Question

If x = sin t, y = sin pt, prove that (1 – x²)d²y/dx²-x dy/dx +p²y = 0.
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Answer 1

Answer:

x=sin t y=sin pt dx/dt=cost=√(1-sin²t)=√(1-x²) dy/dt=pcos pt=p√(1-sin²pt)=p√(1-y²) dy/dx=p√(1-x²)/√(1-y²) ------eq1 again differentiate wrt x: (1-x²)d²y/dx² =p{ √(1-x²) × -2y/2√(1-y²) - √(1-y²)×-2x/2√(1-x²)} (1-x²)d²y/dx²=-py√(1-x²)/√(1-y²) + xp√(1-y²)/√(1-x²)} from eq1: (1-x²)d²y/dx²=-p²y + xdy/dx (1-x²)d²y/dx² - xdy/dx + p²y -------provedRead more on Sarthaks.com - https://www.sarthaks.com/201216/if-x-sin-t-and-y-sin-pt-then-prove-that-1-x-2-d-2y-dx-2-xdy-dx-p-2y-0

Answer 2

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

Given

$\rm :\longmapsto\:x = sint$

and

$\rm :\longmapsto\:y = sinpt$

Consider,

$\rm :\longmapsto\:x = sint$

can be rewritten as

$\rm :\longmapsto\:t = {sin}^{ - 1}x$

and

$\rm :\longmapsto\:y = sinpt$

can be rewritten as

$\rm :\longmapsto\:pt = {sin}^{ - 1}y$

On substituting the value of 't', we get

$\rm :\longmapsto\:p \: {sin}^{ - 1}x = {sin}^{ - 1}y$

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

$\rm :\longmapsto\:\dfrac{d}{dx}p \: {sin}^{ - 1}x = \dfrac{d}{dx} {sin}^{ - 1}y$

$\rm :\longmapsto\:p \: \dfrac{d}{dx} \: {sin}^{ - 1}x = \dfrac{d}{dx} {sin}^{ - 1}y$

We know that,

$\underbrace{ \boxed{ \bf \: \dfrac{d}{dx} {sin}^{ - 1}x = \frac{1}{ \sqrt{1 - {x}^{2} } }}}$

So, on using this identity, we get

$\rm :\longmapsto\:p \times \dfrac{1}{ \sqrt{1 - {x}^{2} } } = \dfrac{1}{ \sqrt{1 - {y}^{2} } }\dfrac{dy}{dx}$

$\rm :\longmapsto\: \sqrt{1 - {x}^{2} }\dfrac{dy}{dx} = p \sqrt{1 - {y}^{2} }$

On squaring both sides, we get

$\rm :\longmapsto\: ({1 - x}^{2}) {\bigg(\dfrac{dy}{dx}\bigg) }^{2} = {p}^{2} (1 - {y}^{2})$

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

$\rm :\longmapsto\: \dfrac{d}{dx}({1 - x}^{2}) {\bigg(\dfrac{dy}{dx}\bigg) }^{2} = \dfrac{d}{dx}{p}^{2} (1 - {y}^{2})$

We know,

$\underbrace{ \boxed{ \bf \: \dfrac{d}{dx}uv =u\dfrac{d}{dx}v + \dfrac{d}{dx}u}}$

and

$\underbrace{ \boxed{ \bf \: \dfrac{d}{dx}kf(x) = k\dfrac{d}{dx}f(x)}}$

So, on using these Identities we get

$\rm :\longmapsto\: {(1 - x}^{2})\dfrac{d}{dx} \bigg(\dfrac{dy}{dx} \bigg)^{2} + \bigg(\dfrac{dy}{dx} \bigg)^{2}\dfrac{d}{dx}(1 - {x}^{2} ) = {p}^{2}\dfrac{d}{dx}(1 - {y}^{2})$

$\rm :\longmapsto\: {(1 - x}^{2})2\dfrac{dy}{dx}\dfrac{ {d}^{2} y}{d {x}^{2} } + \bigg(\dfrac{dy}{dx} \bigg)^{2}(0 - 2x) = - {p}^{2}2y\dfrac{dy}{dx}$

$\rm :\longmapsto\: {(1 - x}^{2})2\dfrac{dy}{dx}\dfrac{ {d}^{2} y}{d {x}^{2} } - 2x \bigg(\dfrac{dy}{dx} \bigg)^{2} = - {p}^{2}2y\dfrac{dy}{dx}$

On cancelation the common terms, we get

$\rm :\longmapsto\: {(1 - x}^{2})\dfrac{ {d}^{2} y}{d {x}^{2} } - x \dfrac{dy}{dx} = - {p}^{2}y$

$\rm :\longmapsto\: {(1 - x}^{2})\dfrac{ {d}^{2} y}{d {x}^{2} } - x \dfrac{dy}{dx} + {p}^{2}y = 0$

Hence, Proved

Additional Information :-

$\boxed{ \bf \: \dfrac{d}{dx}x = 1}$

$\boxed{ \bf \: \dfrac{d}{dx}k= 0}$

$\boxed{ \bf \: \dfrac{d}{dx}logx= \frac{1}{x} }$

$\boxed{ \bf \: \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } }$

$\boxed{ \bf \: \dfrac{d}{dx}sinx= cosx}$

$\boxed{ \bf \: \dfrac{d}{dx}cosx= - \: sinx}$

$\boxed{ \bf \: \dfrac{d}{dx}cosecx= - \: cosecx \: cotx}$

$\boxed{ \bf \: \dfrac{d}{dx}secx= \:secx \: tanx}$