y = Ae2x + Be–2x
As the equating has two different arbitrary constants so, we can differentiate it twice with respect to x. So, on differentiating once with respect to x we get,
\(\frac{dy}{dx}=\) 2Ae2x - 2Be-2x
Again, differentiating it with respect to x, we get
Hence the differential equation corresponding to the curves
y = Ae2x + Be–2x is \(\frac{d^2y}{dx^2}=4y\)