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