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Suppose I have to solve the following boundary value problem:

$$-u''=f\quad\text{and}\quad u(0)=0,\ u(1)=0.\qquad \mbox{Where $f \colon [0,1] \to \mathbb{R}$.}$$

In the solution to the exercise, the professor integrates as follows to solve the equation: $$u(x)=-\int_{0}^{x}\int_{1}^{y}f(s)dsdy+c_{1}x+c_{2}$$

Could someone explain to me why does he use these limits of integration specifically? Why from $1$ to $y$ then from $0$ to $x$?

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up vote 6 down vote accepted

Call $u_1(x)=-\displaystyle\int_{0}^{x}\int_{1}^{y}f(s)dsdy$ and $u_0(x)=-\displaystyle\int_{0}^{x}\int_{0}^{y}f(s)dsdy$. Then $u_1(x)=u_0(x)+cx$ with $c=\displaystyle\int_{0}^{1}f(s)ds$ hence both $u(x)=u_0(x)+c_1x+c_2$ (as you probably have in mind) or $u(x)=u_1(x)+c_1x+c_2$ (as your professor did) are valid forms of the general solution to the equation $-u''=f$.

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Thank you very much. – squ1d Feb 20 '11 at 20:59

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