Question

Plz solve this quickly​

Answer 1

Given :-

$\rm y = {e}^{ {tan}^{ - 1} x}$

To prove :-

$\rm(1 + {x}^{2} ) \dfrac{ {d}^{2} y}{d {x}^{2} } + (2x - 1)\dfrac{ {d} y}{d {x} } = 0$

Proof :-

$\rm \implies y = {e}^{ {tan}^{ - 1} x}$

Now , first we will find dy/dx

$\rm \implies \dfrac{dy}{dx} = \dfrac{d({e}^{ {tan}^{ - 1} x})}{dx}$

We know that :-

=> d(e^x)/dx = e^x

=> d( tan^(-1)x )/dx = 1/( 1+x²)

$\rm \implies \dfrac{dy}{dx} = {e}^{ {tan}^{ - 1} x} \times \dfrac{d({tan}^{ - 1} x)}{dx}$

$\rm \implies \dfrac{dy}{dx} = {e}^{ {tan}^{ - 1} x} \times \dfrac{1}{1 + {x}^{2} }$

$\rm \implies \dfrac{dy}{dx} = \dfrac{{e}^{ {tan}^{ - 1} x}}{1 + {x}^{2} } .......(1)$

Now , multiply (2x-1) in LHS and RHS of equation (1)

$\rm \implies (2x - 1) \dfrac{dy}{dx} = (2x - 1) \dfrac{{e}^{ {tan}^{ - 1} x}}{1 + {x}^{2} } .......(2)$

Now we will find out d²y/dx² using quotient rule of differentiation.

$\rm \implies \dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{d \bigg(\dfrac{{e}^{ {tan}^{ - 1} x}}{1 + {x}^{2} } \bigg )}{d}$

$\rm \implies \dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{{e}^{ {tan}^{ - 1} x} \times \frac{1}{(1 + {x}^{2} ) } \times { ( {x}^{2} + 1)} - 2x \: {e}^{ {tan}^{ - 1} x}}{ {(1 + {x}^{2} )}^{2} }$

$\rm \implies \dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{{e}^{ {tan}^{ - 1} x}(1 - 2x)}{ {(1 + {x}^{2} )}^{2} }$

Now multiply (1+x²) :-

$\rm \implies{(1 + {x}^{2} )} \times \dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{{e}^{ {tan}^{ - 1} x}(1 - 2x)}{ {(1 + {x}^{2} )}^{2} } \times {(1 + {x}^{2} )}$

$\rm \implies{(1 + {x}^{2} )} \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \dfrac{( 2x - 1){e}^{ {tan}^{ - 1} x}}{ {(1 + {x}^{2} )} } .......(3)$

Now add ( 2) and (3) :-

$\rm \implies{(1 + {x}^{2} )} \dfrac{ {d}^{2} y}{d {x}^{2} } + (2x - 1) \dfrac{dy}{dx} = (2x - 1) \dfrac{{e}^{ {tan}^{ - 1} x}}{1 + {x}^{2} } - \dfrac{( 2x - 1){e}^{ {tan}^{ - 1} x}}{ {(1 + {x}^{2} )} }$

$\rm \implies{(1 + {x}^{2} )} \dfrac{ {d}^{2} y}{d {x}^{2} } + (2x - 1) \dfrac{dy}{dx} = (2x - 1) \dfrac{{e}^{ {tan}^{ - 1} x}}{1 + {x}^{2} } - \dfrac{( 2x - 1){e}^{ {tan}^{ - 1} x}}{ {(1 + {x}^{2} )} } = 0$

Hence proved