Assume that $\lambda > \pi$
We know that a differential equation like:
$h''(x) + \lambda h(x) = 0$ has solutions $h = c_1e^{-\sqrt\lambda x}$ + $c_2e^{\sqrt\lambda x} = c_3\sin(\sqrt\lambda x) + c_4\cos(\sqrt\lambda x)$.
However, for an equation like $h''(x) + h(x) + \lambda h = 0 $, how do we split our answer into a combination of exponentials and of sine and cosine?
Solution to the equation $h′′(x)+λh(x)=0$ should have the form: $h(x) = c_1 e^{−i\sqrt\lambda x} + c_2 e^{+i\sqrt\lambda x}$.
Write the other equation as: $h′′(x)+ (1+λ)h(x)=0$. Thus, the solution to this equation has the following form:
$h(x) = c_1 e^{−i\sqrt{ 1+ \lambda} x} + c_2 e^{+i\sqrt{1 +\lambda} x} = c_3 \sin(\sqrt{1 +\lambda} x) + c_4 \cos(\sqrt{1 +\lambda} x)$.