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Consider the differential equation, $y^{2}dx+\left ( x-\frac{1}{y} \right )dy=0.$ If value of y is 1 when $x=1,$ then the value of x for which $y=2,$ is :

Option 1)
$\frac{1}{2}+\frac{1}{\sqrt{e}}$

Option 2)
$\frac{5}{2}+\frac{1}{\sqrt{e}}$

Option 3)
$\frac{3}{2}-\frac{1}{\sqrt{e}}$

Option 4)
$\frac{3}{2}-\sqrt{e}$

Answers (1)

best_answer

$y^{2}dx+\left ( x-\frac{1}{y} \right )dy=0$

$y^{2}dx+xdy=\frac{\mathrm{d} y}{\mathrm{y} }$

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

IF= $e^{\int \frac{1}{y^{2}}dy}=e^{\frac{-1}{y}}$

$e^{\frac{-1}{y}}\cdot x=\int e^{-\frac{1}{y}}\cdot \frac{1}{y^{3}}dy+C$

$xe^{-\frac{1}{y}}=\int e^{-\frac{1}{y}}\cdot \frac{1}{y^{3}}dy+C$

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

$\left ( x,y \right )=\left ( 1,1 \right )$

$C=-\frac{1}{e}$

$x=\frac{3}{2}-\frac{1}{\sqrt{e}}$ when $y=2$

Option 1)

$\frac{1}{2}+\frac{1}{\sqrt{e}}$

Option 2)

$\frac{5}{2}+\frac{1}{\sqrt{e}}$

Option 3)

$\frac{3}{2}-\frac{1}{\sqrt{e}}$

Option 4)

$\frac{3}{2}-\sqrt{e}$

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