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

May 12, 2018

We seek to show that:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + 2 \frac{\mathrm{dy}}{\mathrm{dx}} + \left(1 + {b}^{2}\right) y = 0$ where $y = {e}^{-} x \sin b x$

Using the product rule, In conjunction with the chain rule, then differentiating $y = {e}^{-} x \sin b x$ wrt $x$ we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left({e}^{-} x\right) \left(b \cos b x\right) + \left(- {e}^{-} x\right) \left(\sin b x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = b {e}^{-} x \cos b x - {e}^{-} x \sin b x$
$\setminus \setminus \setminus \setminus \setminus \setminus = {e}^{-} x \left(b \cos b x - \sin b x\right)$

And differentiating a second time, we have:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = {e}^{-} x \left(- {b}^{2} \sin b x - b \cos b x\right) + \left(- {e}^{-} x\right) \left(b \cos b x - \sin b x\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{-} x \left(- {b}^{2} \sin b x - b \cos b x - b \cos b x + \sin b x\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{-} x \left(- {b}^{2} \sin b x - 2 b \cos b x + \sin b x\right)$

And so, considering the LHS of the given expression:

$L H S = \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + 2 \frac{\mathrm{dy}}{\mathrm{dx}} + \left(1 + {b}^{2}\right) y$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left\{{e}^{-} x \left(- {b}^{2} \sin b x - 2 b \cos b x + \sin b x\right)\right\} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus + 2 \left\{{e}^{-} x \left(b \cos b x - \sin b x\right)\right\} + \left(1 + {b}^{2}\right) \left\{{e}^{-} x \sin b x\right\}$

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

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

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 0 \setminus \setminus \setminus$ QED