What is wrong in my solution? 2nd order ODE I have an ODE, however I can't find the solution as it's given in the book.
\begin{align}
\begin{cases}
y''(x)=f(x), \qquad 0<x< 1 \\
y(0)=y(1)=0
\end{cases}
\end{align}
My solution:
\begin{align}
y'(x)&= \int f(x) \, \mathrm{d}x=F(x)+C_1 \\
y(x)&= \int (F(x)+C_1)\, \mathrm{d}x=\int F(x) \, \mathrm{d}x+C_1x + C_2
\end{align}

The solution:
  \begin{align}
y'(x)&=\int_0^x f(y) \, \mathrm{d} y +C_1 \\
y(x)&=\int_0^x \int_0^z f(y) \,  \mathrm{d}y \mathrm{d}z+ C_1x+C_2
\end{align}

Questions:
Why definite integrals in the solution? 
Where are the limits from?
Why is there a double integral in the solution? 
Thanks!

You have no initial condition for $y'$ (I mean $y'(0)$ isn't given) so all you can do for the first integration is an undefinite integral,
$$y'(x)=\int f(t)\,dt+C=F(t)+C.$$
Now for the second integration, you can use the fact that $y(0)=0$ and use a definite integral
$$y(x)=0+\int_0^xy'(u)\,du=\int_0^x(F(u)+C)\,du=\int_0^xF(u)\,du+Cx.$$
then plugging the final condition,
$$\int_0^1F(u)\,du+C=0$$ and
$$y(x)=\int_0^xF(u)\,du-\left(\int_0^1F(u)\,du\right)x.$$

If you don't want to use the intermediate $F$, then a double definite integral can do, such as
$$y(x)=\int_{u=0}^x\int_{t=u_0}^u f(t)\,dt\,du-\left(\int_{u=0}^1\int_{t=u_0}^u f(t)\,dt\,du\right)x,$$
where $u_0$ is arbitrary.
