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The curve satisfying the differential equation, \left ( x^{2}-y^{2} \right )dx+2xydy=0 and passing through the point (1, 1) is :

  • Option 1)

    a circle of radius one.

  • Option 2)

    a hyperbola.

  • Option 3)

    an ellipse

  • Option 4)

    a circle of radius two.

 

Answers (2)

best_answer

As we learned,

 

Homogeneous Differential Equation -

A function

f\left ( x,y \right ) 

is called homogeneous function of  degree n, if

f\left (dx,dy \right )=d^{n}\left ( x,y \right )

- wherein

eg:

f\left ( x,y \right )=x^{2}y^{2}-xy^{3}

 

 and

 

Homogeneous Differential Equation -

Put

\frac{y}{x}=v

\frac{dy}{dx}=v+\frac{xdv}{dx}

-

 

 

 \left ( x^{2}-y^{2} \right )dx+2xydy=0

\frac{dy}{dx}=\frac{y^{2}-x^{2}}{2xy}

If y = kx

\frac{dy}{dx}=\left ( k+\frac{xdk}{dx} \right )

k+\frac{xdk}{dx}=\frac{k^{2}-1}{2k}

\frac{xdk}{dx}=\frac{k^{2}-1}{2k}-k=\frac{k^{2}-1-2k^{2}}{2k}

\frac{xdk}{dx}=\frac{-\left (k^{2}+1 \right )}{2k}

\int \frac{2kdk}{k^{2}+1}=\int -\frac{dx}{x}

\ln \left ( k^{2}+1 \right )=-\ln\left ( x \right )+C

\frac{y^{2}}{x^{2}}+1=kx

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

Value of k=2

The radius of circle = 1


Option 1)

a circle of radius one.

Option 2)

a hyperbola.

Option 3)

an ellipse

Option 4)

a circle of radius two.

Posted by

Himanshu

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