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Find the constant $c$ which solves the following problem: $$-u''=c\quad \text{in}\quad(a,b)$$ $$u'(a)=-1 \quad u'(b)=1$$

One way to solve this is considering the cases when $c>0$ and $c<0$ and writing the solution in terms of $\cos$ , $\sin$ . But that method is not giving the value of $c$

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    $-u'' = c$ does not have either cosine or sine as a solution on any open interval. Is this a typo, or a braino? – 2017-01-19
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    Then maybe $c<0$ – 2017-01-19
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    Uh...it doesn't have that as a solution for *any* value of $c$. – 2017-01-19

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If the second derivative of $u$ is a constant $c$ then we can guess $u$ is a polynomial of second degree of the form $u = dx^2 + ex + f$. Taking the derivatives we get:

$$-u'' = -2d = c \iff d = -\frac{c}{2}\tag{1}\\u' = 2dx + e$$

Because we have two boundary conditions with $u'$ we are able to find the values of $d$ and $e$ as functions of $a$ and $b$. Using that and the equation $(1)$ one is able to find the constant $c$ that gives a solution to this problem.

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    This is a very nice argument , I was actually thinking of making cases $c>0$ and $c<0$, the one we usually do and then write charecteristic equation so here it would be $-r^2=c$ – 2017-01-19
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    @MathMan thanks; please beware that I am not very versed on differential equations; this looks, however, like a fairly decent approach. – 2017-01-19
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    Oh I see, never mind, thanks – 2017-01-19