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Question Number 15797 by tawa tawa last updated on 14/Jun/17

If (a + bx) e^{y / x} = x, where a and b are constant,

prove that, : x^{3} y^{″} = {({xy}^{'} - y)}^{2}

Commented by tawa tawa last updated on 14/Jun/17

please any answer on this ? . please help . Thanks in advamce .

Answered by ajfour last updated on 14/Jun/17

(a + b x) e^{y / x} = x

\frac{d}{d x} [(a + b x) e^{y / x}] = \frac{d}{d x} (x)

(a + b x) e^{y / x} [\frac{y^{'}}{x} - \frac{y}{x^{2}}] + {b e}^{y / x} = 1

\Rightarrow x [\frac{y^{'}}{x} - \frac{y}{x^{2}}] + {b e}^{y / x} = 1

o r y^{'} - \frac{y}{x} + {b e}^{y / x} = 1

(1 - y^{'} + \frac{y}{x}) e^{- y / x} = b

d i f f e r e n t i a t i n g a g a i n :

(- y^{″} + \frac{y^{'}}{x} - \frac{y}{x^{2}}) e^{- y / x} +

(1 - y^{'} + \frac{y}{x}) e^{- y / x} [- \frac{y^{'}}{x} + \frac{y}{x^{2}}] = 0

m u l t i p l y i n g b y e^{y / x} :

- y^{″} + \frac{y^{'}}{x} - \frac{y}{x^{2}} = - (1 - y^{'} + \frac{y}{x}) (- \frac{y^{'}}{x} + \frac{y}{x^{2}})

m u l t i p l y i n g b y (- x^{3}) :

x^{3} y^{″} - x^{2} y^{'} + x y = (x - {x y}^{'} + y) (- {x y}^{'} + y)

x^{3} y^{″} - x^{2} y^{'} + x y = - x^{2} y^{'} + x y + {({x y}^{'})}^{2}

- {x y y}^{'} - {x y y}^{'} + y^{2}

o r x^{3} y^{″} = {({x y}^{'} - y)}^{2} .

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