Having trouble finding solution to a DE $$y^{(4)} - \lambda y = 0, y'(0)=0, y'''(0)=0, y(\pi)=0, y''(\pi)=0$$
let $\lambda = a^4$
my solution: $$r^4-a^4=0 \implies r=\pm a \\ y = c_1e^{ax}+c_2e^{-ax}$$
but my book says the solution is $y=c_1cos(ax)+c_2sin(ax)+c_3cosh(ax)+c_4sinh(ax)$.
What am I doing wrong?
 A: I take it that $\lambda$ is positive. Beside the solutions $r=\pm a$, we have the solutions $r=\pm ia$.
You have essentially taken care of the $\cosh ax$ and $\sinh ax$ part in another way. But the complex roots have not been taken care of. They yield solutions $e^{iax}$ and $e^{-iax}$, which can be written in terms of sines and cosines.
A: The equation
$$
r^4 - a^4 = 0
$$
has 4 solutions, not 2:
$$
r_{1,2} = \pm a \\
r_{3,4} = \pm ai
$$
$(1)$  and $(2)$ give
$$
y_1 = c_1 e^{ax} + c_2 e^{-ax}
$$
WLOG we can take 
$$
c_1 \equiv \frac{c_1 + c_2}2 \\
c_2 \equiv \frac {c_1 - c_2}2
$$
Then solution becomes
$$
y_1 = \frac {c_1 + c_2}2 e^{ax} + \frac {c_1 - c_2}2 e^{-ax} = c_1\frac {e^{ax} + e^{-ax}}2 + c_1\frac {e^{ax} - e^{-ax}}2 = c_1 \cosh ax + c_2 \sinh ax
$$
Analogously for $(3)$ and $(4)$
$$
y_2 = c_3 e^{axi} + c_4 e^{-axi}
$$
WLOG we can take 
$$
c_3 \equiv \frac {c_3 - ic_4}2 \\
c_4 \equiv \frac {c_3 + ic_4}2
$$
so solution becomes
$$
y_2 = \frac {c_3 - ic_4}2 e^{axi} + \frac {c_3 + ic_4}2 e^{-axi} = c_3 \frac {e^{axi} + e^{-axi}}2 - c_4 i\frac {e^{axi} - e^{-axi}}2 = \\
= c_3 \frac {e^{axi} + e^{-axi}}2 + c_4 \frac {e^{axi} - e^{-axi}}{2i} = c_3 \cos ax + c_4 \sin ax
$$
so final solution is
$$
y = c_1 \cosh ax + c_2 \sinh ax + c_3 \cos ax + c_4 \sin ax
$$
A: the solution is $2Ccos x[4]√(λ)+2C((cosπ[4]√(λ))/(e^{π[4]√(λ)}e^{x[4]√(λ)}+(1/(e^{π[4]√(λ)}))e^{x[4]√(λ)}))+2C(cosπ[4]√(λ))((e^{x[4]√(λ)})/(e^{π[4]√(λ)}+(1/(e^{π[4]√(λ)}))))$. A pair operations leads to the solution of the book.
