If the differential equation representing the family of all circles touching x-axis at the origin is (x^{2}-y^{2})\frac{dy}{dx}=g(x)\: y,then\: g(x) equals :

  • Option 1)

    \frac{1}{2}x

  • Option 2)

    2x^{2}

  • Option 3)

    2x

  • Option 4)

    \frac{1}{2}x^{2}

 

Answers (1)
V Vakul

As we learnt in 

Formation of Differential Equations -

A differential equation can be derived from its equation by the process of differentiation and other algebraical process of elimination

-

 Let the equation of circle is 

( x-o) \right )^{2}+(y-a)^{2}=a^{2}

\Rightarrow \left ( x)^{2}+(y-a)^{2}=a^{2}\ \, \, \, .....(i)

\Rightarrow 2x +2(y-a)\frac{dy}{dx}=0

\therefore x+(y-a)\frac{dy}{dx}=0

\therefore \frac{dy}{dx}=\frac{x}{(y-a)}      or     a =\frac{x+y\frac{dy}{dy}}{dy/dx}

Put a in (i)

x^{2}+ \frac{x^{2}}{(dy/dx)^{2}} = (\frac{x+y\frac{dy}{dx}}{(dy/dx)})^{2}

\\ \Rightarrow (x^{2}-y^{2})\frac{dy}{dx}=2xy \\ \therefore g(x) =2x


Option 1)

\frac{1}{2}x

This option is incorrect.

Option 2)

2x^{2}

This option is incorrect.

Option 3)

2x

This option is correct.

Option 4)

\frac{1}{2}x^{2}

This option is incorrect.

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