Question

If y = P cos ux + Q sin ux, show that `(d^2y)/(dx^2) + u^2y = 0`.

Answer 1

y = P cos ux + Q sin ux

First Derivative `((dy)/(dx))`:

Using the chain rule, `d/dx(cos ux) = u sin ux and d/dx (sin ux) = u cos ux`:

`(dy)/(dx) = -Pu sin(ux) + Qu cos(ux)`

Second derivative `(d^2y)/(dx^2)`:

Differentiating again with respect to x:

`(d^2y)/(dx^2) = -pu^2 cos(ux) - Qu^2 sin(ux)`

Factor out −u²:

Notice that −u² is common in both terms:

`(d^2y)/(dx^2) = -u^2 [P cos(ux) + Q sin(ux)]`

Since the term in the parenthesis is the original \(y\), substitute it back:

`(d^2y)/(dx^2) = -u^2y`

Rearrange the equation to match the required format:

`(d^2y)/(dx^2) + u^2y = 0`

Hence Proved.