$\frac{d^2y}{dx^2}=f(x)$ with boundary conditions, how to find integration bounds

Question

Given $$\frac{d^2y}{dx^2}=f(x),\quad y(-1)=y(1)=0,$$ I used $u=y'$ and $u(x_0)=u_0$ to get $$ u(x)=u_0+\int_{x_0}^xf(\xi)d\xi. $$

Then we have $y'=u$, which we can integrate again using $y(x_0)=y_0$ to get

$$ y(x)=y_0+\int_{x_0}^x\Big(u_0+\int_{x_0}^\zeta f(\xi)d\xi\Big)d\zeta=(y_0-u_0x_0)+u_0x+\int_{x_0}^x\Big(\int_{x_0}^\zeta f(\xi)d\xi\Big)d\zeta. $$

Now my question is, what do I do with $y_0,x_0$ and $u_0$ to get it into the normal form using $C_1,C_2$? Wolfram Alpha gives the following expression:

$$ y(x)=C_1+C_2x+\int_{1}^x\Big(\int_{1}^\zeta f(\xi)d\xi\Big)d\zeta, $$

but how did they set $x_0=1$? Because we didn't use any boundary conditions yet. I understand $C_1$ and $C_2$ as just relabeling.

Answer 1

This is because $$\int_{x_0}^x f(t) dt=\int_1^x f(t) dt - \int_1^{x_0} f(t) dt=\int_1^x f(t) dt - k$$ For some $k\in\mathbb{R}$. Hence $$\int_{x_0}^x\Big(\int_{x_0}^\zeta f(\xi)d\xi\Big)d\zeta=\int_{x_0}^x\Big(\int_1^\zeta f(\xi)d\xi-k_1\Big)d\zeta$$ $$=\int_1^x\Big(\int_1^\zeta f(\xi)d\xi-k_1\Big)d\zeta-k_2$$ $$=\int_1^x\Big(\int_1^\zeta f(\xi)d\xi\Big)d\zeta-\int_1^xk_1d\zeta-k_2$$ $$=\int_1^x\Big(\int_1^\zeta f(\xi)d\xi\Big)d\zeta-k_1(x-1)-k_2$$ $$=\int_1^x\Big(\int_1^\zeta f(\xi)d\xi\Big)d\zeta+k_3x+k_4$$ So $$y(x)=y_0-u_0x_0+u_0x+\int_1^x\Big(\int_1^\zeta f(\xi)d\xi\Big)d\zeta+k_3x+k_4$$ $$=C_1+C_2x+\int_1^x\Big(\int_1^\zeta f(\xi)d\xi\Big)d\zeta$$