Question

- y = A cos (4x) + B sin (4x)

Answer 1

Given differential equation is-

$y=Acos4x+Bsin4x$

Differentiate above equation w.r.t. 'x'

$\Rightarrow \dfrac{dy}{dx}=-4Asin4x+4Bcos4x$

Again differentiating w.r.t. 'x'

$\dfrac{d^2y}{dx^2}=-16Acos4x-16Bsin4x$

$\Rightarrow \dfrac{d^2y}{dx^2}=-16(Acos4x+Bsin4x)$

$\Rightarrow \dfrac{d^2y}{dx^2}=-16(y)$

$\Rightarrow \dfrac{d^2y}{dx^2}+16y=0$