To find the differential equation with the following solution: $y=ae^x+be^{-x}+c\cos x + d\sin x$  , where a,b,c,d are parameters.
I had already found till the 4th order differential in succession as the above equation contains four arbitrary constants, but could not find out how to eliminate the arbitrary constants using the four equations, so as the find out the differential equation 
 A: By the  Robert Israel's hint we obtain $y''''-y=0$. It's all!
A: Given
$$y=a \, e^x + b \, e^{-x} + c \, \cos x + d \, \sin x$$
then
\begin{align}
y &= a \, e^x + b \, e^{-x} + c \, \cos x + d \, \sin x \\
y' &= a \, e^x - b \, e^{-x} - c \, \sin x + d \, \cos x \\
y'' &= a \, e^x + b \, e^{-x} - c \, \cos x - d \, \sin x \\
y''' &= a \, e^x - b \, e^{-x} + c \, \sin x - d \, \cos x \\
y'''' &= a \, e^x + b \, e^{-x} + c \, \cos x + d \, \sin x
\end{align}
Now it is fairly easy to spot that:
$$y'''' - y =0.$$
As a checking method consider the equation $f'''' - f = 0$ with a solution of the form $f(x) = e^{r \, x}$ which leads to $r^4 - 1 = 0$ which can be seen in the form $(r-1)(r+1)(r-i)(r+i)=0$ which yields $r \in \{ 1, -1, i, -i\}$ and 
$$f(x) = \alpha \, e^{x} + \beta \, e^{-x} + \gamma \, e^{i \, x} + \delta \, e^{- i \, x}.$$
By rearranging some of the constants $f(x)$ can have the forms:
\begin{align}
f(x) &= a \, e^x + b \, e^{-x} + c \, \cos x + d \, \sin x \\
f(x) &= c_{0} \, \cosh x + c_{1} \, \sinh x + c_{2} \, \cos x + c_{3} \, \sin x
\end{align}
