Solving Fourth-Order ODE BVP using dsolve() I'm attempting to solve the following ODE using Matlab's built-in function dsolve
$$\frac{d^4w}{dx^4}=w$$
with the following boundary conditions:
\begin{align}
\frac{dw}{dx}(0)&=0\\
\frac{d^3w}{dx^3}(0)&=0\\
\frac{d^2w}{dx^2}(L)&=0\\
\frac{d^3w}{dx^3}(L)&=0    
\end{align}
I first tried to use the following code
syms w(x)
a = 1;
L = 1;
Dw = diff(w,1);
D2w = diff(w,2);
D3w = diff(w,3);
w = dsolve(diff(w,4) - a*w == 0, Dw(0) == 0, D3w(0) == 0,D2w(L) == 0, D3w(L) == 0)
But, this only lead to the trivial solution $u=0$, which I think was due to having the boundary condition $\frac{dw}{dx}(0)=0$. I tried to remedy this by setting the zero boundary conditions equal to $eps == 2.2204e-16$, which for all practically purposes is equal to zero. 
I did get a nontrivial solution which solved the ODE, but I want to see if this is a reasonable approach to match the boundary condition.
Thanks for any feedback, and sorry for the poor Matlab code formatting!
 A: $$\frac{d^4w}{dx^4}=w$$
Obviously, $e^x$ , $e^{-x}$ , $e^{ix} , e^{-ix}$ are independent particular solutions. So, the general solution can be expressed with linear combinations of them, for example :
$$w=c_1\cosh(x)+c_2\sinh(x)+c_3\cos(x)+c_4\sin(x)$$
$$\frac{dw}{dx}=c_1\sinh(x)+c_2\cosh(x)-c_3\sin(x)+c_4\cos(x)$$
$$\frac{d^2w}{dx^2}=c_1\cosh(x)+c_2\sinh(x)-c_3\cos(x)-c_4\sin(x)$$
$$\frac{d^3w}{dx^3}=c_1\sinh(x)+c_2\cosh(x)+c_3\sin(x)-c_4\cos(x)$$
The conditions are :
$$\begin{cases}
\left(\frac{dw}{dx}\right)_{x=0}=c_2+c_4=0\\
\left(\frac{d^3w}{dx^3}\right)_{x=0}=c_2-c_4=0\\
\left(\frac{d^2w}{dx^2}\right)_{x=L}=c_1\cosh(L)+c_2\sinh(L)-c_3\cos(L)-c_4\sin(L)=0\\
\left(\frac{d^3w}{dx^3}\right)_{x=L}=c_1\sinh(L)+c_2\cosh(L)+c_3\sin(L)-c_4\cos(L)=0
\end{cases}$$
$$\begin{cases}
c_2=0\\
c_4=0\\
\left(\frac{d^2w}{dx^2}\right)_{x=L}=c_1\cosh(L)-c_3\cos(L)=0\\
\left(\frac{d^3w}{dx^3}\right)_{x=L}=c_1\sinh(L)+c_3\sin(L)=0
\end{cases}$$
$$\frac{c_1}{c_3}=\frac{\cos(L)}{\cosh(L)}=-\frac{\sin(L)}{\sinh(L)} \quad\to\quad \tanh(L)=-\tan(L) $$
General case (not particular value of $L$) :
$$\tanh(L)\neq\tan(L) \quad\to\quad \text{ony the trivial solution } w=0$$
Particular cases of $L=L_p$ so that $\tanh(L_p)=-\tan(L_p)$ :
$$w=c_3\left(\frac{c_1}{c_3}\cosh(x)+\cos(x)\right)=c_3\left(\frac{\cos(L_p)}{\cosh(L_p)}\cosh(x)+\cos(x)\right)$$
and with $C=\frac{c_3}{\cosh(L_p)}$
$$w(x)=C\left(\cos(L_p)\cosh(x)+\cosh(L_p)\cos(x)\right)$$
