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  • Help me please, - Ordinary Differential Equations - BITSAT-2

The solution of the equation (1-x^{2})dy+xydx=xy^{2}dx is

  • Option 1)

    (y-1)^{2} \: (1-x^{2})=0

  • Option 2)

    (y-1)^{2} \: (1-x^{2})=c^{2}y^{2}

  • Option 3)

    (y-1)^{2} \: (1+x^{2})=c^{2}y^{2}

  • Option 4)

    None of these

 

Answers (1)

best_answer

 

Solution of Differential Equation -

\frac{\mathrm{d}y }{\mathrm{d} x} =f\left ( ax+by+c \right )

put

 Z =ax+by+c

 

 

- wherein

Equation with convert to

\int \frac{dz}{bf\left ( z \right )+a} =x+c

 

 

 

 (1-x^{2})dy+xydx=xy^{2}dx

(1-x^{2})dy=xy(y-1)dx

\int \frac{dy}{y(y-1)}=\int \frac{x}{1-x^{2}}dx

\int (\frac{1}{y-1}-\frac{1}{y})dy=\int \frac{x}{1-x^{2}}dx

log(y-1)-logy=-\frac{1}{2}log(1-x^{2})+c\\ 2log\frac{y-1}{y}+log(1-x^{2})=c\\ log(\frac{y-1}{y})^{2}.(1-x^{2})=c

\therefore (y-1)^{2}(1-x^{2})=c^{2}y^{2}

 


Option 1)

(y-1)^{2} \: (1-x^{2})=0

This option is incorrect 

Option 2)

(y-1)^{2} \: (1-x^{2})=c^{2}y^{2}

This option is correct 

Option 3)

(y-1)^{2} \: (1+x^{2})=c^{2}y^{2}

This option is incorrect 

Option 4)

None of these

This option is incorrect 

Posted by

divya.saini

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