If y - a cos ln x - b sin ln x- then x2 d2y-dx2 - x dy-dx is equal to

We have, y = a cos (ln x) + b sin (ln x) Differentiating both sides w.r.t. x, we get ⇒dy/dx = a sin (ln x)/x + b cos (ln x)/x ⇒ xdy/dx = a sin(ln x) + bcos (l ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Mathematics

Higher Order Derivatives

If y = a cos ...

Question

If $y = a cos (ln x) + b sin (ln x)$ , then $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x}$ is equal to

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- y

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $- y$
We have,
$y = a cos (ln x) + b sin (ln x)$
Differentiating both sides w.r.t. $x$ , we get
$\Rightarrow \frac{d y}{d x} = \frac{- a sin (ln x)}{x} + \frac{b cos (ln x)}{x}$
$\Rightarrow x \frac{d y}{d x} = - a sin (ln x) + b cos (ln x)$
$\Rightarrow x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = \frac{- a cos (ln x)}{x} + \frac{- b sin (ln x)}{x}$
$\Rightarrow x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = - y$

Suggest Corrections

Similar questions

Q. If

y = a cos (ln x) + b sin (ln x)

, then

x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x}

is equal to

Q. If

y = l n {(\frac{x}{a + b x})}^{x}

,then

x^{3} \frac{d^{2} y}{d x^{2}}

is equal to

Q. If

y = {(x + \sqrt{1 + x^{2}})}^{n}

, then

(1 + x^{2}) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x}

is equal to

Q. If

y = a cos (log x) - b sin (log x)

, then the value of

x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y

Q. If

y = f (x) = (x + \sqrt{1 + x^{2}})^{n}

, then

(1 + x^{2}) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x}

equals

Join BYJU'S Learning Program

If y=acos (lnx)+bsin (lnx), then x2d2ydx2+xdydx is equal to

The correct option is C −yWe have, y=acos (lnx)+bsin (lnx) Differentiating both sides w.r.t. x, we get ⇒dydx=−a sin (lnx)x+b cos (lnx)x ⇒xdydx=−a sin(lnx)+bcos (lnx) ⇒xd2ydx2+dydx=− a cos (lnx)x+−b sin(lnx)x ⇒x2d2ydx2+xdydx=−y

If $y = a cos (ln x) + b sin (ln x)$ , then $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x}$ is equal to