Question

If x = sint and y = cospt, prove that

prove that

$(1 - {x}^{2})y_2 - xy_1 + {p}^{2} y = 0$

where symbols have usual meanings.​

Answer 1

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

$\rm \: x = sint$

and

$\rm \: y = cos \: pt$

Now,

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

Also,

$\rm \: y = cos \: pt$

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

On substituting the value of t from above, we get

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

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

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

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

can be further rewritten as

$\rm \: \dfrac{p}{ \sqrt{1 - {x}^{2} } } = - \dfrac{1}{ \sqrt{1 - {y}^{2} } }y_1$

$\rm \: y_1 \sqrt{1 - {x}^{2} } = - p \: \sqrt{1 - {y}^{2} }$

On squaring both sides, we get

$\rm \: {y_1}^{2}(1 - {x}^{2}) = {p}^{2}(1 - {y}^{2})$

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

$\rm \: \dfrac{d}{dx}{y_1}^{2}(1 - {x}^{2}) = \dfrac{d}{dx}{p}^{2}(1 - {y}^{2})$

We know,

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

So, using this result, we get

$\rm \: {y_1}^{2}\dfrac{d}{dx}(1 - {x}^{2}) + (1 - {x}^{2})\dfrac{d}{dx} {y_1}^{2} = {p}^{2}(0 - 2yy_1)$

$\rm \: {y_1}^{2}(0 - 2x) + (1 - {x}^{2})(2y_1y_2) = {p}^{2}( - 2yy_1)$

$\rm \: {y_1}^{2}( - 2x) + (1 - {x}^{2})(2y_1y_2) = {p}^{2}( - 2yy_1)$

$\rm \: 2y_1( - xy_1 + (1 - {x}^{2} )y_2) = - 2 {p}^{2}yy_1$

$\rm \: - xy_1 + (1 - {x}^{2} )y_2 = - {p}^{2}y$

$\rm\implies \:\: (1 - {x}^{2} )y_2 - xy_1 + {p}^{2}y = 0$

Hence, Proved

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Formulae Used

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

$\boxed{\tt{ \dfrac{d}{dx} {cos}^{ - 1}x = - \: \frac{1}{ \sqrt{1 - {x}^{2} } } \: }}$

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

$\boxed{\tt{ \: \dfrac{d}{dx} {x}^{n} \: = \: n {x}^{n - 1} \: }}$

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ADDITIONAL INFORMATION

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

Answer 2

Solution:

Here,

x = sint

y = cospt

So,

x = sint

=> t = sin^-1x

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

p d/dx sin^-1x

= d/dx cos^-1y

= p × 1/1-x²

= - 1/√1-y² dy/dx

Now,

$\rm \leadsto \: y _{1 ^{2} }(1 - {x}^{2} ) \\ \\ \leadsto \rm {p}^{2} (1 - {y}^{2} ) \\ \\ \sf \: on \: \: diffrentating \: w.r.t.x \: \: we \: \: get \\ \\ \implies \sf \: y_{1} \:^{2} \: \frac{d}{dx} \: (1 - {x}^{2} ) \: + \: (1 - {x}^{2}) \: \frac{d}{dx} y_{1} \:^{2} = p ^{2} (0 - 2yy _{1} )$

=> y₁²(0 - 2x) + (1 - x²) (y₁y₂) = (p²(0-2yy₁)

=> y₁²(0 - 2x) + (1 - x²)(2y₁y₂) = (p²(-2yy₁)

=> 2y₁( -2x) + ( 1 - x²) y₂) = -2p²(-2yy₁)

=> - xy₁ + (1 - x²) y₂ = - p²y

=> (1 - x²) y₂-xy₁ + p²y = 0