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I would like to solve the boundary value problem $\frac{d^2 y}{dx^2} = f(x)$ with initial conditions $$y(-1)=y(1)=0$$ My thoughts were to try to find the solution without the conditions, and then filling in this solution in the differential equation to find all the right constants. However, I'm stuck.

I thought the following: $$y(x) = A + Bx + \int_{x_0}^x \left(\int_{x_0}^\eta f(\xi)d\xi \right)d\eta$$

But how can I proceed?

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    $\begingroup$ Simply plug the initial conditions to get the values of $A$ and$B$. $\endgroup$ – Yves Daoust Mar 30 '17 at 11:52
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    $\begingroup$ If you have power series representation of $f(x)$ then yes A and B like that should do nicely because there will be nothing left from $f$ for those coefficients in the power series. $\endgroup$ – mathreadler Mar 30 '17 at 11:53
  • $\begingroup$ @YvesDaoust, Yes I know I'm supposed to, but I don't know how to plug this in in the double integral $\endgroup$ – Di-lemma Mar 30 '17 at 12:15
  • $\begingroup$ Replace $x$ by its value. $\endgroup$ – Yves Daoust Mar 30 '17 at 12:29
  • $\begingroup$ $$A=-\frac{(F(-1)+F(1))}{2}$$ $$B=F(1)-A=F(1)+\frac{(F(-1)+F(1))}{2}=F(1)+\frac{F(-1)}{2}$$ With $F(x)=\int_{x_{0}}^{x}\Big(\int_{x_{0}}^{\nu}f(\xi)d\xi\Big)d\nu$ $\endgroup$ – Kiryl Pesotski Mar 30 '17 at 12:31

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