Question

$\tt{\pmb {if \quad y^{2}=a\left(b^{2}-x^{2}\right) \quad show\quad that x y \dfrac{d^{2} y}{d x^{2}}+x\left(\dfrac{d y}{d x}\right)^{2}=y \dfrac{d y}{d x}}}$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: y^{2}=a\bigg(b^{2}-x^{2}\bigg)$

On differentiating both sides w .r. t. x, we get

$\rm \:\dfrac{d}{dx} y^{2}=\dfrac{d}{dx}a\bigg(b^{2}-x^{2}\bigg)$

We know,

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

and

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx}k \: f(x) \: = \: k \: \dfrac{d}{dx}f(x) \: \: }}$

So, on using these results, we get

$\rm \: 2y\dfrac{dy}{dx} = a\dfrac{d}{dx}\bigg( {b}^{2} - {x}^{2}\bigg)$

$\rm \: 2y\dfrac{dy}{dx} = a(0 - 2x)$

$\rm \: 2y\dfrac{dy}{dx} = - 2ax$

On dividing both sides by 2, we get

$\rm \: y\dfrac{dy}{dx} = - ax - - - (1)$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}\bigg(y\dfrac{dy}{dx}\bigg) = - a\dfrac{d}{dx}x$

$\rm \: y\dfrac{d}{dx}\bigg(\dfrac{dy}{dx}\bigg) + \dfrac{dy}{dx} \times \dfrac{d}{dx}y= - a$

$\rm \: y\dfrac{ {d}^{2} y}{d {x}^{2} } + {\bigg(\dfrac{dy}{dx}\bigg)}^{2} = - a$

Multiply both sides by x, we get

$\rm \: xy\dfrac{ {d}^{2} y}{d {x}^{2} } + x{\bigg(\dfrac{dy}{dx}\bigg)}^{2} = - ax$

On substituting the value of ax from equation (1), we get

$\rm \: xy\dfrac{ {d}^{2} y}{d {x}^{2} } + x{\bigg(\dfrac{dy}{dx}\bigg)}^{2} = y\dfrac{dy}{dx}$

Hence,

$\rm\implies \:\boxed{\pmb{  \:\rm \: xy\dfrac{ {d}^{2} y}{d {x}^{2} } + x{\bigg(\dfrac{dy}{dx}\bigg)}^{2} = y\dfrac{dy}{dx} \: \: }}$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$