Question

If y=sin5x, show that d²y/dx²+25=0

Answer 1

Correct question---->

$if \: y \: = sin5x \: \\ show \: that \: \: \\ \dfrac{ {d}^{2}y }{d {x}^{2} } + 25y \: = 0$

Concept used ---->

1)

$first \: order \: derivative \: = \dfrac{dy}{dx} \\ second \: derivative \: = \dfrac{ {d}^{2}y }{d {x}^{2} }$

2)

$\dfrac{d}{dx} \:(sinx) = cosx \\ \dfrac{d}{dx} \: (cosx) = - sinx \\ \dfrac{d}{dx} (x) = 1$

Solution--->

$y \: = sin5x$

$differentiating \: with \: respect \: to \: x$

$= > \dfrac{dy}{dx} = \dfrac{d}{dx} (sin5x)$

$= > \dfrac{dy}{dx} \: = cos5x \: \dfrac{d}{dx} (5x)$

$= > \dfrac{dy}{dx} \: = cos5x \: ( \: 5 \: )$

$= > \dfrac{dy}{dx} \: = 5 \: cos5x$

$differentiating \: with \: respect \: to \: x \:$

$= > \dfrac{d}{dx} ( \dfrac{dy}{dx} ) \: = 5 \: \dfrac{d}{dx} \: ( \: cos5x \: )$

$= > \dfrac{d ^{2} y}{dx^{2} } \: = 5 \: ( - sin5x \: ) \: \dfrac{d}{dx} (5x)$

$= > \dfrac{d ^{2}y }{d {x}^{2} } \: = - 5 \: sin5x \: ( \: 5 \: )$

$= > \dfrac{ {d}^{2}y }{d {x}^{2} } = - 25 \: sin5x$

$= > \dfrac{ {d}^{2} y}{d {x}^{2} } \: = - 25 \: y$

$= > \dfrac{ {d}^{2}y }{d {x}^{2} } \: + \: 25 \: y \: = \: 0$

Answer 2

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