Question

Solve the very important 12th board VVI question ​

Answer 1

$\huge\mathfrak{solution}$

Given:

$y = 5cos \: x - 3sinx \:$

To prove:

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

Differentiate Y w.r.t x , we get

$\frac{dy}{dx} = \frac{d}{dx} 5cosx - 3sinx \\ \\ = - 5sinx - 3cosx$

Again differentiate w.r.t. X

$\frac{ {d}^{2}y }{d {x}^{2} } = \frac{d}{dx} ( - 5sinx - 3cosx) \\ \\ = \: - 5cosx + 3sinx \\ \\ \: = - (5 \: cosx - 3 \: sinx) \\ \\ = - y$

So,

$\implies \: \frac{ {d}^{2} y}{d {x}^{2} } = - y \\ \\ \implies \frac{ {d}^{2}y }{d {x}^{2} } + y = 0$

Proved .

Formula used :

$\frac{d}{dx} sin \: x = cosx \\ \\ \frac{d}{dx} cosx = - sinx$

Answer 2

❏ Used Formulas

$\sf\longrightarrow \frac{d \sin x}{dx}=\cos x$

$\sf\longrightarrow \frac{d \cos x}{dx}=-\sin x$

$\sf\longrightarrow \frac{d \tan x}{dx}=\sec{}^{2} x$

$\sf\longrightarrow \frac{d \cot x}{dx}=-\cosec{}^{2} x$

$\sf\longrightarrow \frac{d \sec x}{dx}=\sec x\tan x$

$\sf\longrightarrow \frac{d \cosec x}{dx}=-\cosec x\cot x$

❏ Solution

Q)

if $y=5\cos x-3\sin x$ then,

prove that, $\frac{d{}^{2}y}{dx{}^{2}}+y=0$

Ans)

$\sf\longrightarrow y=5\cos x-3\sin x$

$\sf\longrightarrow \frac{dy}{dx}=5(-\sin x)-3(\cos x)$

$\sf\longrightarrow \frac{d{}^{2}y}{dx{}^{2}}= -5\cos x+3\sin x$

$\sf\longrightarrow \frac{d{}^{2}y}{dx{}^{2}}+( 5\cos x-3\sin x)=0$

$\sf\longrightarrow\boxed{ \frac{d{}^{2}y}{dx{}^{2}}+y=0\:(proved)}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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