# 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