Question

If y=(cos⁻¹ x)², prove (1-x²)y² -xy₁=2

Answer 1

Step-by-step explanation:

If

$y = ({ {cos}^{ - 1}x })^{2}$

differentiate y with respect to x

$\frac{dy}{dx} = {y}^{1} = - 2 {cos}^{ - 1} x. \frac{1}{ \sqrt{1 - {x}^{2} }} .. eq1 \\ \\ \frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - 2\sqrt{1 - {x}^{2} } . \frac{ - 1}{ \sqrt{1 - {x}^{2} } } + 2 {cos}^{ - 1} x. \frac{1}{ 2\sqrt{1 - {x}^{2} } }( - 2x) }{1 - {x}^{2} } \\ \\ \frac{ {d}^{2} y}{d {x}^{2} } = \frac{2 - \frac{2x \: {cos}^{ - 1} x}{ \sqrt{1 - {x}^{2} } } }{1 - {x}^{2} } \\ \\ \frac{ {d}^{2} y}{d {x}^{2} } = {y}^{11} = \frac{2 \sqrt{1 - {x}^{2} } -2x \: {cos}^{ - 1} x \: }{ {(1 - {x}^{2} )}^{ \frac{3}{2} } }$

Put the value of y' and y'' in the equation

$(1 - {x}^{2} ) {y}^{''} - x {y}^{'} = 2 \\ \\ (1 - {x}^{2} )\Bigg( \frac{2 \sqrt{1 - {x}^{2} } -2x \: {cos}^{ - 1} x \: }{ {(1 - {x}^{2} )}^{ \frac{3}{2} } }\Bigg) - x \Bigg(- 2 {cos}^{ - 1} x. \frac{1}{ \sqrt{1 - {x}^{2} }}\Bigg) \\ \\ \frac{2( \sqrt{1 - {x}^{2} } - x \: {cos}^{ - 1}x) }{ \sqrt{1 - {x}^{2} } } + \frac{2x. {cos}^{ - 1}x }{ \sqrt{1 - {x}^{2} } } \\ \\ split \: the \: LCM \: in \: first \: term \\ \\ 2 \frac{ \sqrt{1 - {x}^{2} } }{ \sqrt{1 - {x}^{2} } } - \frac{2x. {cos}^{ - 1}x }{ \sqrt{1 - {x}^{2} } } + \frac{2x. {cos}^{ - 1}x }{ \sqrt{1 - {x}^{2} } } \\ \\ = 2\\\\ (1 - {x}^{2} ) {y}^{''} - x {y}^{'}=2\\\\ =RHS$

Hence proved.

Hope it helps you.