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)

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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