Question

(b) If y = sin (log x), then show that x? y2 + xy1 + y = 0.​

Answer 1

Appropriate Question :-

$\sf\:If \: y =\sin(logx) \: then \: show \: that \: {x}^{2}y_2 + xy_1 + y = 0.$

Solution :-

Identities Used :-

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

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

$\blue{\boxed{\sf\:\dfrac{d}{dx}logx = \dfrac{1}{x}}}$

$\blue{\boxed{\sf\:\dfrac{d}{dx}y = y_1}}$

$\blue{\boxed{\sf\:\dfrac{d}{dx}y_1 = y_2}}$

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

Let's solve the problem now!!

Consider,

$\rm :\longmapsto\:y = \sin(logx)$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}\sin(logx)$

$\rm :\longmapsto\:y_1 = \cos(logx)\dfrac{d}{dx}logx$

$\rm :\longmapsto\:y_1 = \cos(logx)\dfrac{1}{x}$

$\rm :\longmapsto\:x \: y_1 = \cos(logx)$

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

$\rm :\longmapsto\:\dfrac{d}{dx} \: x \: y_1 = \dfrac{d}{dx}\cos(logx)$

$\rm :\longmapsto\:x\dfrac{d}{dx}y_1 + y_1\dfrac{d}{dx}x = - \sin(logx)\dfrac{d}{dx}logx$

$\rm :\longmapsto\:xy_2 + y_1 \times 1 = - y \times \dfrac{1}{x} \: \: \: \: \: \red{ \because\{ y = \sin(logx) \}}$

$\rm :\longmapsto\:xy_2 + y_1 = - \dfrac{y}{x}$

$\rm :\longmapsto\: {x}^{2}y_2 + xy_1 = - y$

$\bf\implies \: {x}^{2}y_2 + xy_1 + y = 0$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\blue{\boxed{\sf\:\dfrac{d}{dx} x = 1}}$

$\blue{\boxed{\sf\:\dfrac{d}{dx}k= 0}}$

$\blue{\boxed{\sf\:\dfrac{d}{dx}tanx= {sec}^{2}x}}$

$\blue{\boxed{\sf\:\dfrac{d}{dx}cotx= { - \: cosec}^{2}x}}$

$\blue{\boxed{\sf\:\dfrac{d}{dx}kf(x)= k\dfrac{d}{dx}f(x)}}$

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

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

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