Question

Given that `y= e^-x sinbx`, where `b` is a constant, show that `(d^2y)/(dx^2) + 2dy/dx + ( 1 + b^2) y = 0` ?

Answer 1

We seek to show that:

`(d^2y)/(dx^2) + 2dy/dx + ( 1 + b^2) y = 0` where `y= e^-x sinbx`

Using the product rule, In conjunction with the chain rule, then differentiating `y= e^-x sinbx` wrt `x` we have:

`dy/dx = (e^-x)(bcosbx) + (-e^-x)(sinbx)`
`\ \ \ \ \ \ = be^-x cosbx - e^-x sinbx`
`\ \ \ \ \ \ = e^-x (bcosbx - sinbx)`

And differentiating a second time, we have:

`(d^2y)/(dx^2) = e^-x (-b^2sinbx - bcosbx) + (-e^-x) (bcosbx - sinbx)`

`\ \ \ \ \ \ \ = e^-x (-b^2sinbx - bcosbx - bcosbx + sinbx)`

`\ \ \ \ \ \ \ = e^-x (-b^2sinbx - 2bcosbx + sinbx)`

And so, considering the LHS of the given expression:

`LHS = (d^2y)/(dx^2) + 2dy/dx + ( 1 + b^2) y`

`\ \ \ \ \ \ \ \ = {e^-x (-b^2sinbx - 2bcosbx + sinbx)} +`
`\ \ \ \ \ \ \ \ \ \ \ \ + 2{e^-x (bcosbx - sinbx)} + ( 1 + b^2) {e^-x sinbx}`

`\ \ \ \ \ \ \ \ = e^-x {-b^2sinbx - 2bcosbx + sinbx + 2bcosbx`
`\ \ \ \ \ \ \ \ \ \ \ \ - 2sinbx + ( 1 + b^2)sinbx}`

`\ \ \ \ \ \ \ \ = e^-x {-b^2sinbx - 2bcosbx + sinbx + 2bcosbx`
`\ \ \ \ \ \ \ \ \ \ \ \ - 2sinbx + sinbx + b^2sinbx}`

`\ \ \ \ \ \ \ \ = 0 \ \ \` QED