) $$
$$2xf'(x)+f(x)=1$$
Solve ODE
$$2x\frac{df(x)}{dx}+f(x)=1$$
$$\frac{df(x)}{dx}=\frac{1-f(x)}{2x}$$
$$\frac{2\frac{df(x)}{dx}}{1-f(x)}=\frac1x$$
Right side becomes $ln(x)+C$, since $x>0$, not sure about the left side
 A: Hint:
for $x\ne 0$ write the equation as:
$$
x[f(x)]^2= \int_0^xf(t)dt
$$
and derive:
$$
[f(x)]^2+2xf(x)f'(x)=f(x)
$$
that for $f(x)\ne 0$
becomes:
$$
2xf'(x)+f(x)=1
$$
can you solve this? ( it's a separable ode)
A: Let $F(x)$ be the integral of $f(x)$. Then the equation is rewritten as the ODE
$$xF'^2(x)=F(x)$$ or
$$\frac{F'(x)}{\sqrt{F(x)}}=\pm\frac1{\sqrt x}$$ with the initial condition $F(0)=0$.
After integration,
$$\sqrt{F(x)}=C\pm\sqrt x$$ and as $C=0$,
$$F(x)=\left(\pm\sqrt x\right)^2=x,\\
f(x)=1.$$
Unfortunately all this development is flawed, as the values at $x=0$ are undefined.
A: When you divide by f(x),  you have f(x)=0 
Or 
The ode f'(x) +f(x) /2x=1/2x
And if you solve it with an analysis,  it seems to be f(x) =1 the solution. 
You know with the first equation that f is continuous.  So 2 solutions f is the zero function or the one functions. 
A: using both side derivative and leibniz rule you will get {f(x)-1}*{f(x)+2f'(x).x}=0 so ,in this second case you have a function f(x) which is a function of its derivative f'(x) and -2x so the function can be [- e^x * x^2]
