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$$x'' + 3x= 7$$
Given conditions $x'(0)=x'(5)=0$.

I checked the list and I went through three books. I am doing intro to differential equations. I just don't know how to get the extensions... I was told if there is a derivative use $\cos$, but I can't hack it. If there is any simple explanation. Please feel free to expound. This is to simply find the formal solution of the said ODE using Fourier Series. What do I substitute at right hand side? Since I think left hand is simply summation $\cos n \pi/L$.

......I just know that $L=5$? And $7$ is $f(t)$ or $f(x)$? Maybe?


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One thing to note is that the substitution $x=\frac{1}{3}(y + 7)$ reduces the equation to $y'' + 3y = 0$. – Mr. G May 18 '13 at 15:44

I will use the idea of @Mr.G and consider the equation for $y$. The equation is $$ y''+3y=0, y'(0)=y'(5)=0. $$ Obviously $y=0$ is a solution. Thus $x=7/3$ is a solution of the original problem.

The idea behind all this is that you can look for solutions which are trigonometric polinomials $$ y_k(t)=a_k \cos(2\pi kt/5)+b_k\sin(2\pi kt/5). $$ If you insert this ansatz into the boundary conditions you get $$ y'_k(t)=0\Rightarrow b_k=0 $$ Thus, Neumann BC gives you cosines. As the ODE is linear you can have linear combination of it $$ y(t)=\sum a_k \cos(2\pi kt/5). $$ $$ \sum \left(-\left(2\pi k/5\right)^2+3\right)a_k\cos(2\pi kt/5)=0 $$ Now you can multply by $a_k\cos(2\pi kt/5)$ and integrate. You have a series with nonnegative terms equals to zero. You conclude that $a_k$ should be zero.

The Theorem behind this is that $\{e^{i\lambda t}\}$ form a basis of the Hilbert space $$ L^2([0,5])=\{f| \int_0^5 |f(x)|^2 dx<\infty\} $$ You can read the great book by Prof. Walter Strauss 'An introduction to Partial Differential Equations'.

I hope that this helps you

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