Question

if y= 5 cosx -3 sinx,prove that y2 + y= 0

Answer 1

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

$\rm :\longmapsto\:y = 5cosx - 3sinx$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:y_2 + y = 0$

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

Given that

$\rm :\longmapsto\:y = 5cosx - 3sinx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}( 5cosx - 3sinx)$

We know,

$\boxed{ \rm{ \dfrac{d}{dx}y = y_1}}$

So, using this we get

$\rm :\longmapsto\:y_1 = \dfrac{d}{dx}5cosx - \dfrac{d}{dx}3sinx$

$\rm :\longmapsto\:y_1 =5 \dfrac{d}{dx}cosx - 3\dfrac{d}{dx}sinx$

We know that,

$\boxed{ \rm{ \dfrac{d}{dx}sinx = cosx}}$

and

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

So, using this, we get

$\red{\rm :\longmapsto\:y_1 = - 5sinx - 3cosx}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y_1 = \dfrac{d}{dx}( - 5sinx - 3cosx)$

We know that

$\boxed{ \rm{ \dfrac{d}{dx}y_1 = y_2}}$

So, using this we get

$\rm :\longmapsto\:y_2 = \dfrac{d}{dx}( - 5sinx) - \dfrac{d}{dx}3cosx$

$\rm :\longmapsto\:y_2 = - 5\dfrac{d}{dx}sinx -3 \dfrac{d}{dx}cosx$

$\rm :\longmapsto\:y_2 = - 5cosx + 3sinx$

$\rm :\longmapsto\:y_2 = - (5cosx - 3sinx)$

$\rm :\longmapsto\:y_2 = - y$

$\red{\bigg \{ \because \: y = 5cosx - 3sinx\bigg \}}$

$\rm :\longmapsto\:y_2 + y = 0$

Hence, proved

Additional Information :-

$\boxed{ \rm{ \dfrac{d}{dx}tanx = {sec}^{2}x}}$

$\boxed{ \rm{ \dfrac{d}{dx}cotx = { - cosec}^{2}x}}$

$\boxed{ \rm{ \dfrac{d}{dx}cosecx = - \: cosecx \: cotx}}$

$\boxed{ \rm{ \dfrac{d}{dx}secx \: = \: secx \: tanx}}$

$\boxed{ \rm{ \dfrac{d}{dx}logx = \frac{1}{x}}}$

$\boxed{ \rm{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}}$

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

$\boxed{ \rm{ \dfrac{d}{dx} {a}^{x} = {a}^{x} \: loga}}$