Consider the differential equation y2dx+x−1ydy=0 . If y1=1 then x is given by

dxdy+1y2x=1y3I.F=e−1/yG.S is xe−1/y=∫1y3 e−1/ydy+cy1=1⇒c=−1/eHence xe−1/y=−e−1/y−1/y−1−1ex=1+1y−e1/ye

Consider the differential equation y2dx+x−1ydy=0 . If y1=1 then x is given by

dxdy+1y2x=1y3I.F=e−1/yG.S is xe−1/y=∫1y3 e−1/ydy+cy1=1⇒c=−1/eHence xe−1/y=−e−1/y−1/y−1−1ex=1+1y−e1/ye

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Consider the differential equation $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If $y (1) = 1$ then x is given by

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$3 - \frac{1}{y} + \frac{e^{1 / y}}{e}$

$1 + \frac{1}{y} - \frac{e^{1 / y}}{e}$

$1 - \frac{1}{y} + \frac{e^{1 / y}}{e}$

$4 - \frac{2}{y} - \frac{e^{1 / y}}{e}$

answer is B.

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Detailed Solution

$\frac{d x}{d y} + \frac{1}{y^{2}} x = \frac{1}{y^{3}}$

$I . F = e^{- 1 / y}$

G.S is $x e^{- 1 / y} = \int \frac{1}{y^{3}} e^{- 1 / y} d y + c$

$y (1) = 1 \Rightarrow c = - 1 / e$

Hence $x e^{- 1 / y} = - e^{- 1 / y} (- 1 / y - 1) - \frac{1}{e}$

$x = 1 + \frac{1}{y} - \frac{e^{1 / y}}{e}$

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Consider the differential equation y2dx+x−1ydy=0 . If y1=1 then x is given by

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Consider the differential equation $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If $y (1) = 1$ then x is given by