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I am stuck on solving the following differential equation which contains a definite integral that I don't know how to deal with:

$$ f^{\prime\prime} + a^2 f - b\int_0^L f(t) \, dt = c$$

The boundary condition is $f(0)=0$ and $f(L)=R$. Anyone help me out of here? Thank you in advance.

Gonçalo
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wallen
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    $$ f^{\prime\prime}(t) + a^2 f(t) - b\int_0^L f(t) , dt = c $$ It's a bit confusing to use the same letter, $t,$ to refer both to the independent variable in the differential equation and to the bound variable of integration. So let's write this as $$ f^{\prime\prime}(t) + a^2 f(t) - b\int_0^L f(u) , du = c. $$ This says $$ f''(t) + a^2 f(t) = \text{constant} $$ but then it also says something about that "constant", and "constant" means not depending on the independent variable $t.$ – Michael Hardy Feb 06 '18 at 17:13
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    @MichaelHardy : $f_p=Ct^2+Dt+E$ gives $f_p''+a^2f_p=2C+a^2Ct^2+a^2Dt+a^2E$, which is not constant. – Lutz Lehmann Feb 06 '18 at 17:16
  • Any development from there? I think the equation it's under the category "integro-differential equation" – Giorgio Pastasciutta May 29 '22 at 20:21

3 Answers3

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Your solution is $$f(t)=A\cos(at)+B\sin(at)+C$$ where the boundary conditions and the original equation lead to 3 linear equations in the 3 parameters $A,B,C$, \begin{align} 0&=f(0)=A+C\\ R&=f(L)=A\cos(aL)+B\sin(aL)+C\\ c&=a^2C-b\left(\frac Aa\sin(AL)+\frac Ba(1-\cos(aL))+CL\right) \end{align} This linear system might be without solution, but then the problem is not solvable for the given boundaries and boundary conditions.

Lutz Lehmann
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Note that $\int_0^L f(t) dt$ is a constant depending on $L$ and call it $d$. Now you just need to integrate twice. I write $f(t)$ as $f$.

$$f' \frac{df'}{df}=c+bd-a^2f$$

King Tut
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    The last step is an unnecessary complication. The oscillation equation $f''+a^2f=k$ can be solved with standard methods. – Lutz Lehmann Feb 06 '18 at 17:13
  • Good point. Thanks a lot – wallen Feb 06 '18 at 17:13
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    I do not know if that method is correct. It is not clear that the function obtained will satisfy the relation given by the first equation with the integral. (since the constant described as $d$ in this post is in fact a function of $f$ ) – nicomezi Feb 06 '18 at 17:20
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Hint.

As $b\int_0^L f(t)dt = c_f$ we have

$$ f(t) = \frac{c+c_f}{a^2}+c_1 \cos (a t)+c_2 \sin (a t)\ \ \ \ \ (*) $$

and we have that

$$ b\int_0^L f(t)dt = c_f\Rightarrow c_f = \frac{a b c_1 \sin (a L)-a b c_2 \cos (a L)+a b c_2+b c L}{a^2-b L} $$

now after substituting $c_f$ in $(*)$ we can proceed to determine $\{c_1,c_2\}$ using the boundary conditions.

NOTE

The final solution gives

$$ f(t) = \frac{b R \sin \left(\frac{a L}{2}\right)+a c \cos \left(\frac{a L}{2}\right)}{a \left(a^2-b L\right) \cos \left(\frac{a L}{2}\right)+2 b \sin \left(\frac{a L}{2}\right)}+\frac{\left(a^3 R-a b L R+b R \sin (a L)+a c \cos (a L)-a c\right)}{a \left(a^2-b L\right) \sin (a L)-2 b \cos (a L)+2 b}\sin(a t)-\frac{ \left(b R \sin \left(\frac{a L}{2}\right)+a c \cos \left(\frac{a L}{2}\right)\right)}{a \left(a^2-b L\right) \cos \left(\frac{a L}{2}\right)+2 b \sin \left(\frac{a L}{2}\right)}\cos(a t) $$

Cesareo
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