Question

y=e^x sinx show that y''-2y'+2y'=0
(plz answer this one)​

Answer 1

Appropriate Question

$\rm :\longmapsto\:If \: y = {e}^{x}sinx ,\: show \: that \: y'' - 2y' + 2y = 0$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:y = {e}^{x} \: sinx - - - (1)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx}{e}^{x} \: sinx$

We know

$\boxed{ \sf{ \:\dfrac{d}{dx}uv = u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u}}$

On applying this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {e}^{x}\dfrac{d}{dx}sinx + sinx\dfrac{d}{dx}{e}^{x}$

We know,

$\boxed{ \sf{ \:\dfrac{d}{dx}{e}^{x} = {e}^{x}}}$

and

$\boxed{ \sf{ \:\dfrac{d}{dx}sinx = cosx}}$

On applying these,

$\rm :\longmapsto\:y'= {e}^{x} \: cosx + sinx \: {e}^{x}$

$\rm :\longmapsto\:y'= {e}^{x} \: cosx + y \: \: \: \: \: \: \: \: \: \: \: \{using \: (1) \}$

$\rm :\implies\:y' = {e}^{x}cosx + y - - - (2)$

On differentiating w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y' = \dfrac{d}{dx}{e}^{x} \: cosx \: + \dfrac{d}{dx}y$

$\rm :\longmapsto\:y'' = {e}^{x}\dfrac{d}{dx}cosx + cosx\dfrac{d}{dx}{e}^{x} + y'$

$\rm :\longmapsto\:y'' = - {e}^{x}sinx + cosx \: {e}^{x} + y'$

$\rm :\longmapsto\:y'' = - y + \red{cosx \: {e}^{x}} + y' \: \: \: \: \: \{ \: using \: (1) \}$

$\rm :\longmapsto\:y'' = - y + y' - y + y'$

$\: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: y' = {e}^{x}cosx + y\bigg \}}$

$\rm :\longmapsto\:y'' = - 2y + 2y'$

$\bf :\longmapsto\:y'' - 2y' + 2y = 0$

Hence, Proved