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  • I have a doubt, kindly clarify. - Ordinary Differential Equations - BITSAT-2

Equation of curve through point(1,0) which satisfies the differential equation (1+y^{2})dx-xydy=0 is

  • Option 1)

    x^{2}+y^{2}=1

  • Option 2)

    x^{2}-y^{2}=1

  • Option 3)

    2x^{2}-y^{2}=2

  • Option 4)

    None of these

 

Answers (1)

best_answer

As we discussed in concept

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+y^{2})dx=xydy

\int \frac{dx}{x}\:=\:\int \frac{y}{1+y^{2}}\:dy

logx+c=\frac{1}{2}\:log(1+y^{2})

log1+c=\frac{1}{2}\:log(1+0)

\therefore\:c=0

\therefore\:2logx=log(1+y^{2})

=> logx^{2}=log(1+y^{2})\:\:\:=>x^{2}=1+y^{2}

                                                     \therefore\:x^{2}-y^{2}\:=\:1


Option 1)

x^{2}+y^{2}=1

This option is incorrect.

Option 2)

x^{2}-y^{2}=1

This option is correct.

Option 3)

2x^{2}-y^{2}=2

This option is incorrect.

Option 4)

None of these

This option is incorrect.

Posted by

Aadil

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