Question

If ay=sin(x+y) prove that d2y/dx2 + y(1+dy/dx)3

Answer 1

Answer:

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Answer 2

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Q1) If ay=sin(x+y) prove that d²y/dx²+y(1+dy/dx)³

$\: \: \underline {\boxed{ \bf{ \ddot \smile \: answers}}}$

$\divideontimes \boxed{ \sf{to \: prove \to \: \frac{ {d}^{2}y }{d {x}^{2} } + y {(1 + \frac{dy}{dx} )}^{3} }}$

$\: \implies \displaystyle \tt{differentiating \: both \: side \: with \: respect \: to \: x}$

$\implies \displaystyle \tt{ ay = sin(x + y)}$

$\: \implies \displaystyle \tt{a \frac{dy}{dx} = cosx(x + y) \times 1.......by \: using \: chain \: rule}$

$\: \implies \displaystyle \tt{a \frac{dy}{dx} = cos(x + y)}$

$\: \implies \displaystyle \tt{ \frac{dy}{dx} = \frac{cos(x + y)}{a} }$

$\bullet \underline{\displaystyle \bf{ \:double \: differentiating \: the \: equation \: with \: respect \: to \: x}}$

$\: \: \bullet \boxed{ \bf{by \: using \: quotient \: rule}}$

$\implies \displaystyle \tt{ \frac{ {d}^{2}y }{ {d} {x}^{2} } = \frac{a \frac{dy}{dx}cos(x + y) + \frac{dy}{dx} a \times cos(x + y)}{ {a}^{2} } }$

$\implies \displaystyle \tt{ \frac{ {d}^{2}y }{d {x}^{2} } = \frac{a( - sin(x + y) \times 1) + 0 \times cos(x + y)}{ {a}^{2} } }$

$\: \implies \displaystyle \tt{ \frac{ {d}^{2}y }{ {dx}^{2} } = \frac{ - asin(x + y) + 0}{ {a}^{2} } }$

$\: \implies \displaystyle \tt{ \frac{ {d}^{2}y }{ {dx}^{2} } = \frac{ - asin(x + y)}{ {a}^{2} } }$

$\: \bullet \displaystyle\bf{substiute \: the \: value \: in \: equation}$

$\: \: \implies \displaystyle \tt{ \frac{ {d}^{2} y}{ {dx}^{2} } + y\bigg(({1 + \frac{dy}{dx} )}^{3} \bigg)}$

$\implies \displaystyle \tt{ \frac{ - asin(x + y)}{ {a}^{2} } + y {(1 + \frac{cos(x + y)}{a} }^{3} )}$

$\: \implies \displaystyle \tt{ \frac{ - sin(x + y)}{a} + y \bigg({(1 + \frac{cos(x + y)}{a} }^{3} \bigg) }$