If y-acoslog x-bsinlog x then prove that x2d2ydx2-xdydx-y-0-

Given, y=acos(log x)+bsin(log x) .......(1).Now differentiating both sides we get, dydx= asin (log x)+bcos (log x)x or, xdydx= asin (log x)+bsin(lo ...

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Byju's Answer

Standard XII

Mathematics

Formation of a Differential Equation from a General Solution

If y=acoslo...

Question

If $y = a cos (log x) + b sin (log x)$ then prove that $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0$ .

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Solution

Given,
$y = a cos (log x) + b sin (log x)$ .......(1).

Now differentiating both sides we get,

$\frac{d y}{d x} = \frac{- a sin (log x) + b cos (log x)}{x}$

or, $x \frac{d y}{d x} = - a sin (log x) + b sin (log x)$

Now differentiating both sides we get,

$x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = - \frac{a cos (log x) + a sin (log x)}{x}$

or, $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = - y$ [ Using (1)]

or, $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0$ .

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Similar questions

Q. Prove that

y = a c o s (l o g x) + b s i n (l o g x)

is the solution of

x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0.

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. Let

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

is a solution of the differential equation

x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0

.Prove.

Q. If

y = {sin}^{- 1} x

, then prove that

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 0

If y = ({sin}^{- 1} x)^{2},

prove that

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - 2 = 0.

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If y=acos(logx)+bsin(logx) then prove that x2d2ydx2+xdydx+y=0.

Given,y=acos(logx)+bsin(logx).......(1).Now differentiating both sides we get,dydx=−asin(logx)+bcos(logx)xor, xdydx=−asin(logx)+bsin(logx)Now differentiating both sides we get,xd2ydx2+dydx=−acos(logx)+asin(logx)xor, x2d2ydx2+xdydx=−y [ Using (1)]or, x2d2ydx2+xdydx+y=0.

If $y = a cos (log x) + b sin (log x)$ then prove that $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0$ .