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  • Can someone explain - Ordinary Differential Equations - BITSAT-2

The solution of (x\sqrt{1+y^{2}})dx+\left ( y\sqrt{1+x^{2}} \right )dy=0 is,

 

  • Option 1)

    \sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c

  • Option 2)

    \sqrt{1+x^{2}}-\sqrt{1+y^{2}}=c

  • Option 3)

    (1+x^{2})^{\frac{3}{2}}+\left ( 1+y^{2} \right )^{\frac{3}{2}}=c

  • Option 4)

    None

 

Answers (1)

best_answer

Use the concept of

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

 

 

 

 (x\sqrt{1+y^{2}})dx+(y\sqrt{1+x^{2}})dy=0

=> x\sqrt{1+y^{2}}dx=-y\sqrt{1+x^{2}}dy

=> \int \frac{x}{\sqrt{1+x^{2}}}dx=-\int \frac{y}{\sqrt{1+y^2}}dy

=\sqrt{1+x^{2}}=-\sqrt{1+y^2}+C

=\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=C


Option 1)

\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c

Option is correct

Option 2)

\sqrt{1+x^{2}}-\sqrt{1+y^{2}}=c

Option is incorrect

Option 3)

(1+x^{2})^{\frac{3}{2}}+\left ( 1+y^{2} \right )^{\frac{3}{2}}=c

Option is incorrect

Option 4)

None

Option is incorrect

Posted by

Plabita

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