Form the differential equation by eliminating the arbitrary constants from the equation $y = a cos (2 x + b)$ is

\frac{d^{2} y}{d x^{2}} + 4 y = 0

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\frac{d^{2} y}{d x^{2}} - 4 y = 0

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

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

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Solution

The correct option is A $\frac{d^{2} y}{d x^{2}} + 4 y = 0$
$y = a cos (2 x + b)$

$\Rightarrow \frac{y}{cos (2 x + b)} = a$

$\Rightarrow \frac{\frac{d y}{d x} \times cos (2 x + b) + 2 y sin (2 x + b)}{{cos}^{2} (2 x + b)} = 0$

$\Rightarrow \frac{d y}{d x} = \frac{- 2 y sin (2 x + b)}{cos (2 x + b)}$

$\Rightarrow \frac{d y}{d x} = - 2 y tan (2 x + b)$ --(i)

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - 2 [\frac{d y}{d x} tan (2 x + b) + 2 y {sec}^{2} (2 x + b)]$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - 2 [- 2 y {tan}^{2} (2 x + b) + 2 y {sec}^{2} (2 x + b)]$ (from (i))

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - 4 y$

$\Rightarrow \frac{d^{2} y}{d x^{2}} + 4 y = 0$

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Form the differential equation by eliminating the arbitrary constants from the equation y=acos(2x+b) is

The correct option is A d2ydx2+4y=0y=acos(2x+b)⇒ycos(2x+b)=a⇒dydx×cos(2x+b)+2ysin(2x+b)cos2(2x+b)=0⇒dydx=−2ysin(2x+b)cos(2x+b)⇒dydx=−2ytan(2x+b) --(i)⇒d2ydx2=−2[dydxtan(2x+b)+2ysec2(2x+b)]⇒d2ydx2=−2[−2ytan2(2x+b)+2ysec2(2x+b)] (from (i))⇒d2ydx2=−4y⇒d2ydx2+4y=0

Form the differential equation by eliminating the arbitrary constants from the equation $y = a cos (2 x + b)$ is