Question

y=aex+be-x,then show that dsquarey ÷dxsquare-y=1​

Answer 1

Answer:

Incorrect. The given expression equals zero, not one.

Step-by-step explanation:

$y = ae^x + be^{-x}$

$\frac{dy}{dx} = ae^x - be^{-x}$

$\frac{d^2y}{dx^2} = ae^x + be^{-x}$

So, $\frac{d^2y}{dx^2} = y$

And: $\frac{d^2y}{dx^2} - y = 0$