Integral Equation-Volterra 2nd kind Given 
$$
f(x) = \sqrt{x} + \lambda\int_0^x\sqrt{xy}f(y)dy.
$$
I found the derivative of $f$ to be
$$
f'(x) = \frac{1}{2\sqrt{x}} + \lambda\left(xf(x) + \frac{1}{2\sqrt{x}}\int_0^x\sqrt{y}f(y)dy\right).
$$
I then solved for 
$$
\int_0^x\sqrt{y}f(y)dy = \frac{f(x) - \sqrt{x}}{\lambda\sqrt{x}}
$$
and plugged it into to $f'(x)$ to get
$$
f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{2\lambda\sqrt{x}} + \left(\lambda x + \frac{1}{2\lambda x}\right)f(x).
$$
Now what can I do to solve the problem?
 A: The equation
$f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{2\lambda\sqrt{x}} + \left(\lambda x + \frac{1}{2\lambda x}\right)f(x), \tag{1}$
being a first order, linear ordinary differential equation, is susceptible to standard methods.  First re-arrange things a little bit:
$f'(x) - \left(\lambda x + \frac{1}{2\lambda x}\right)f(x)= \frac{1}{2\sqrt{x}} - \frac{1}{2\lambda\sqrt{x}} , \tag{2}$
then write down the solution to the homogeneous part
$f'(x) - \left(\lambda x + \frac{1}{2\lambda x}\right)f(x)= 0; \tag{3}$
it is
$f(x) = f(x_0) \exp(\int_{x_0}^x (\lambda s + \frac{1}{2 \lambda s})ds); \tag{4}$
making the assumption that that $0 < x_0 \le x$ allows the integral to be evaluated explicitly:
$\int_{x_0}^x (\lambda s + \frac{1}{2 \lambda s})ds  = \frac{1}{2} \lambda (x^2 - x_0^2) + \frac{1}{2 \lambda} (\ln x - \ln x_0)$
$= \frac{1}{2} \lambda (x^2 - x_0^2) + (\ln x^{(2 \lambda)^{-1}} - \ln x_0^{(2 \lambda)^{-1}}), \tag{5}$
whence
$f(x) = f(x_0) (\frac{x}{x_0})^{(2 \lambda)^{-1}}\exp{(\frac{1}{2} \lambda (x^2 - x_0^2))}; \tag{6}$
such is the solution to the homogeneous equation (3).  Note that I am being somewhat cavalier insofar as I am neglecting to thoroughly analyze the ranges of $x$ and the parameter $\lambda$ for which these mathematical machinations are valid; but it seems at first glance that $0 < x_0 \le x$ and $\lambda \ne 0$ are sufficient; I am leaving the finer details of such points to my readers.  We next obtain the solution to the inhomogeneous equation (2) under the assumption that $f(x_0) = 0$; one can in fact verify by direct differentiation that
$f(x) = \exp{(\int_{x_0}^x (\lambda s + \frac{1}{2 \lambda s})ds)}\int_{x_0}^x \exp{(-\int_{x_0}^r (\lambda s + \frac{1}{2 \lambda s})ds)}(\frac{1}{2\sqrt{r}} - \frac{1}{2\lambda\sqrt{r}})dr; \tag{7}$
the complete solution satisfying (1), (2) with initial value $f(x_0)$ is thus
$f(x)$
$= \exp{(\int_{x_0}^x (\lambda s + \frac{1}{2 \lambda s})ds)}(f(x_0) + \int_{x_0}^x \exp{(-\int_{x_0}^r (\lambda s + \frac{1}{2 \lambda s})ds)}(\frac{1}{2\sqrt{r}} - \frac{1}{2\lambda\sqrt{r}})dr), \tag{8}$
as may be readily verified by direct differentiation.  Having taken it thus far, I'm going to leave the evaluation of 
$\int_{x_0}^x \exp{(-\int_{x_0}^r (\lambda s + \frac{1}{2 \lambda s})ds)}(\frac{1}{2\sqrt{r}} - \frac{1}{2\lambda\sqrt{r}})dr \tag{9}$
to my audience; definite integrals and antiderivatives are not my strong point, and if it can be written in closed form, it is probably in some book or table someplace.  Like
Gradshteyn and Ryzhik's famous tome, which I just found out can be freely downloaded from here.  Or just write a C++ program for your ipad or 'droid!  Note that the right hand side of (5) may be substituted for 
$\int_{x_0}^x (\lambda s + \frac{1}{2 \lambda s})ds \tag{10}$
wherever it occurs; I carried this through integral form in the above to make the typing a little easier.
The derivation given above is of course the standard, variation of parameters approach applied to the first order, linear system (1), (2).  The texts are full of it; a derivation may also be found in my answer to this question; see especially equations (5)-(12).
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
