The following exercise is a part of a bigger exercise. Therefore I first give you the setting of the whole exercise and then (in the grey box below) the part of the exercise which I mean here.
Setting of the whole exercise:
Consider $-\infty<a<b< +\infty, F\in C([a,b]^2\times\mathbb{R}), f\in C([a,b])$ with $$ \exists L\geq 0~\forall x,y\in[a,b]~\forall u,v\in\mathbb{R}:~~\lvert F(x,y,u)-F(x,y,v)\rvert\leq L~\lvert u-v\rvert. $$ Furthermore consider the non-linear operator $T\colon C([a,b])\to C([a,b])$ given by $$ T(u)(x):=\int_a^xF(x,y,u(y))\, dy+f(x)~~\text{for}~~x\in [a,b]. $$
Consider the special case $a:=0, b:=1, F(x,y,u):=xyu, f(x):=1$ and the integral equation $$ u(x)-\int_a^x F(x,y,u(y))\, dy+f(x)~~~~~\text{for }x\in [a,b]. $$ Transfer the integral equation for the given special case in an equivalent initial value problem of an ODE of degree 2 on $[0,1]$.
Could you please explain me, how I can make an initial value problem out of the given integral equation that is in the special case $$ u(x)-\int_0^x xyu(y)\, dy=1~~~~~\text{for }x\in [0,1]? $$
What is searched is something like
$$ y''=F(x,y(x),y'(x)), y(x_0)=y_0, y'(x_0)=y_0' $$
but I do not know how to reach that.
For your $u$, provided $u$ is well-behaved near $y=0$, one has $u(0)=0$, and $$ u' = \frac{d}{dx}\left(x\int_{0}^{x}yu(y)\,dy\right). $$ Keep going with that and see where it leads. Remember, you can use $u=1+x\int_{0}^{x}yu(y)dy$ if you need to.