An example of Sturm-Liouville eigenvalue problem I'm not sure if this has been asked before but I couldn't find it.
I want to solve the Sturm-Liouville eigenvalue problem
$$ u''+\lambda u=0,\ \ u'(0)=u(1)=0.$$
Let the SL differential operator be $Lu=u''.$ I want to find the eigenvalues/functions of $L$ by solving the boundary value ODE. (Hint: The solution of the ODE is 
$$ u(x)=\lambda\int_0^1 g(x,s)u(s)ds$$
where $g(x,s)=
     \begin{cases} 
         s(1-x),\ 0\leq s\leq x\leq 1\\
         x(1-s),\ 0\leq x\leq s\leq 1
     \end{cases}.$)
Here's my solution. 
To solve the ODE, let $u''=\frac{\partial u' }{\partial x}$. Then
$$ \int \partial u' = -\lambda \int u(t) \partial t$$
$$ u'(x)  = -\lambda \int_a^x u(t) \partial t + c_1$$
Now, let $u'(x) = \frac{\partial u(x) }{\partial x}$. Then,
$$\int \partial u(x) = \int^x_b \big( -\lambda \int^x_a  u(t) \partial t + c_1\big)\ dx  $$
$$ u(x) =-\lambda \int^x_b \big( \int^x_a  u(t) \partial t + c_1\big)\ dx +c_2$$
Can someone help me with this?

A simpler approach.
The general solution for the DE $u''+\lambda u = 0$ is
$$
u = c_1\sin(\sqrt \lambda t)+c_2\cos(\sqrt\lambda t)
$$
the boundary conditions are
$$
\cases{u'(0) = \sqrt\lambda c_1\cos(\sqrt\lambda 0)-\sqrt\lambda c_2\sin(\sqrt\lambda 0) = 0\\
u(1) = c_1\sin(\sqrt \lambda)+c_2\cos(\sqrt\lambda) = 0}
$$
or
$$
\left(\begin{array}{cc}1 & 0\\ \sin(\sqrt\lambda) & \cos(\sqrt\lambda)\end{array}\right)\left(\begin{array}{c}c_1\\ c_2\end{array}\right) = \left(\begin{array}{c}0\\ 0\end{array}\right)
$$
now to have non trivial solutions we need $\cos(\sqrt\lambda) = 0$ or
$$
\sqrt\lambda = (2k+1)\frac{\pi}{2}
$$
then the eigenvalues are
$$
\lambda_k =  \left((2k+1)\frac{\pi}{2}\right)^2
$$
and the eigenfunctions are
$$
u_k = \cos\left((2k+1)\frac{\pi}{2}t\right)
$$
Those eigenfunctions  can be used to solve non-homogeneous problems with the same boundary conditions.
