Question

3. If y sin (msin x) prove (1-x²) v-xy₁+m²y =0​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

$\bf :\longmapsto\:If \: y = sin[m {sin}^{ - 1}x], \: prove \: that$

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

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

Given that,

$\rm :\longmapsto\:y = sin[m {sin}^{ - 1}x]$

can be rewritten as

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

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

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

We know,

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

So, using this, we get

$\rm :\longmapsto\: \dfrac{1}{ \sqrt{1 - {y}^{2} } } \dfrac{d}{dx}y =m\dfrac{d}{dx}\: {sin}^{ - 1}x$

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

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

On squaring both sides, we get

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

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

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

We know,

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

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

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

So, using this, we get

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

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

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

$\rm :\longmapsto\: 2y_1\bigg[ - xy_1+ {(1 - x}^{2})y_2 \bigg]= - 2 {m}^{2}yy_1$

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

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

Additional Information :-

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